try? shouldn't work on non-method-call


(Sikhapol Saijit) #1

Hi all,

Yesterday I tried this code:

func couldFailButWillNot() throws -> Any {
    return 42
}

if let a = try? couldFailButWillNot() as? Int {
    print(a)
}

And was surprised that the output was Optional(42) on both Swift 2 and Swift 3.
I always have the impression that when a variable is resolved with if let it will never be optional.

So, with a little investigation, I found out that it happens because as? has higher precedence than try? and is evaluated first.
And the whole expression `try? couldFailButWillNot() as? Int` evaluated as Optional(Optional(42)).

Also, I’m surprised that try? can be used with non-method-call.
This code: `print(try? 42)` will print Optional(42).

So, the questions are:

1. Is it intentional that try? can be used with a "non-method-call" and return an optional of the type that follows?

2. Should we design try? to have higher precedence than as? or any operators at all?
My intuition tells me that
let a = try? couldFailButWillNot() as? Int
should be equivalent to
let a = (try? couldFailButWillNot()) as? Int

3. Do you think that doubly-nested optional (or multi-level-nested optional) is confusing and should be removed from Swift? (Yes, I’ve seen this blog post Optionals Case Study: valuesForKeys <https://developer.apple.com/swift/blog/?id=12>).
For me Optional(nil) (aka Optional.Some(Optional.None))) doesn’t make much sense.
Maybe, one of the solution is to always have optional of optional merged into a single level optional? Like Optional(Optional(Optional(42))) should be the merged to and evaluated as Optional(42).

BTW, the code above is merely for a demonstration. The actual code was more of something like this:

func parse(JSON: Data) throws -> Any {
    // …
}

if let dict = try? parse(JSON: json) as? [String: Any] {
    // assume dict is a valid [String: Any] dictionary
    // …
}

I’m new to this mailing list so I’m not sure if this belongs here. I’m sorry in advance if it doesn’t.

Thank you,
Sam


(Slava Pestov) #2

Opinions inline:

Hi all,

Yesterday I tried this code:

func couldFailButWillNot() throws -> Any {
    return 42
}

if let a = try? couldFailButWillNot() as? Int {
    print(a)
}

And was surprised that the output was Optional(42) on both Swift 2 and Swift 3.
I always have the impression that when a variable is resolved with if let it will never be optional.

So, with a little investigation, I found out that it happens because as? has higher precedence than try? and is evaluated first.
And the whole expression `try? couldFailButWillNot() as? Int` evaluated as Optional(Optional(42)).

Also, I’m surprised that try? can be used with non-method-call.
This code: `print(try? 42)` will print Optional(42).

So, the questions are:

1. Is it intentional that try? can be used with a "non-method-call" and return an optional of the type that follows?

I think this is the real solution. try and try? should not be allowed on non-throwing functions or expressions.

This is a warning right now — do you think it should be an error?

Slavas-MacBook-Pro:~ slava$ cat ttt.swift
func f() {}

func g() {
  try f()
  try? f()
}

Slavas-MacBook-Pro:~ slava$ swiftc ttt.swift
ttt.swift:4:3: warning: no calls to throwing functions occur within 'try' expression
  try f()
  ^
ttt.swift:5:8: warning: no calls to throwing functions occur within 'try' expression
  try? f()
       ^

···

On Aug 18, 2016, at 12:52 AM, David Hart via swift-evolution <swift-evolution@swift.org> wrote:

On 18 Aug 2016, at 07:43, Sikhapol Saijit via swift-evolution <swift-evolution@swift.org <mailto:swift-evolution@swift.org>> wrote:

2. Should we design try? to have higher precedence than as? or any operators at all?
My intuition tells me that
let a = try? couldFailButWillNot() as? Int
should be equivalent to
let a = (try? couldFailButWillNot()) as? Int

That’s worth considering. try feels like it should tie very strongly with the throwing expression.

3. Do you think that doubly-nested optional (or multi-level-nested optional) is confusing and should be removed from Swift? (Yes, I’ve seen this blog post Optionals Case Study: valuesForKeys <https://developer.apple.com/swift/blog/?id=12>).
For me Optional(nil) (aka Optional.Some(Optional.None))) doesn’t make much sense.
Maybe, one of the solution is to always have optional of optional merged into a single level optional? Like Optional(Optional(Optional(42))) should be the merged to and evaluated as Optional(42).

I don’t think this is the solution. Even if it was, how would you expect to “remove” them from Swift? Optionals are simply an enum with an associated value. We’d have to introduce a language feature to restrict values that can be stored in enum cases? It sounds awfully complicated.

BTW, the code above is merely for a demonstration. The actual code was more of something like this:

func parse(JSON: Data) throws -> Any {
    // …
}

if let dict = try? parse(JSON: json) as? [String: Any] {
    // assume dict is a valid [String: Any] dictionary
    // …
}

I’m new to this mailing list so I’m not sure if this belongs here. I’m sorry in advance if it doesn’t.

Thank you,
Sam
_______________________________________________
swift-evolution mailing list
swift-evolution@swift.org <mailto:swift-evolution@swift.org>
https://lists.swift.org/mailman/listinfo/swift-evolution

_______________________________________________
swift-evolution mailing list
swift-evolution@swift.org <mailto:swift-evolution@swift.org>
https://lists.swift.org/mailman/listinfo/swift-evolution


(Slava Pestov) #3

Opinions inline:

Hi all,

Yesterday I tried this code:

func couldFailButWillNot() throws -> Any {
    return 42
}

if let a = try? couldFailButWillNot() as? Int {
    print(a)
}

And was surprised that the output was Optional(42) on both Swift 2 and Swift 3.
I always have the impression that when a variable is resolved with if let it will never be optional.

So, with a little investigation, I found out that it happens because as? has higher precedence than try? and is evaluated first.
And the whole expression `try? couldFailButWillNot() as? Int` evaluated as Optional(Optional(42)).

Also, I’m surprised that try? can be used with non-method-call.
This code: `print(try? 42)` will print Optional(42).

So, the questions are:

1. Is it intentional that try? can be used with a "non-method-call" and return an optional of the type that follows?

I think this is the real solution. try and try? should not be allowed on non-throwing functions or expressions.

This is a warning right now — do you think it should be an error?

Slavas-MacBook-Pro:~ slava$ cat ttt.swift
func f() {}

func g() {
  try f()
  try? f()
}

Slavas-MacBook-Pro:~ slava$ swiftc ttt.swift
ttt.swift:4:3: warning: no calls to throwing functions occur within 'try' expression
  try f()
  ^
ttt.swift:5:8: warning: no calls to throwing functions occur within 'try' expression
  try? f()
       ^

… Replying to myself: I also think it would be more logical for this to be an error, but I vaguely recall some internal discussion where we agreed that removing ‘throws’ from a function should not break source or binary compatibility (I don’t know if the current implementation makes good on the latter).

Jordan, do you remember the reasoning for this being a warning and not an error?

···

On Aug 18, 2016, at 1:42 AM, Slava Pestov via swift-evolution <swift-evolution@swift.org> wrote:

On Aug 18, 2016, at 12:52 AM, David Hart via swift-evolution <swift-evolution@swift.org <mailto:swift-evolution@swift.org>> wrote:

On 18 Aug 2016, at 07:43, Sikhapol Saijit via swift-evolution <swift-evolution@swift.org <mailto:swift-evolution@swift.org>> wrote:

2. Should we design try? to have higher precedence than as? or any operators at all?
My intuition tells me that
let a = try? couldFailButWillNot() as? Int
should be equivalent to
let a = (try? couldFailButWillNot()) as? Int

That’s worth considering. try feels like it should tie very strongly with the throwing expression.

3. Do you think that doubly-nested optional (or multi-level-nested optional) is confusing and should be removed from Swift? (Yes, I’ve seen this blog post Optionals Case Study: valuesForKeys <https://developer.apple.com/swift/blog/?id=12>).
For me Optional(nil) (aka Optional.Some(Optional.None))) doesn’t make much sense.
Maybe, one of the solution is to always have optional of optional merged into a single level optional? Like Optional(Optional(Optional(42))) should be the merged to and evaluated as Optional(42).

I don’t think this is the solution. Even if it was, how would you expect to “remove” them from Swift? Optionals are simply an enum with an associated value. We’d have to introduce a language feature to restrict values that can be stored in enum cases? It sounds awfully complicated.

BTW, the code above is merely for a demonstration. The actual code was more of something like this:

func parse(JSON: Data) throws -> Any {
    // …
}

if let dict = try? parse(JSON: json) as? [String: Any] {
    // assume dict is a valid [String: Any] dictionary
    // …
}

I’m new to this mailing list so I’m not sure if this belongs here. I’m sorry in advance if it doesn’t.

