Can anyone please explain this behavior?


(Jens Persson) #1

// This little Swift program compiles (and runs) fine:

func foo() -> Int { return a }
let a = 1
let b = 2
print(foo())

But if `foo()` returns `b` instead of `a`, I get this compile time error:
"Use of unresolved identifier `b`"


(Marco S Hyman) #2

Interesting. Seems to have something to do with ordering as if you code it this way:

let a = 1
let b = 2
func foo() -> Int { return b }
print(foo())

it works fine (Xcode 8 playground).

···

On Sep 21, 2016, at 2:22 PM, Jens Persson via swift-users <swift-users@swift.org> wrote:

// This little Swift program compiles (and runs) fine:

func foo() -> Int { return a }
let a = 1
let b = 2
print(foo())

But if `foo()` returns `b` instead of `a`, I get this compile time error:
"Use of unresolved identifier `b`”


(Joe Groff) #3

This looks like a bug to me (cc-ing Jordan, who's thought about global scoping issues more than me). In "script mode", it shouldn't be possible to refer to a variable before its initialization is executed. However, the way this is currently modeled is…problematic, to say the least, among other reasons because script globals are still visible to "library" files in the same module.

-Joe

···

On Sep 21, 2016, at 2:22 PM, Jens Persson via swift-users <swift-users@swift.org> wrote:

// This little Swift program compiles (and runs) fine:

func foo() -> Int { return a }
let a = 1
let b = 2
print(foo())

But if `foo()` returns `b` instead of `a`, I get this compile time error:
"Use of unresolved identifier `b`"


(Zhao Xin) #4

I suggest you defining the variables before using them like Marco does.
Although from the global variable point of view, both way should be fine.
But the glitch does exist, especially in playground.

Zhaoxin

···

On Thu, Sep 22, 2016 at 5:59 AM, Marco S Hyman via swift-users < swift-users@swift.org> wrote:

> On Sep 21, 2016, at 2:22 PM, Jens Persson via swift-users < > swift-users@swift.org> wrote:
>
> // This little Swift program compiles (and runs) fine:
>
> func foo() -> Int { return a }
> let a = 1
> let b = 2
> print(foo())
>
> But if `foo()` returns `b` instead of `a`, I get this compile time error:
> "Use of unresolved identifier `b`”

Interesting. Seems to have something to do with ordering as if you code
it this way:

let a = 1
let b = 2
func foo() -> Int { return b }
print(foo())

it works fine (Xcode 8 playground).
_______________________________________________
swift-users mailing list
swift-users@swift.org
https://lists.swift.org/mailman/listinfo/swift-users


(Jens Persson) #5

Did you see the other code examples that came up in that twitter
conversations?
For example:

This worrying little program compiles:
func f() -> Int {
    return a
}
let a = f()

It also compiles if you print(a) at the end, and it will print 0.

If we replace Int with [Int] it will still compile but crash when run.

And also this:

AnotherFile.swift containing:
func f() -> Int {
    return a
}
let a = f()

main.swift containing
print(a)

Compile, run (for eternity, at 0% CPU).

/Jens

···

On Thu, Sep 22, 2016 at 3:13 AM, Joe Groff <jgroff@apple.com> wrote:

> On Sep 21, 2016, at 2:22 PM, Jens Persson via swift-users < > swift-users@swift.org> wrote:
>
> // This little Swift program compiles (and runs) fine:
>
> func foo() -> Int { return a }
> let a = 1
> let b = 2
> print(foo())
>
> But if `foo()` returns `b` instead of `a`, I get this compile time error:
> "Use of unresolved identifier `b`"

This looks like a bug to me (cc-ing Jordan, who's thought about global
scoping issues more than me). In "script mode", it shouldn't be possible to
refer to a variable before its initialization is executed. However, the way
this is currently modeled is…problematic, to say the least, among other
reasons because script globals are still visible to "library" files in the
same module.

-Joe


(Jens Persson) #6

I asked the same question on twitter, and it resulted in some interesting
finds:
https://twitter.com/bitCycle/status/778697998893142016

···

On Thu, Sep 22, 2016 at 12:41 AM, Zhao Xin <owenzx@gmail.com> wrote:

I suggest you defining the variables before using them like Marco does.
Although from the global variable point of view, both way should be fine.
But the glitch does exist, especially in playground.