Thank you,
Sam
_______________________________________________
swift-evolution mailing list
swift-evolution@swift.org <mailto:swift-evolution@swift.org>
https://lists.swift.org/mailman/listinfo/swift-evolution

_______________________________________________
swift-evolution mailing list
swift-evolution@swift.org <mailto:swift-evolution@swift.org>
https://lists.swift.org/mailman/listinfo/swift-evolution

_______________________________________________
swift-evolution mailing list
swift-evolution@swift.org <mailto:swift-evolution@swift.org>
https://lists.swift.org/mailman/listinfo/swift-evolution


(Charles Srstka) #4

1. Is it intentional that try? can be used with a "non-method-call" and return an optional of the type that follows?

a. I’m not sure what you mean by “non-method-call”, since the thing you called in your example *was* a function that was marked with “throws”. Using try? on a non-throwing function or method does indeed produce a warning.

b. I’m not a member of the development team, but I think it probably is intentional. There are two levels of optionality going on here;

2. Should we design try? to have higher precedence than as? or any operators at all?
My intuition tells me that
let a = try? couldFailButWillNot() as? Int
should be equivalent to
let a = (try? couldFailButWillNot()) as? Int

This is more debatable whether it *should* be the case, but it’s worth pointing out that try/try?/try! work on the entire rest of the line, which means you can include more throwing calls and not have to put an ! each time:

func foo() throws -> Int { return 3 }

func bar() throws -> Int { return 5 }

if let i = try? foo() + bar() { // not try foo() + try bar()
    print("i is \(i)")
}

Or this:

func foo() throws -> Int { return 3 }

func bar(_ i: Int) throws -> Int { return i + 2 }

if let i = try? bar(foo()) { // not bar(try foo())
    print("i is \(i)")
}

So multiple throwing statements can be used on a single line without having to throw “try” all over the place. Now, whether that’s worth the admittedly confusing behavior you noted above is probably a decent topic for debate.

Charles

···

On Aug 18, 2016, at 12:43 AM, Sikhapol Saijit via swift-evolution <swift-evolution@swift.org> wrote:


(Charles Srstka) #5

Sorry for the resend! Some of my previous e-mail was left out for some reason. Here it is again, as it should have been:

1. Is it intentional that try? can be used with a "non-method-call" and return an optional of the type that follows?

a. I’m not sure what you mean by “non-method-call”, since the thing you called in your example *was* a function that was marked with “throws”. Using try? on a non-throwing function or method does indeed produce a warning.

b. I’m not an Apple employee or a member of the development team, but I think it probably is intentional. There are two levels of optionality going on here; one, the possibility that the throwing method could have failed, and the other, that the optional result may or may not contain a value. Either of those two things being nil could mean quite different things, so the double-optional makes sense.

2. Should we design try? to have higher precedence than as? or any operators at all?
My intuition tells me that
let a = try? couldFailButWillNot() as? Int
should be equivalent to
let a = (try? couldFailButWillNot()) as? Int

This is more debatable whether it *should* be the case, but it’s worth pointing out that try/try?/try! work on the entire rest of the line, which means you can include more throwing calls and not have to put an ! each time:

func foo() throws -> Int { return 3 }

func bar() throws -> Int { return 5 }

if let i = try? foo() + bar() { // not try foo() + try bar()
    print("i is \(i)")
}

Or this:

func foo() throws -> Int { return 3 }

func bar(_ i: Int) throws -> Int { return i + 2 }

if let i = try? bar(foo()) { // not bar(try foo())
    print("i is \(i)")
}

So multiple throwing statements can be used on a single line without having to throw “try” all over the place. Now, whether that’s worth the admittedly confusing behavior you noted above is probably a decent topic for debate.

Charles

···

On Aug 18, 2016, at 12:43 AM, Sikhapol Saijit via swift-evolution <swift-evolution@swift.org <mailto:swift-evolution@swift.org>> wrote:


(David Hart) #6

Opinions inline:

Hi all,

Yesterday I tried this code:

func couldFailButWillNot() throws -> Any {
    return 42
}

if let a = try? couldFailButWillNot() as? Int {
    print(a)
}

And was surprised that the output was Optional(42) on both Swift 2 and Swift 3.
I always have the impression that when a variable is resolved with if let it will never be optional.

So, with a little investigation, I found out that it happens because as? has higher precedence than try? and is evaluated first.
And the whole expression `try? couldFailButWillNot() as? Int` evaluated as Optional(Optional(42)).

Also, I’m surprised that try? can be used with non-method-call.
This code: `print(try? 42)` will print Optional(42).

So, the questions are:

1. Is it intentional that try? can be used with a "non-method-call" and return an optional of the type that follows?

I think this is the real solution. try and try? should not be allowed on non-throwing functions or expressions.

2. Should we design try? to have higher precedence than as? or any operators at all?
My intuition tells me that
let a = try? couldFailButWillNot() as? Int
should be equivalent to
let a = (try? couldFailButWillNot()) as? Int

That’s worth considering. try feels like it should tie very strongly with the throwing expression.

3. Do you think that doubly-nested optional (or multi-level-nested optional) is confusing and should be removed from Swift? (Yes, I’ve seen this blog post Optionals Case Study: valuesForKeys <https://developer.apple.com/swift/blog/?id=12>).
For me Optional(nil) (aka Optional.Some(Optional.None))) doesn’t make much sense.
Maybe, one of the solution is to always have optional of optional merged into a single level optional? Like Optional(Optional(Optional(42))) should be the merged to and evaluated as Optional(42).

I don’t think this is the solution. Even if it was, how would you expect to “remove” them from Swift? Optionals are simply an enum with an associated value. We’d have to introduce a language feature to restrict values that can be stored in enum cases? It sounds awfully complicated.

···

On 18 Aug 2016, at 07:43, Sikhapol Saijit via swift-evolution <swift-evolution@swift.org> wrote:

BTW, the code above is merely for a demonstration. The actual code was more of something like this:

func parse(JSON: Data) throws -> Any {
    // …
}

if let dict = try? parse(JSON: json) as? [String: Any] {
    // assume dict is a valid [String: Any] dictionary
    // …
}

I’m new to this mailing list so I’m not sure if this belongs here. I’m sorry in advance if it doesn’t.

Thank you,
Sam
_______________________________________________
swift-evolution mailing list
swift-evolution@swift.org
https://lists.swift.org/mailman/listinfo/swift-evolution


(Sikhapol Saijit) #7

Sorry for the resend! Some of my previous e-mail was left out for some reason. Here it is again, as it should have been:

1. Is it intentional that try? can be used with a "non-method-call" and return an optional of the type that follows?

a. I’m not sure what you mean by “non-method-call”, since the thing you called in your example *was* a function that was marked with “throws”. Using try? on a non-throwing function or method does indeed produce a warning.

In my sample code It did look like that try? was tied to the throwing function call but it actually operate on the result of the following as? operator.

So
if let a = try? couldFailButWillNot() as? Int { … }
was actually work like this:
if let a = try? (couldFailButWillNot() as? Int /* evaluated to Optional(42)) */) { … }

Try typing `print(try? 42)` in the playground and you will understand what I mean.

···

On Aug 18, 2016, at 3:15 PM, Charles Srstka <cocoadev@charlessoft.com> wrote:

On Aug 18, 2016, at 12:43 AM, Sikhapol Saijit via swift-evolution <swift-evolution@swift.org <mailto:swift-evolution@swift.org>> wrote:

b. I’m not an Apple employee or a member of the development team, but I think it probably is intentional. There are two levels of optionality going on here; one, the possibility that the throwing method could have failed, and the other, that the optional result may or may not contain a value. Either of those two things being nil could mean quite different things, so the double-optional makes sense.

2. Should we design try? to have higher precedence than as? or any operators at all?
My intuition tells me that
let a = try? couldFailButWillNot() as? Int
should be equivalent to
let a = (try? couldFailButWillNot()) as? Int

This is more debatable whether it *should* be the case, but it’s worth pointing out that try/try?/try! work on the entire rest of the line, which means you can include more throwing calls and not have to put an ! each time:

func foo() throws -> Int { return 3 }

func bar() throws -> Int { return 5 }

if let i = try? foo() + bar() { // not try foo() + try bar()
    print("i is \(i)")
}

Or this:

func foo() throws -> Int { return 3 }

func bar(_ i: Int) throws -> Int { return i + 2 }

if let i = try? bar(foo()) { // not bar(try foo())
    print("i is \(i)")
}

So multiple throwing statements can be used on a single line without having to throw “try” all over the place. Now, whether that’s worth the admittedly confusing behavior you noted above is probably a decent topic for debate.

Charles


(Sikhapol Saijit) #8

Opinions inline:

Hi all,

Yesterday I tried this code:

func couldFailButWillNot() throws -> Any {
    return 42
}

if let a = try? couldFailButWillNot() as? Int {
    print(a)
}

And was surprised that the output was Optional(42) on both Swift 2 and Swift 3.
I always have the impression that when a variable is resolved with if let it will never be optional.

So, with a little investigation, I found out that it happens because as? has higher precedence than try? and is evaluated first.
And the whole expression `try? couldFailButWillNot() as? Int` evaluated as Optional(Optional(42)).

Also, I’m surprised that try? can be used with non-method-call.
This code: `print(try? 42)` will print Optional(42).

So, the questions are:

1. Is it intentional that try? can be used with a "non-method-call" and return an optional of the type that follows?

I think this is the real solution. try and try? should not be allowed on non-throwing functions or expressions.

This is a warning right now — do you think it should be an error?

Slavas-MacBook-Pro:~ slava$ cat ttt.swift
func f() {}

func g() {
  try f()
  try? f()
}

Slavas-MacBook-Pro:~ slava$ swiftc ttt.swift
ttt.swift:4:3: warning: no calls to throwing functions occur within 'try' expression
  try f()
  ^
ttt.swift:5:8: warning: no calls to throwing functions occur within 'try' expression
  try? f()
       ^

Thank you Slava,

While I think using try/try? on anything but a throwing function call should be an error, right now it even works with anything. `try? 42` will just wrap 42 in an optional and give some warning now.

···

On Aug 18, 2016, at 3:42 PM, Slava Pestov <spestov@apple.com> wrote:

On Aug 18, 2016, at 12:52 AM, David Hart via swift-evolution <swift-evolution@swift.org <mailto:swift-evolution@swift.org>> wrote:

On 18 Aug 2016, at 07:43, Sikhapol Saijit via swift-evolution <swift-evolution@swift.org <mailto:swift-evolution@swift.org>> wrote:

2. Should we design try? to have higher precedence than as? or any operators at all?
My intuition tells me that
let a = try? couldFailButWillNot() as? Int
should be equivalent to
let a = (try? couldFailButWillNot()) as? Int

That’s worth considering. try feels like it should tie very strongly with the throwing expression.

3. Do you think that doubly-nested optional (or multi-level-nested optional) is confusing and should be removed from Swift? (Yes, I’ve seen this blog post Optionals Case Study: valuesForKeys <https://developer.apple.com/swift/blog/?id=12>).
For me Optional(nil) (aka Optional.Some(Optional.None))) doesn’t make much sense.
Maybe, one of the solution is to always have optional of optional merged into a single level optional? Like Optional(Optional(Optional(42))) should be the merged to and evaluated as Optional(42).

I don’t think this is the solution. Even if it was, how would you expect to “remove” them from Swift? Optionals are simply an enum with an associated value. We’d have to introduce a language feature to restrict values that can be stored in enum cases? It sounds awfully complicated.

BTW, the code above is merely for a demonstration. The actual code was more of something like this:

func parse(JSON: Data) throws -> Any {
    // …
}

if let dict = try? parse(JSON: json) as? [String: Any] {
    // assume dict is a valid [String: Any] dictionary
    // …
}

I’m new to this mailing list so I’m not sure if this belongs here. I’m sorry in advance if it doesn’t.

Thank you,
Sam
_______________________________________________
swift-evolution mailing list
swift-evolution@swift.org <mailto:swift-evolution@swift.org>
https://lists.swift.org/mailman/listinfo/swift-evolution

_______________________________________________
swift-evolution mailing list
swift-evolution@swift.org <mailto:swift-evolution@swift.org>
https://lists.swift.org/mailman/listinfo/swift-evolution


(John McCall) #9

Lots of interesting points here. I do think there's an improvement possible here, but it's actually along the lines of Sam's original suggestion #3 (not vis-a-vis all of Swift, but specifically for how try? composes with as?):

A. I'm in favor of the current behavior where try prefixes an entire statement: it solves the precise issue of multiple nested optionals or multiple unwrapping of optionals in the situation where one statement has calls to many throwing functions. It says instead, I want nil if anything in this statement throws, otherwise, give me .some(value).

Sam--I think you may have misunderstood Charles's explanation. He's not saying "try?" attaches with lower or higher precedence as compared to "as?". Rather, I think the mental model is that "try?" prefixes the whole right-hand side (rhs), and if *any* call on the rhs throws, the whole rhs evaluates to nil, but if *any* call could potentially throw but doesn't, "try?" wraps the entire rhs and gives you .some(value). IMO, this is pretty sensible for the reason he gives.

B. I'm in favor of warning instead of error, for precisely the internal discussion rationale communicated by Slava. I'm willing to live with "try? 42" being only a warning if that means my code won't stop compiling when someone decides a library function doesn't need to throw.

Sam--here, changing warning to error would not solve your original problem, because in that example "try?" does prefix at least one throwing function, so you wouldn't get an error anyway.

C. However, given the thinking in (A), I do think how "try?" composes with "as?" is a little counterintuitive or at least overly ceremonious, though technically it is possible to reason through.

It's true that currently you can use the multiple nested optionals to figure out whether either a throwing function threw (but not which throwing function out of potentially more than one) or whether the cast did not succeed. But, since "try?" after all means "give me nil if anything throws," it kind of makes less sense that you get all this nesting and detailed information when it composes with "as?". If you really wanted that level of detail, you could always evaluate "try?" and "as?" in separate statements. What I'd propose instead is this:

If "try?" is composed with "as?", and "as?" yields "nil", then "try?" should not wrap that value in another optional.

We can't make the typing decision dependent on a dynamic property like whether the cast fails. And I don't like the idea of changing its typing rule based on the form of the nested expression. But we could make "try? foo()" avoid adding an extra level of optionality, the same way that "a?.foo()" does.

John.

···

On Aug 18, 2016, at 8:19 AM, Xiaodi Wu via swift-evolution <swift-evolution@swift.org> wrote:

Does that sound sensible?

On Thu, Aug 18, 2016 at 3:54 AM, Sikhapol Saijit via swift-evolution <swift-evolution@swift.org <mailto:swift-evolution@swift.org>> wrote:

On Aug 18, 2016, at 3:42 PM, Slava Pestov <spestov@apple.com <mailto:spestov@apple.com>> wrote:

On Aug 18, 2016, at 12:52 AM, David Hart via swift-evolution <swift-evolution@swift.org <mailto:swift-evolution@swift.org>> wrote:

Opinions inline:

On 18 Aug 2016, at 07:43, Sikhapol Saijit via swift-evolution <swift-evolution@swift.org <mailto:swift-evolution@swift.org>> wrote:

Hi all,

Yesterday I tried this code:

func couldFailButWillNot() throws -> Any {
    return 42
}

if let a = try? couldFailButWillNot() as? Int {
    print(a)
}

And was surprised that the output was Optional(42) on both Swift 2 and Swift 3.
I always have the impression that when a variable is resolved with if let it will never be optional.

So, with a little investigation, I found out that it happens because as? has higher precedence than try? and is evaluated first.
And the whole expression `try? couldFailButWillNot() as? Int` evaluated as Optional(Optional(42)).

Also, I’m surprised that try? can be used with non-method-call.
This code: `print(try? 42)` will print Optional(42).

So, the questions are:

1. Is it intentional that try? can be used with a "non-method-call" and return an optional of the type that follows?

I think this is the real solution. try and try? should not be allowed on non-throwing functions or expressions.

This is a warning right now — do you think it should be an error?

Slavas-MacBook-Pro:~ slava$ cat ttt.swift
func f() {}

func g() {
  try f()
  try? f()
}

Slavas-MacBook-Pro:~ slava$ swiftc ttt.swift
ttt.swift:4:3: warning: no calls to throwing functions occur within 'try' expression
  try f()
  ^
ttt.swift:5:8: warning: no calls to throwing functions occur within 'try' expression
  try? f()
       ^

Thank you Slava,

While I think using try/try? on anything but a throwing function call should be an error, right now it even works with anything. `try? 42` will just wrap 42 in an optional and give some warning now.

2. Should we design try? to have higher precedence than as? or any operators at all?
My intuition tells me that
let a = try? couldFailButWillNot() as? Int
should be equivalent to
let a = (try? couldFailButWillNot()) as? Int

That’s worth considering. try feels like it should tie very strongly with the throwing expression.

3. Do you think that doubly-nested optional (or multi-level-nested optional) is confusing and should be removed from Swift? (Yes, I’ve seen this blog post Optionals Case Study: valuesForKeys <https://developer.apple.com/swift/blog/?id=12>).
For me Optional(nil) (aka Optional.Some(Optional.None))) doesn’t make much sense.
Maybe, one of the solution is to always have optional of optional merged into a single level optional? Like Optional(Optional(Optional(42))) should be the merged to and evaluated as Optional(42).

I don’t think this is the solution. Even if it was, how would you expect to “remove” them from Swift? Optionals are simply an enum with an associated value. We’d have to introduce a language feature to restrict values that can be stored in enum cases? It sounds awfully complicated.

BTW, the code above is merely for a demonstration. The actual code was more of something like this:

func parse(JSON: Data) throws -> Any {
    // …
}

if let dict = try? parse(JSON: json) as? [String: Any] {
    // assume dict is a valid [String: Any] dictionary
    // …
}

I’m new to this mailing list so I’m not sure if this belongs here. I’m sorry in advance if it doesn’t.

Thank you,
Sam
_______________________________________________
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swift-evolution@swift.org <mailto:swift-evolution@swift.org>
https://lists.swift.org/mailman/listinfo/swift-evolution

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(John McCall) #10

The issue would be that, in the case of "try? foo()", nil and .some(nil) might mean very different things.

This is true of a?.foo() as well. But yes, I think it is more likely that someone would want to treat them differently for try?.

John.

···

On Aug 18, 2016, at 8:46 AM, Xiaodi Wu <xiaodi.wu@gmail.com> wrote:

On Thu, Aug 18, 2016 at 10:40 John McCall <rjmccall@apple.com <mailto:rjmccall@apple.com>> wrote:

On Aug 18, 2016, at 8:19 AM, Xiaodi Wu via swift-evolution <swift-evolution@swift.org <mailto:swift-evolution@swift.org>> wrote:

Lots of interesting points here. I do think there's an improvement possible here, but it's actually along the lines of Sam's original suggestion #3 (not vis-a-vis all of Swift, but specifically for how try? composes with as?):

A. I'm in favor of the current behavior where try prefixes an entire statement: it solves the precise issue of multiple nested optionals or multiple unwrapping of optionals in the situation where one statement has calls to many throwing functions. It says instead, I want nil if anything in this statement throws, otherwise, give me .some(value).

Sam--I think you may have misunderstood Charles's explanation. He's not saying "try?" attaches with lower or higher precedence as compared to "as?". Rather, I think the mental model is that "try?" prefixes the whole right-hand side (rhs), and if *any* call on the rhs throws, the whole rhs evaluates to nil, but if *any* call could potentially throw but doesn't, "try?" wraps the entire rhs and gives you .some(value). IMO, this is pretty sensible for the reason he gives.