Zhaoxin

On Thu, Sep 22, 2016 at 5:59 AM, Marco S Hyman via swift-users < > swift-users@swift.org> wrote:

> On Sep 21, 2016, at 2:22 PM, Jens Persson via swift-users < >> swift-users@swift.org> wrote:
>
> // This little Swift program compiles (and runs) fine:
>
> func foo() -> Int { return a }
> let a = 1
> let b = 2
> print(foo())
>
> But if `foo()` returns `b` instead of `a`, I get this compile time
error:
> "Use of unresolved identifier `b`”

Interesting. Seems to have something to do with ordering as if you code
it this way:

let a = 1
let b = 2
func foo() -> Int { return b }
print(foo())

it works fine (Xcode 8 playground).
_______________________________________________
swift-users mailing list
swift-users@swift.org
https://lists.swift.org/mailman/listinfo/swift-users


(Joe Groff) #7

Did you see the other code examples that came up in that twitter conversations?

These all look like the same problem. Whatever you're seeing is an accident of undefined behavior due to the hole in our semantic checking.

-Joe

···

On Sep 21, 2016, at 11:04 PM, Jens Persson <jens@bitcycle.com> wrote:

For example:

This worrying little program compiles:
func f() -> Int {
    return a
}
let a = f()

It also compiles if you print(a) at the end, and it will print 0.

If we replace Int with [Int] it will still compile but crash when run.

And also this:

AnotherFile.swift containing:
func f() -> Int {
    return a
}
let a = f()

main.swift containing
print(a)

Compile, run (for eternity, at 0% CPU).

/Jens

On Thu, Sep 22, 2016 at 3:13 AM, Joe Groff <jgroff@apple.com> wrote:

> On Sep 21, 2016, at 2:22 PM, Jens Persson via swift-users <swift-users@swift.org> wrote:
>
> // This little Swift program compiles (and runs) fine:
>
> func foo() -> Int { return a }
> let a = 1
> let b = 2
> print(foo())
>
> But if `foo()` returns `b` instead of `a`, I get this compile time error:
> "Use of unresolved identifier `b`"

This looks like a bug to me (cc-ing Jordan, who's thought about global scoping issues more than me). In "script mode", it shouldn't be possible to refer to a variable before its initialization is executed. However, the way this is currently modeled is…problematic, to say the least, among other reasons because script globals are still visible to "library" files in the same module.

-Joe


(Jordan Rose) #8

Yep, it really is a long-standing bug. Script-mode top-level locals are treated as globals (module-scope bindings) by the compiler, but their initial bindings are evaluated eagerly instead of lazily (as you’d want in a script). Taken together, this means that you can get this completely unsafe behavior.

So, why is ‘a’ accepted but ‘b’ not in your original example?

func foo() -> Int { return b }
let a = 1
let b = 2
print(foo())

The secret to the current behavior is that script mode is executed interactively, instead of parsing it all up front. To make things a little better, it actually parses any number of declarations until it sees something it actually needs to execute—a statement or a declaration with an initial value expression. This allows for recursive functions while still being “live”.

The consequence here is that one top-level binding after a series of functions may be visible. This is obviously not optimal.

To fix this, we should:

- Distinguish between script-mode top-level locals and module-scope variables that happen to be declared. My personal preference is to treat anything with explicit access control as a normal lazy global and anything without access as a top-level local.

- Consider parsing everything up front, even if we don’t type-check it, so that we can say “use of ‘b’ before it’s initialized” instead of “undeclared name ‘b’”

Note that we do need to be conservative here. This code should continue to be rejected, even though ‘f’ doesn’t refer to ‘local’ directly, because calling ‘f' would be dangerous before the initialization of ‘local':

internal func f() -> Int {
  return g()
}
// more code here

let local = 42
private func g() -> Int {
  return local
}

Thanks for bringing this up, if only so I have an opportunity to write out the issue. :slight_smile:
Jordan

···

On Sep 21, 2016, at 23:04, Jens Persson <jens@bitcycle.com> wrote:

Did you see the other code examples that came up in that twitter conversations?
For example:

This worrying little program compiles:
func f() -> Int {
    return a
}
let a = f()

It also compiles if you print(a) at the end, and it will print 0.

If we replace Int with [Int] it will still compile but crash when run.

And also this:

AnotherFile.swift containing:
func f() -> Int {
    return a
}
let a = f()

main.swift containing
print(a)

Compile, run (for eternity, at 0% CPU).

/Jens

On Thu, Sep 22, 2016 at 3:13 AM, Joe Groff <jgroff@apple.com <mailto:jgroff@apple.com>> wrote:

> On Sep 21, 2016, at 2:22 PM, Jens Persson via swift-users <swift-users@swift.org <mailto:swift-users@swift.org>> wrote:
>
> // This little Swift program compiles (and runs) fine:
>
> func foo() -> Int { return a }
> let a = 1
> let b = 2
> print(foo())
>
> But if `foo()` returns `b` instead of `a`, I get this compile time error:
> "Use of unresolved identifier `b`"

This looks like a bug to me (cc-ing Jordan, who's thought about global scoping issues more than me). In "script mode", it shouldn't be possible to refer to a variable before its initialization is executed. However, the way this is currently modeled is…problematic, to say the least, among other reasons because script globals are still visible to "library" files in the same module.