B. I'm in favor of warning instead of error, for precisely the internal discussion rationale communicated by Slava. I'm willing to live with "try? 42" being only a warning if that means my code won't stop compiling when someone decides a library function doesn't need to throw.

Sam--here, changing warning to error would not solve your original problem, because in that example "try?" does prefix at least one throwing function, so you wouldn't get an error anyway.

C. However, given the thinking in (A), I do think how "try?" composes with "as?" is a little counterintuitive or at least overly ceremonious, though technically it is possible to reason through.

It's true that currently you can use the multiple nested optionals to figure out whether either a throwing function threw (but not which throwing function out of potentially more than one) or whether the cast did not succeed. But, since "try?" after all means "give me nil if anything throws," it kind of makes less sense that you get all this nesting and detailed information when it composes with "as?". If you really wanted that level of detail, you could always evaluate "try?" and "as?" in separate statements. What I'd propose instead is this:

If "try?" is composed with "as?", and "as?" yields "nil", then "try?" should not wrap that value in another optional.

We can't make the typing decision dependent on a dynamic property like whether the cast fails. And I don't like the idea of changing its typing rule based on the form of the nested expression. But we could make "try? foo()" avoid adding an extra level of optionality, the same way that "a?.foo()" does.

John.

Does that sound sensible?

On Thu, Aug 18, 2016 at 3:54 AM, Sikhapol Saijit via swift-evolution <swift-evolution@swift.org <mailto:swift-evolution@swift.org>> wrote:

On Aug 18, 2016, at 3:42 PM, Slava Pestov <spestov@apple.com <mailto:spestov@apple.com>> wrote:

On Aug 18, 2016, at 12:52 AM, David Hart via swift-evolution <swift-evolution@swift.org <mailto:swift-evolution@swift.org>> wrote:

Opinions inline:

On 18 Aug 2016, at 07:43, Sikhapol Saijit via swift-evolution <swift-evolution@swift.org <mailto:swift-evolution@swift.org>> wrote:

Hi all,

Yesterday I tried this code:

func couldFailButWillNot() throws -> Any {
    return 42
}

if let a = try? couldFailButWillNot() as? Int {
    print(a)
}

And was surprised that the output was Optional(42) on both Swift 2 and Swift 3.
I always have the impression that when a variable is resolved with if let it will never be optional.

So, with a little investigation, I found out that it happens because as? has higher precedence than try? and is evaluated first.
And the whole expression `try? couldFailButWillNot() as? Int` evaluated as Optional(Optional(42)).

Also, I’m surprised that try? can be used with non-method-call.
This code: `print(try? 42)` will print Optional(42).

So, the questions are:

1. Is it intentional that try? can be used with a "non-method-call" and return an optional of the type that follows?

I think this is the real solution. try and try? should not be allowed on non-throwing functions or expressions.

This is a warning right now — do you think it should be an error?

Slavas-MacBook-Pro:~ slava$ cat ttt.swift
func f() {}

func g() {
  try f()
  try? f()
}

Slavas-MacBook-Pro:~ slava$ swiftc ttt.swift
ttt.swift:4:3: warning: no calls to throwing functions occur within 'try' expression
  try f()
  ^
ttt.swift:5:8: warning: no calls to throwing functions occur within 'try' expression
  try? f()
       ^

Thank you Slava,

While I think using try/try? on anything but a throwing function call should be an error, right now it even works with anything. `try? 42` will just wrap 42 in an optional and give some warning now.

2. Should we design try? to have higher precedence than as? or any operators at all?
My intuition tells me that
let a = try? couldFailButWillNot() as? Int
should be equivalent to
let a = (try? couldFailButWillNot()) as? Int

That’s worth considering. try feels like it should tie very strongly with the throwing expression.

3. Do you think that doubly-nested optional (or multi-level-nested optional) is confusing and should be removed from Swift? (Yes, I’ve seen this blog post Optionals Case Study: valuesForKeys <https://developer.apple.com/swift/blog/?id=12>).
For me Optional(nil) (aka Optional.Some(Optional.None))) doesn’t make much sense.
Maybe, one of the solution is to always have optional of optional merged into a single level optional? Like Optional(Optional(Optional(42))) should be the merged to and evaluated as Optional(42).

I don’t think this is the solution. Even if it was, how would you expect to “remove” them from Swift? Optionals are simply an enum with an associated value. We’d have to introduce a language feature to restrict values that can be stored in enum cases? It sounds awfully complicated.

BTW, the code above is merely for a demonstration. The actual code was more of something like this:

func parse(JSON: Data) throws -> Any {
    // …
}

if let dict = try? parse(JSON: json) as? [String: Any] {
    // assume dict is a valid [String: Any] dictionary
    // …
}

I’m new to this mailing list so I’m not sure if this belongs here. I’m sorry in advance if it doesn’t.

Thank you,
Sam
_______________________________________________
swift-evolution mailing list
swift-evolution@swift.org <mailto:swift-evolution@swift.org>
https://lists.swift.org/mailman/listinfo/swift-evolution

_______________________________________________
swift-evolution mailing list
swift-evolution@swift.org <mailto:swift-evolution@swift.org>
https://lists.swift.org/mailman/listinfo/swift-evolution

_______________________________________________
swift-evolution mailing list
swift-evolution@swift.org <mailto:swift-evolution@swift.org>
https://lists.swift.org/mailman/listinfo/swift-evolution

_______________________________________________
swift-evolution mailing list
swift-evolution@swift.org <mailto:swift-evolution@swift.org>
https://lists.swift.org/mailman/listinfo/swift-evolution


(Xiaodi Wu) #11

Lots of interesting points here. I do think there's an improvement possible
here, but it's actually along the lines of Sam's original suggestion #3
(not vis-a-vis all of Swift, but specifically for how try? composes with
as?):

A. I'm in favor of the current behavior where try prefixes an entire
statement: it solves the precise issue of multiple nested optionals or
multiple unwrapping of optionals in the situation where one statement has
calls to many throwing functions. It says instead, I want nil if anything
in this statement throws, otherwise, give me .some(value).

Sam--I think you may have misunderstood Charles's explanation. He's not
saying "try?" attaches with lower or higher precedence as compared to
"as?". Rather, I think the mental model is that "try?" prefixes the whole
right-hand side (rhs), and if *any* call on the rhs throws, the whole rhs
evaluates to nil, but if *any* call could potentially throw but doesn't,
"try?" wraps the entire rhs and gives you .some(value). IMO, this is pretty
sensible for the reason he gives.

B. I'm in favor of warning instead of error, for precisely the internal
discussion rationale communicated by Slava. I'm willing to live with "try?
42" being only a warning if that means my code won't stop compiling when
someone decides a library function doesn't need to throw.

Sam--here, changing warning to error would not solve your original problem,
because in that example "try?" does prefix at least one throwing function,
so you wouldn't get an error anyway.

C. However, given the thinking in (A), I do think how "try?" composes with
"as?" is a little counterintuitive or at least overly ceremonious, though
technically it is possible to reason through.

It's true that currently you can use the multiple nested optionals to
figure out whether either a throwing function threw (but not which throwing
function out of potentially more than one) or whether the cast did not
succeed. But, since "try?" after all means "give me nil if anything
throws," it kind of makes less sense that you get all this nesting and
detailed information when it composes with "as?". If you really wanted that
level of detail, you could always evaluate "try?" and "as?" in separate
statements. What I'd propose instead is this:

If "try?" is composed with "as?", and "as?" yields "nil", then "try?"
should not wrap that value in another optional.

Does that sound sensible?

···

On Thu, Aug 18, 2016 at 3:54 AM, Sikhapol Saijit via swift-evolution < swift-evolution@swift.org> wrote:

On Aug 18, 2016, at 3:42 PM, Slava Pestov <spestov@apple.com> wrote:

On Aug 18, 2016, at 12:52 AM, David Hart via swift-evolution < > swift-evolution@swift.org> wrote:

Opinions inline:

On 18 Aug 2016, at 07:43, Sikhapol Saijit via swift-evolution < > swift-evolution@swift.org> wrote:

Hi all,

Yesterday I tried this code:

func couldFailButWillNot() throws -> Any {
    return 42
}

if let a = try? couldFailButWillNot() as? Int {
    print(a)
}

And was surprised that the output was *Optional(42)* on both Swift 2 and
Swift 3.
I always have the impression that when a variable is resolved with if let it
will never be optional.

So, with a little investigation, I found out that it happens because as? has
higher precedence than try? and is evaluated first.
And the whole expression `try? couldFailButWillNot() as? Int` evaluated
as *Optional(Optional(42))*.

Also, I’m surprised that try? can be used with non-method-call.
This code: `print(try? 42)` will print *Optional(42)*.

So, the questions are:

1. Is it intentional that try? can be used with a "non-method-call" and
return an optional of the type that follows?

I think this is the real solution. try and try? should not be allowed on
non-throwing functions or expressions.

This is a warning right now — do you think it should be an error?

Slavas-MacBook-Pro:~ slava$ cat ttt.swift
func f() {}

func g() {
  try f()
  try? f()
}

Slavas-MacBook-Pro:~ slava$ swiftc ttt.swift
*ttt.swift:4:3: **warning: **no calls to throwing functions occur within
'try' expression*
  try f()
* ^*
*ttt.swift:5:8: **warning: **no calls to throwing functions occur within
'try' expression*
  try? f()
* ^*

Thank you Slava,

While I think using try/try? on anything but a throwing function call
should be an error, right now it even works with anything. `try? 42` will
just wrap 42 in an optional and give some warning now.

2. Should we design try? to have higher precedence than as? or any
operators at all?
My intuition tells me that
let a = try? couldFailButWillNot() as? Int
should be equivalent to
let a = (try? couldFailButWillNot()) as? Int

That’s worth considering. try feels like it should tie very strongly with
the throwing expression.

3. Do you think that doubly-nested optional (or multi-level-nested
optional) is confusing and should be removed from Swift? (Yes, I’ve seen
this blog post Optionals Case Study: valuesForKeys
<https://developer.apple.com/swift/blog/?id=12>).
For me *Optional(nil)* (aka *Optional.Some(Optional.None))*) doesn’t make
much sense.
Maybe, one of the solution is to always have optional of optional merged
into a single level optional? Like *Optional(Optional(Optional(42)))* should
be the merged to and evaluated as *Optional(42)*.

I don’t think this is the solution. Even if it was, how would you expect
to “remove” them from Swift? Optionals are simply an enum with an
associated value. We’d have to introduce a language feature to restrict
values that can be stored in enum cases? It sounds awfully complicated.

BTW, the code above is merely for a demonstration. The actual code was
more of something like this:

func parse(JSON: Data) throws -> Any {
    // …
}

if let dict = try? parse(JSON: json) as? [String: Any] {
    // assume dict is a valid [String: Any] dictionary
    // …
}

I’m new to this mailing list so I’m not sure if this belongs here. I’m
sorry in advance if it doesn’t.

Thank you,
Sam
_______________________________________________
swift-evolution mailing list
swift-evolution@swift.org
https://lists.swift.org/mailman/listinfo/swift-evolution

_______________________________________________
swift-evolution mailing list
swift-evolution@swift.org
https://lists.swift.org/mailman/listinfo/swift-evolution

_______________________________________________
swift-evolution mailing list
swift-evolution@swift.org
https://lists.swift.org/mailman/listinfo/swift-evolution


(Xiaodi Wu) #12

Lots of interesting points here. I do think there's an improvement
possible here, but it's actually along the lines of Sam's original
suggestion #3 (not vis-a-vis all of Swift, but specifically for how try?
composes with as?):

A. I'm in favor of the current behavior where try prefixes an entire
statement: it solves the precise issue of multiple nested optionals or
multiple unwrapping of optionals in the situation where one statement has
calls to many throwing functions. It says instead, I want nil if anything
in this statement throws, otherwise, give me .some(value).

Sam--I think you may have misunderstood Charles's explanation. He's not
saying "try?" attaches with lower or higher precedence as compared to
"as?". Rather, I think the mental model is that "try?" prefixes the whole
right-hand side (rhs), and if *any* call on the rhs throws, the whole rhs
evaluates to nil, but if *any* call

(Here, I meant to say, if *all* calls that could potentially throw don't
throw...)

···

On Thu, Aug 18, 2016 at 10:19 AM, Xiaodi Wu <xiaodi.wu@gmail.com> wrote:

could potentially throw but doesn't, "try?" wraps the entire rhs and gives
you .some(value). IMO, this is pretty sensible for the reason he gives.

B. I'm in favor of warning instead of error, for precisely the internal
discussion rationale communicated by Slava. I'm willing to live with "try?
42" being only a warning if that means my code won't stop compiling when
someone decides a library function doesn't need to throw.

Sam--here, changing warning to error would not solve your original
problem, because in that example "try?" does prefix at least one throwing
function, so you wouldn't get an error anyway.

C. However, given the thinking in (A), I do think how "try?" composes with
"as?" is a little counterintuitive or at least overly ceremonious, though
technically it is possible to reason through.

It's true that currently you can use the multiple nested optionals to
figure out whether either a throwing function threw (but not which throwing
function out of potentially more than one) or whether the cast did not
succeed. But, since "try?" after all means "give me nil if anything
throws," it kind of makes less sense that you get all this nesting and
detailed information when it composes with "as?". If you really wanted that
level of detail, you could always evaluate "try?" and "as?" in separate
statements. What I'd propose instead is this:

If "try?" is composed with "as?", and "as?" yields "nil", then "try?"
should not wrap that value in another optional.

Does that sound sensible?

On Thu, Aug 18, 2016 at 3:54 AM, Sikhapol Saijit via swift-evolution < > swift-evolution@swift.org> wrote:

On Aug 18, 2016, at 3:42 PM, Slava Pestov <spestov@apple.com> wrote:

On Aug 18, 2016, at 12:52 AM, David Hart via swift-evolution < >> swift-evolution@swift.org> wrote:

Opinions inline:

On 18 Aug 2016, at 07:43, Sikhapol Saijit via swift-evolution < >> swift-evolution@swift.org> wrote:

Hi all,

Yesterday I tried this code:

func couldFailButWillNot() throws -> Any {
    return 42
}

if let a = try? couldFailButWillNot() as? Int {
    print(a)
}

And was surprised that the output was *Optional(42)* on both Swift 2 and
Swift 3.
I always have the impression that when a variable is resolved with if let
it will never be optional.

So, with a little investigation, I found out that it happens because as? has
higher precedence than try? and is evaluated first.
And the whole expression `try? couldFailButWillNot() as? Int` evaluated
as *Optional(Optional(42))*.

Also, I’m surprised that try? can be used with non-method-call.
This code: `print(try? 42)` will print *Optional(42)*.

So, the questions are:

1. Is it intentional that try? can be used with a "non-method-call" and
return an optional of the type that follows?

I think this is the real solution. try and try? should not be allowed on
non-throwing functions or expressions.

This is a warning right now — do you think it should be an error?

Slavas-MacBook-Pro:~ slava$ cat ttt.swift
func f() {}

func g() {
  try f()
  try? f()
}

Slavas-MacBook-Pro:~ slava$ swiftc ttt.swift
*ttt.swift:4:3: **warning: **no calls to throwing functions occur within
'try' expression*
  try f()
* ^*
*ttt.swift:5:8: **warning: **no calls to throwing functions occur within
'try' expression*
  try? f()
* ^*

Thank you Slava,

While I think using try/try? on anything but a throwing function call
should be an error, right now it even works with anything. `try? 42` will
just wrap 42 in an optional and give some warning now.

2. Should we design try? to have higher precedence than as? or any
operators at all?
My intuition tells me that
let a = try? couldFailButWillNot() as? Int
should be equivalent to
let a = (try? couldFailButWillNot()) as? Int

That’s worth considering. try feels like it should tie very strongly with
the throwing expression.

3. Do you think that doubly-nested optional (or multi-level-nested
optional) is confusing and should be removed from Swift? (Yes, I’ve seen
this blog post Optionals Case Study: valuesForKeys
<https://developer.apple.com/swift/blog/?id=12>).
For me *Optional(nil)* (aka *Optional.Some(Optional.None))*) doesn’t
make much sense.
Maybe, one of the solution is to always have optional of optional merged
into a single level optional? Like *Optional(Optional(Optional(42)))* should
be the merged to and evaluated as *Optional(42)*.

I don’t think this is the solution. Even if it was, how would you expect
to “remove” them from Swift? Optionals are simply an enum with an
associated value. We’d have to introduce a language feature to restrict
values that can be stored in enum cases? It sounds awfully complicated.

BTW, the code above is merely for a demonstration. The actual code was
more of something like this:

func parse(JSON: Data) throws -> Any {
    // …
}

if let dict = try? parse(JSON: json) as? [String: Any] {
    // assume dict is a valid [String: Any] dictionary
    // …
}

I’m new to this mailing list so I’m not sure if this belongs here. I’m
sorry in advance if it doesn’t.

Thank you,
Sam
_______________________________________________
swift-evolution mailing list
swift-evolution@swift.org
https://lists.swift.org/mailman/listinfo/swift-evolution

_______________________________________________
swift-evolution mailing list
swift-evolution@swift.org
https://lists.swift.org/mailman/listinfo/swift-evolution

_______________________________________________
swift-evolution mailing list
swift-evolution@swift.org
https://lists.swift.org/mailman/listinfo/swift-evolution


(Xiaodi Wu) #13

The issue would be that, in the case of "try? foo()", nil and .some(nil)
might mean very different things.

···

On Thu, Aug 18, 2016 at 10:40 John McCall <rjmccall@apple.com> wrote:

On Aug 18, 2016, at 8:19 AM, Xiaodi Wu via swift-evolution < > swift-evolution@swift.org> wrote:

Lots of interesting points here. I do think there's an improvement
possible here, but it's actually along the lines of Sam's original
suggestion #3 (not vis-a-vis all of Swift, but specifically for how try?
composes with as?):

A. I'm in favor of the current behavior where try prefixes an entire
statement: it solves the precise issue of multiple nested optionals or
multiple unwrapping of optionals in the situation where one statement has
calls to many throwing functions. It says instead, I want nil if anything
in this statement throws, otherwise, give me .some(value).

Sam--I think you may have misunderstood Charles's explanation. He's not
saying "try?" attaches with lower or higher precedence as compared to
"as?". Rather, I think the mental model is that "try?" prefixes the whole
right-hand side (rhs), and if *any* call on the rhs throws, the whole rhs
evaluates to nil, but if *any* call could potentially throw but doesn't,
"try?" wraps the entire rhs and gives you .some(value). IMO, this is pretty
sensible for the reason he gives.

B. I'm in favor of warning instead of error, for precisely the internal
discussion rationale communicated by Slava. I'm willing to live with "try?
42" being only a warning if that means my code won't stop compiling when
someone decides a library function doesn't need to throw.

Sam--here, changing warning to error would not solve your original
problem, because in that example "try?" does prefix at least one throwing
function, so you wouldn't get an error anyway.

C. However, given the thinking in (A), I do think how "try?" composes with
"as?" is a little counterintuitive or at least overly ceremonious, though
technically it is possible to reason through.

It's true that currently you can use the multiple nested optionals to
figure out whether either a throwing function threw (but not which throwing
function out of potentially more than one) or whether the cast did not
succeed. But, since "try?" after all means "give me nil if anything
throws," it kind of makes less sense that you get all this nesting and
detailed information when it composes with "as?". If you really wanted that
level of detail, you could always evaluate "try?" and "as?" in separate
statements. What I'd propose instead is this:

If "try?" is composed with "as?", and "as?" yields "nil", then "try?"
should not wrap that value in another optional.

We can't make the typing decision dependent on a dynamic property like
whether the cast fails. And I don't like the idea of changing its typing
rule based on the form of the nested expression. But we could make "try?
foo()" avoid adding an extra level of optionality, the same way that
"a?.foo()" does.

John.

Does that sound sensible?

On Thu, Aug 18, 2016 at 3:54 AM, Sikhapol Saijit via swift-evolution < > swift-evolution@swift.org> wrote:

On Aug 18, 2016, at 3:42 PM, Slava Pestov <spestov@apple.com> wrote:

On Aug 18, 2016, at 12:52 AM, David Hart via swift-evolution < >> swift-evolution@swift.org> wrote:

Opinions inline:

On 18 Aug 2016, at 07:43, Sikhapol Saijit via swift-evolution < >> swift-evolution@swift.org> wrote:

Hi all,

Yesterday I tried this code:

func couldFailButWillNot() throws -> Any {
    return 42
}

if let a = try? couldFailButWillNot() as? Int {
    print(a)
}

And was surprised that the output was *Optional(42)* on both Swift 2 and
Swift 3.
I always have the impression that when a variable is resolved with if let
it will never be optional.

So, with a little investigation, I found out that it happens because as? has
higher precedence than try? and is evaluated first.
And the whole expression `try? couldFailButWillNot() as? Int` evaluated
as *Optional(Optional(42))*.

Also, I’m surprised that try? can be used with non-method-call.
This code: `print(try? 42)` will print *Optional(42)*.

So, the questions are:

1. Is it intentional that try? can be used with a "non-method-call" and
return an optional of the type that follows?

I think this is the real solution. try and try? should not be allowed on
non-throwing functions or expressions.

This is a warning right now — do you think it should be an error?

Slavas-MacBook-Pro:~ slava$ cat ttt.swift
func f() {}

func g() {
  try f()
  try? f()
}

Slavas-MacBook-Pro:~ slava$ swiftc ttt.swift
*ttt.swift:4:3: **warning: **no calls to throwing functions occur within
'try' expression*
  try f()
* ^*
*ttt.swift:5:8: **warning: **no calls to throwing functions occur within
'try' expression*
  try? f()
* ^*

Thank you Slava,

While I think using try/try? on anything but a throwing function call
should be an error, right now it even works with anything. `try? 42` will
just wrap 42 in an optional and give some warning now.

2. Should we design try? to have higher precedence than as? or any
operators at all?
My intuition tells me that
let a = try? couldFailButWillNot() as? Int
should be equivalent to
let a = (try? couldFailButWillNot()) as? Int

That’s worth considering. try feels like it should tie very strongly with
the throwing expression.

3. Do you think that doubly-nested optional (or multi-level-nested
optional) is confusing and should be removed from Swift? (Yes, I’ve seen
this blog post Optionals Case Study: valuesForKeys
<https://developer.apple.com/swift/blog/?id=12>).
For me *Optional(nil)* (aka *Optional.Some(Optional.None))*) doesn’t
make much sense.
Maybe, one of the solution is to always have optional of optional merged
into a single level optional? Like *Optional(Optional(Optional(42)))* should
be the merged to and evaluated as *Optional(42)*.

I don’t think this is the solution. Even if it was, how would you expect
to “remove” them from Swift? Optionals are simply an enum with an
associated value. We’d have to introduce a language feature to restrict
values that can be stored in enum cases? It sounds awfully complicated.

BTW, the code above is merely for a demonstration. The actual code was
more of something like this:

func parse(JSON: Data) throws -> Any {
    // …
}

if let dict = try? parse(JSON: json) as? [String: Any] {
    // assume dict is a valid [String: Any] dictionary
    // …
}

I’m new to this mailing list so I’m not sure if this belongs here. I’m
sorry in advance if it doesn’t.

Thank you,
Sam
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(Xiaodi Wu) #14

The issue would be that, in the case of "try? foo()", nil and .some(nil)
might mean very different things.

This is true of a?.foo() as well. But yes, I think it is more likely that
someone would want to treat them differently for try?.

Agreed.

My proposed solution was half-baked, but it may be workable--I'm not
suggesting typing decisions based on a dynamic property, of course. It'd be
something like this:

`as?` would produce a result of a type named something like
CastingOptional<T>, which on assignment or essentially any other operation
is promoted/bridged/[insert much more correct term here] to an Optional<T>
like how T is automatically promoted to Optional<T>. However, `try?` will
not wrap a CastingOptional<T> into an Optional<Optional<T>>.

···

On Thu, Aug 18, 2016 at 11:30 AM, John McCall <rjmccall@apple.com> wrote:

On Aug 18, 2016, at 8:46 AM, Xiaodi Wu <xiaodi.wu@gmail.com> wrote:

John.

On Thu, Aug 18, 2016 at 10:40 John McCall <rjmccall@apple.com> wrote:

On Aug 18, 2016, at 8:19 AM, Xiaodi Wu via swift-evolution < >> swift-evolution@swift.org> wrote:

Lots of interesting points here. I do think there's an improvement
possible here, but it's actually along the lines of Sam's original
suggestion #3 (not vis-a-vis all of Swift, but specifically for how try?
composes with as?):

A. I'm in favor of the current behavior where try prefixes an entire
statement: it solves the precise issue of multiple nested optionals or
multiple unwrapping of optionals in the situation where one statement has
calls to many throwing functions. It says instead, I want nil if anything
in this statement throws, otherwise, give me .some(value).

Sam--I think you may have misunderstood Charles's explanation. He's not
saying "try?" attaches with lower or higher precedence as compared to
"as?". Rather, I think the mental model is that "try?" prefixes the whole
right-hand side (rhs), and if *any* call on the rhs throws, the whole rhs
evaluates to nil, but if *any* call could potentially throw but doesn't,
"try?" wraps the entire rhs and gives you .some(value). IMO, this is pretty
sensible for the reason he gives.

B. I'm in favor of warning instead of error, for precisely the internal
discussion rationale communicated by Slava. I'm willing to live with "try?
42" being only a warning if that means my code won't stop compiling when
someone decides a library function doesn't need to throw.

Sam--here, changing warning to error would not solve your original
problem, because in that example "try?" does prefix at least one throwing
function, so you wouldn't get an error anyway.

C. However, given the thinking in (A), I do think how "try?" composes
with "as?" is a little counterintuitive or at least overly ceremonious,
though technically it is possible to reason through.

It's true that currently you can use the multiple nested optionals to
figure out whether either a throwing function threw (but not which throwing
function out of potentially more than one) or whether the cast did not
succeed. But, since "try?" after all means "give me nil if anything
throws," it kind of makes less sense that you get all this nesting and
detailed information when it composes with "as?". If you really wanted that
level of detail, you could always evaluate "try?" and "as?" in separate
statements. What I'd propose instead is this:

If "try?" is composed with "as?", and "as?" yields "nil", then "try?"
should not wrap that value in another optional.

We can't make the typing decision dependent on a dynamic property like
whether the cast fails. And I don't like the idea of changing its typing
rule based on the form of the nested expression. But we could make "try?
foo()" avoid adding an extra level of optionality, the same way that
"a?.foo()" does.

John.

Does that sound sensible?

On Thu, Aug 18, 2016 at 3:54 AM, Sikhapol Saijit via swift-evolution < >> swift-evolution@swift.org> wrote:

On Aug 18, 2016, at 3:42 PM, Slava Pestov <spestov@apple.com> wrote:

On Aug 18, 2016, at 12:52 AM, David Hart via swift-evolution < >>> swift-evolution@swift.org> wrote:

Opinions inline:

On 18 Aug 2016, at 07:43, Sikhapol Saijit via swift-evolution < >>> swift-evolution@swift.org> wrote:

Hi all,

Yesterday I tried this code:

func couldFailButWillNot() throws -> Any {
    return 42
}

if let a = try? couldFailButWillNot() as? Int {
    print(a)
}

And was surprised that the output was *Optional(42)* on both Swift 2
and Swift 3.
I always have the impression that when a variable is resolved with if
let it will never be optional.

So, with a little investigation, I found out that it happens because as? has
higher precedence than try? and is evaluated first.
And the whole expression `try? couldFailButWillNot() as? Int` evaluated
as *Optional(Optional(42))*.

Also, I’m surprised that try? can be used with non-method-call.
This code: `print(try? 42)` will print *Optional(42)*.

So, the questions are:

1. Is it intentional that try? can be used with a "non-method-call" and
return an optional of the type that follows?

I think this is the real solution. try and try? should not be allowed on
non-throwing functions or expressions.

This is a warning right now — do you think it should be an error?

Slavas-MacBook-Pro:~ slava$ cat ttt.swift
func f() {}

func g() {
  try f()
  try? f()
}

Slavas-MacBook-Pro:~ slava$ swiftc ttt.swift
*ttt.swift:4:3: **warning: **no calls to throwing functions occur
within 'try' expression*
  try f()
* ^*
*ttt.swift:5:8: **warning: **no calls to throwing functions occur
within 'try' expression*
  try? f()
* ^*

Thank you Slava,

While I think using try/try? on anything but a throwing function call
should be an error, right now it even works with anything. `try? 42` will
just wrap 42 in an optional and give some warning now.

2. Should we design try? to have higher precedence than as? or any
operators at all?
My intuition tells me that
let a = try? couldFailButWillNot() as? Int
should be equivalent to
let a = (try? couldFailButWillNot()) as? Int

That’s worth considering. try feels like it should tie very strongly
with the throwing expression.

3. Do you think that doubly-nested optional (or multi-level-nested
optional) is confusing and should be removed from Swift? (Yes, I’ve seen
this blog post Optionals Case Study: valuesForKeys
<https://developer.apple.com/swift/blog/?id=12>).
For me *Optional(nil)* (aka *Optional.Some(Optional.None))*) doesn’t
make much sense.
Maybe, one of the solution is to always have optional of optional merged
into a single level optional? Like *Optional(Optional(Optional(42)))* should
be the merged to and evaluated as *Optional(42)*.

I don’t think this is the solution. Even if it was, how would you expect
to “remove” them from Swift? Optionals are simply an enum with an
associated value. We’d have to introduce a language feature to restrict
values that can be stored in enum cases? It sounds awfully complicated.

BTW, the code above is merely for a demonstration. The actual code was
more of something like this:

func parse(JSON: Data) throws -> Any {
    // …
}

if let dict = try? parse(JSON: json) as? [String: Any] {
    // assume dict is a valid [String: Any] dictionary
    // …
}

I’m new to this mailing list so I’m not sure if this belongs here. I’m
sorry in advance if it doesn’t.

Thank you,
Sam
_______________________________________________
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https://lists.swift.org/mailman/listinfo/swift-evolution

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https://lists.swift.org/mailman/listinfo/swift-evolution

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https://lists.swift.org/mailman/listinfo/swift-evolution

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https://lists.swift.org/mailman/listinfo/swift-evolution


(John McCall) #15

The issue would be that, in the case of "try? foo()", nil and .some(nil) might mean very different things.

This is true of a?.foo() as well. But yes, I think it is more likely that someone would want to treat them differently for try?.

Agreed.

My proposed solution was half-baked, but it may be workable--I'm not suggesting typing decisions based on a dynamic property, of course. It'd be something like this:

`as?` would produce a result of a type named something like CastingOptional<T>, which on assignment or essentially any other operation is promoted/bridged/[insert much more correct term here] to an Optional<T> like how T is automatically promoted to Optional<T>. However, `try?` will not wrap a CastingOptional<T> into an Optional<Optional<T>>.

The way this is done for ?-chaining is that the result of the chain is coerced to T?, for a fresh unbound type T. If the result is already of type U?, T will be bound to U and there's no "stacking" of optionals; if the result is a non-optional type V, T will be bound to V and therefore the chain gains a level of optionality. I think that is simpler and more consistent than inventing a new flavor of Optional with complex conversion and defaulting rules.

John.

···

On Aug 18, 2016, at 10:11 AM, Xiaodi Wu <xiaodi.wu@gmail.com> wrote:
On Thu, Aug 18, 2016 at 11:30 AM, John McCall <rjmccall@apple.com <mailto:rjmccall@apple.com>> wrote:

On Aug 18, 2016, at 8:46 AM, Xiaodi Wu <xiaodi.wu@gmail.com <mailto:xiaodi.wu@gmail.com>> wrote:

John.

On Thu, Aug 18, 2016 at 10:40 John McCall <rjmccall@apple.com <mailto:rjmccall@apple.com>> wrote:

On Aug 18, 2016, at 8:19 AM, Xiaodi Wu via swift-evolution <swift-evolution@swift.org <mailto:swift-evolution@swift.org>> wrote:

Lots of interesting points here. I do think there's an improvement possible here, but it's actually along the lines of Sam's original suggestion #3 (not vis-a-vis all of Swift, but specifically for how try? composes with as?):

A. I'm in favor of the current behavior where try prefixes an entire statement: it solves the precise issue of multiple nested optionals or multiple unwrapping of optionals in the situation where one statement has calls to many throwing functions. It says instead, I want nil if anything in this statement throws, otherwise, give me .some(value).

Sam--I think you may have misunderstood Charles's explanation. He's not saying "try?" attaches with lower or higher precedence as compared to "as?". Rather, I think the mental model is that "try?" prefixes the whole right-hand side (rhs), and if *any* call on the rhs throws, the whole rhs evaluates to nil, but if *any* call could potentially throw but doesn't, "try?" wraps the entire rhs and gives you .some(value). IMO, this is pretty sensible for the reason he gives.

B. I'm in favor of warning instead of error, for precisely the internal discussion rationale communicated by Slava. I'm willing to live with "try? 42" being only a warning if that means my code won't stop compiling when someone decides a library function doesn't need to throw.

Sam--here, changing warning to error would not solve your original problem, because in that example "try?" does prefix at least one throwing function, so you wouldn't get an error anyway.

C. However, given the thinking in (A), I do think how "try?" composes with "as?" is a little counterintuitive or at least overly ceremonious, though technically it is possible to reason through.

It's true that currently you can use the multiple nested optionals to figure out whether either a throwing function threw (but not which throwing function out of potentially more than one) or whether the cast did not succeed. But, since "try?" after all means "give me nil if anything throws," it kind of makes less sense that you get all this nesting and detailed information when it composes with "as?". If you really wanted that level of detail, you could always evaluate "try?" and "as?" in separate statements. What I'd propose instead is this:

If "try?" is composed with "as?", and "as?" yields "nil", then "try?" should not wrap that value in another optional.

We can't make the typing decision dependent on a dynamic property like whether the cast fails. And I don't like the idea of changing its typing rule based on the form of the nested expression. But we could make "try? foo()" avoid adding an extra level of optionality, the same way that "a?.foo()" does.

John.

Does that sound sensible?

On Thu, Aug 18, 2016 at 3:54 AM, Sikhapol Saijit via swift-evolution <swift-evolution@swift.org <mailto:swift-evolution@swift.org>> wrote:

On Aug 18, 2016, at 3:42 PM, Slava Pestov <spestov@apple.com <mailto:spestov@apple.com>> wrote:

On Aug 18, 2016, at 12:52 AM, David Hart via swift-evolution <swift-evolution@swift.org <mailto:swift-evolution@swift.org>> wrote:

Opinions inline:

On 18 Aug 2016, at 07:43, Sikhapol Saijit via swift-evolution <swift-evolution@swift.org <mailto:swift-evolution@swift.org>> wrote:

Hi all,

Yesterday I tried this code:

func couldFailButWillNot() throws -> Any {
    return 42
}

if let a = try? couldFailButWillNot() as? Int {
    print(a)
}

And was surprised that the output was Optional(42) on both Swift 2 and Swift 3.
I always have the impression that when a variable is resolved with if let it will never be optional.

So, with a little investigation, I found out that it happens because as? has higher precedence than try? and is evaluated first.
And the whole expression `try? couldFailButWillNot() as? Int` evaluated as Optional(Optional(42)).

Also, I’m surprised that try? can be used with non-method-call.
This code: `print(try? 42)` will print Optional(42).

So, the questions are:

1. Is it intentional that try? can be used with a "non-method-call" and return an optional of the type that follows?

I think this is the real solution. try and try? should not be allowed on non-throwing functions or expressions.

This is a warning right now — do you think it should be an error?

Slavas-MacBook-Pro:~ slava$ cat ttt.swift
func f() {}

func g() {
  try f()
  try? f()
}

Slavas-MacBook-Pro:~ slava$ swiftc ttt.swift
ttt.swift:4:3: warning: no calls to throwing functions occur within 'try' expression
  try f()
  ^
ttt.swift:5:8: warning: no calls to throwing functions occur within 'try' expression
  try? f()
       ^

Thank you Slava,

While I think using try/try? on anything but a throwing function call should be an error, right now it even works with anything. `try? 42` will just wrap 42 in an optional and give some warning now.

2. Should we design try? to have higher precedence than as? or any operators at all?
My intuition tells me that
let a = try? couldFailButWillNot() as? Int
should be equivalent to
let a = (try? couldFailButWillNot()) as? Int

That’s worth considering. try feels like it should tie very strongly with the throwing expression.

3. Do you think that doubly-nested optional (or multi-level-nested optional) is confusing and should be removed from Swift? (Yes, I’ve seen this blog post Optionals Case Study: valuesForKeys <https://developer.apple.com/swift/blog/?id=12>).
For me Optional(nil) (aka Optional.Some(Optional.None))) doesn’t make much sense.
Maybe, one of the solution is to always have optional of optional merged into a single level optional? Like Optional(Optional(Optional(42))) should be the merged to and evaluated as Optional(42).

I don’t think this is the solution. Even if it was, how would you expect to “remove” them from Swift? Optionals are simply an enum with an associated value. We’d have to introduce a language feature to restrict values that can be stored in enum cases? It sounds awfully complicated.

BTW, the code above is merely for a demonstration. The actual code was more of something like this:

func parse(JSON: Data) throws -> Any {
    // …
}

if let dict = try? parse(JSON: json) as? [String: Any] {
    // assume dict is a valid [String: Any] dictionary
    // …
}

I’m new to this mailing list so I’m not sure if this belongs here. I’m sorry in advance if it doesn’t.

Thank you,
Sam
_______________________________________________
swift-evolution mailing list
swift-evolution@swift.org <mailto:swift-evolution@swift.org>
https://lists.swift.org/mailman/listinfo/swift-evolution

_______________________________________________
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swift-evolution@swift.org <mailto:swift-evolution@swift.org>
https://lists.swift.org/mailman/listinfo/swift-evolution

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swift-evolution@swift.org <mailto:swift-evolution@swift.org>
https://lists.swift.org/mailman/listinfo/swift-evolution

_______________________________________________
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swift-evolution@swift.org <mailto:swift-evolution@swift.org>
https://lists.swift.org/mailman/listinfo/swift-evolution


(Xiaodi Wu) #16

The issue would be that, in the case of "try? foo()", nil and .some(nil)
might mean very different things.

This is true of a?.foo() as well. But yes, I think it is more likely
that someone would want to treat them differently for try?.

Agreed.

My proposed solution was half-baked, but it may be workable--I'm not
suggesting typing decisions based on a dynamic property, of course. It'd be
something like this:

`as?` would produce a result of a type named something like
CastingOptional<T>, which on assignment or essentially any other operation
is promoted/bridged/[insert much more correct term here] to an Optional<T>
like how T is automatically promoted to Optional<T>. However, `try?` will
not wrap a CastingOptional<T> into an Optional<Optional<T>>.

The way this is done for ?-chaining is that the result of the chain is
coerced to T?, for a fresh unbound type T. If the result is already of
type U?, T will be bound to U and there's no "stacking" of optionals; if
the result is a non-optional type V, T will be bound to V and therefore the
chain gains a level of optionality. I think that is simpler and more
consistent than inventing a new flavor of Optional with complex conversion
and defaulting rules.

Yeah, that's definitely much simpler.

I'm just not sure I'd be comfortable with `try?` refusing to stack
optionals for all arbitrary functions `(...) -> Optional<T>` just like it
works for ?-chaining, for the reasons we just discussed. So, if we're to
stick to consistent rules, I'd rather that `try?` continue to stack
optionals all the time.

The status quo isn't unteachable: if you have try? and as?, that's two
question marks, so you get back two stacked optionals. That's livable.

···

On Thu, Aug 18, 2016 at 1:20 PM, John McCall <rjmccall@apple.com> wrote:

On Aug 18, 2016, at 10:11 AM, Xiaodi Wu <xiaodi.wu@gmail.com> wrote:
On Thu, Aug 18, 2016 at 11:30 AM, John McCall <rjmccall@apple.com> wrote:

On Aug 18, 2016, at 8:46 AM, Xiaodi Wu <xiaodi.wu@gmail.com> wrote:

John.

John.

On Thu, Aug 18, 2016 at 10:40 John McCall <rjmccall@apple.com> wrote:

On Aug 18, 2016, at 8:19 AM, Xiaodi Wu via swift-evolution < >>> swift-evolution@swift.org> wrote:

Lots of interesting points here. I do think there's an improvement
possible here, but it's actually along the lines of Sam's original
suggestion #3 (not vis-a-vis all of Swift, but specifically for how try?
composes with as?):

A. I'm in favor of the current behavior where try prefixes an entire
statement: it solves the precise issue of multiple nested optionals or
multiple unwrapping of optionals in the situation where one statement has
calls to many throwing functions. It says instead, I want nil if anything
in this statement throws, otherwise, give me .some(value).

Sam--I think you may have misunderstood Charles's explanation. He's not
saying "try?" attaches with lower or higher precedence as compared to
"as?". Rather, I think the mental model is that "try?" prefixes the whole
right-hand side (rhs), and if *any* call on the rhs throws, the whole rhs
evaluates to nil, but if *any* call could potentially throw but doesn't,
"try?" wraps the entire rhs and gives you .some(value). IMO, this is pretty
sensible for the reason he gives.

B. I'm in favor of warning instead of error, for precisely the internal
discussion rationale communicated by Slava. I'm willing to live with "try?
42" being only a warning if that means my code won't stop compiling when
someone decides a library function doesn't need to throw.

Sam--here, changing warning to error would not solve your original
problem, because in that example "try?" does prefix at least one throwing
function, so you wouldn't get an error anyway.

C. However, given the thinking in (A), I do think how "try?" composes
with "as?" is a little counterintuitive or at least overly ceremonious,
though technically it is possible to reason through.

It's true that currently you can use the multiple nested optionals to
figure out whether either a throwing function threw (but not which throwing
function out of potentially more than one) or whether the cast did not
succeed. But, since "try?" after all means "give me nil if anything
throws," it kind of makes less sense that you get all this nesting and
detailed information when it composes with "as?". If you really wanted that
level of detail, you could always evaluate "try?" and "as?" in separate
statements. What I'd propose instead is this:

If "try?" is composed with "as?", and "as?" yields "nil", then "try?"
should not wrap that value in another optional.

We can't make the typing decision dependent on a dynamic property like
whether the cast fails. And I don't like the idea of changing its typing
rule based on the form of the nested expression. But we could make "try?
foo()" avoid adding an extra level of optionality, the same way that
"a?.foo()" does.

John.

Does that sound sensible?

On Thu, Aug 18, 2016 at 3:54 AM, Sikhapol Saijit via swift-evolution < >>> swift-evolution@swift.org> wrote:

On Aug 18, 2016, at 3:42 PM, Slava Pestov <spestov@apple.com> wrote:

On Aug 18, 2016, at 12:52 AM, David Hart via swift-evolution < >>>> swift-evolution@swift.org> wrote:

Opinions inline:

On 18 Aug 2016, at 07:43, Sikhapol Saijit via swift-evolution < >>>> swift-evolution@swift.org> wrote:

Hi all,

Yesterday I tried this code:

func couldFailButWillNot() throws -> Any {
    return 42
}

if let a = try? couldFailButWillNot() as? Int {
    print(a)
}

And was surprised that the output was *Optional(42)* on both Swift 2
and Swift 3.
I always have the impression that when a variable is resolved with if
let it will never be optional.

So, with a little investigation, I found out that it happens because as
? has higher precedence than try? and is evaluated first.
And the whole expression `try? couldFailButWillNot() as? Int` evaluated
as *Optional(Optional(42))*.

Also, I’m surprised that try? can be used with non-method-call.
This code: `print(try? 42)` will print *Optional(42)*.

So, the questions are:

1. Is it intentional that try? can be used with a "non-method-call"
and return an optional of the type that follows?

I think this is the real solution. try and try? should not be allowed
on non-throwing functions or expressions.

This is a warning right now — do you think it should be an error?

Slavas-MacBook-Pro:~ slava$ cat ttt.swift
func f() {}

func g() {
  try f()
  try? f()
}

Slavas-MacBook-Pro:~ slava$ swiftc ttt.swift
*ttt.swift:4:3: **warning: **no calls to throwing functions occur
within 'try' expression*
  try f()
* ^*
*ttt.swift:5:8: **warning: **no calls to throwing functions occur
within 'try' expression*
  try? f()
* ^*

Thank you Slava,

While I think using try/try? on anything but a throwing function call
should be an error, right now it even works with anything. `try? 42` will
just wrap 42 in an optional and give some warning now.

2. Should we design try? to have higher precedence than as? or any
operators at all?
My intuition tells me that
let a = try? couldFailButWillNot() as? Int
should be equivalent to
let a = (try? couldFailButWillNot()) as? Int

That’s worth considering. try feels like it should tie very strongly
with the throwing expression.

3. Do you think that doubly-nested optional (or multi-level-nested
optional) is confusing and should be removed from Swift? (Yes, I’ve seen
this blog post Optionals Case Study: valuesForKeys
<https://developer.apple.com/swift/blog/?id=12>).
For me *Optional(nil)* (aka *Optional.Some(Optional.None))*) doesn’t
make much sense.
Maybe, one of the solution is to always have optional of optional
merged into a single level optional? Like
*Optional(Optional(Optional(42)))* should be the merged to and
evaluated as *Optional(42)*.

I don’t think this is the solution. Even if it was, how would you
expect to “remove” them from Swift? Optionals are simply an enum with an
associated value. We’d have to introduce a language feature to restrict
values that can be stored in enum cases? It sounds awfully complicated.

BTW, the code above is merely for a demonstration. The actual code was
more of something like this:

func parse(JSON: Data) throws -> Any {
    // …
}

if let dict = try? parse(JSON: json) as? [String: Any] {
    // assume dict is a valid [String: Any] dictionary
    // …
}

I’m new to this mailing list so I’m not sure if this belongs here. I’m
sorry in advance if it doesn’t.

Thank you,
Sam
_______________________________________________
swift-evolution mailing list
swift-evolution@swift.org
https://lists.swift.org/mailman/listinfo/swift-evolution

_______________________________________________
swift-evolution mailing list
swift-evolution@swift.org
https://lists.swift.org/mailman/listinfo/swift-evolution

_______________________________________________
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swift-evolution@swift.org
https://lists.swift.org/mailman/listinfo/swift-evolution

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