-Joe


(Jens Persson) #9

Thank you for the thorough explanation!
/Jens

···

On Thu, Sep 22, 2016 at 7:28 PM, Jordan Rose <jordan_rose@apple.com> wrote:

Yep, it really is a long-standing bug. Script-mode top-level locals are
treated as globals (module-scope bindings) by the compiler, but their
initial bindings are evaluated eagerly instead of lazily (as you’d want in
a script). Taken together, this means that you can get this completely
unsafe behavior.

So, why is ‘a’ accepted but ‘b’ not in your original example?

func foo() -> Int { return b }
let a = 1
let b = 2
print(foo())

The secret to the current behavior is that script mode is executed
interactively, instead of parsing it all up front. To make things a little
better, it *actually* parses any number of declarations until it sees
something it actually needs to execute—a statement or a declaration with an
initial value expression. This allows for recursive functions while still
being “live”.

The consequence here is that *one* top-level binding after a series of
functions may be visible. This is obviously not optimal.

To fix this, we should:

- Distinguish between script-mode top-level locals and module-scope
variables that happen to be declared. My personal preference is to treat
anything with explicit access control as a normal lazy global and anything
without access as a top-level local.

- Consider parsing everything up front, even if we don’t type-check it, so
that we can say “use of ‘b’ before it’s initialized” instead of “undeclared
name ‘b’”

Note that we *do* need to be conservative here. This code should continue
to be rejected, even though ‘f’ doesn’t refer to ‘local’ directly, because
*calling* ‘f' would be dangerous before the initialization of ‘local':

internal func f() -> Int {
  return g()
}
*// more code here*

let local = 42
private func g() -> Int {
  return local
}

Thanks for bringing this up, if only so I have an opportunity to write out
the issue. :slight_smile:
Jordan

On Sep 21, 2016, at 23:04, Jens Persson <jens@bitcycle.com> wrote:

Did you see the other code examples that came up in that twitter
conversations?
For example:

This worrying little program compiles:
func f() -> Int {
    return a
}
let a = f()

It also compiles if you print(a) at the end, and it will print 0.

If we replace Int with [Int] it will still compile but crash when run.

And also this:

AnotherFile.swift containing:
func f() -> Int {
    return a
}
let a = f()

main.swift containing
print(a)

Compile, run (for eternity, at 0% CPU).

/Jens

On Thu, Sep 22, 2016 at 3:13 AM, Joe Groff <jgroff@apple.com> wrote:

> On Sep 21, 2016, at 2:22 PM, Jens Persson via swift-users < >> swift-users@swift.org> wrote:
>
> // This little Swift program compiles (and runs) fine:
>
> func foo() -> Int { return a }
> let a = 1
> let b = 2
> print(foo())
>
> But if `foo()` returns `b` instead of `a`, I get this compile time
error:
> "Use of unresolved identifier `b`"

This looks like a bug to me (cc-ing Jordan, who's thought about global
scoping issues more than me). In "script mode", it shouldn't be possible to
refer to a variable before its initialization is executed. However, the way
this is currently modeled is…problematic, to say the least, among other
reasons because script globals are still visible to "library" files in the
same module.

-Joe


(Jens Persson) #10

Oh, but how can the following (earlier mentioned) example have anything to
do with Script-mode top-level locals being treated as globals?

Create "AnotherFile.swift" containing:
func f() -> Int { return a }
let a = f()

Create "main.swift" containing:
print(a)

Compile. Run. For ever. At zero % CPU.

/Jens

···

On Thu, Sep 22, 2016 at 8:03 PM, Jens Persson <jens@bitcycle.com> wrote:

Thank you for the thorough explanation!
/Jens

On Thu, Sep 22, 2016 at 7:28 PM, Jordan Rose <jordan_rose@apple.com> > wrote:

Yep, it really is a long-standing bug. Script-mode top-level locals are
treated as globals (module-scope bindings) by the compiler, but their
initial bindings are evaluated eagerly instead of lazily (as you’d want in
a script). Taken together, this means that you can get this completely
unsafe behavior.

So, why is ‘a’ accepted but ‘b’ not in your original example?

func foo() -> Int { return b }
let a = 1
let b = 2
print(foo())

The secret to the current behavior is that script mode is executed
interactively, instead of parsing it all up front. To make things a little
better, it *actually* parses any number of declarations until it sees
something it actually needs to execute—a statement or a declaration with an
initial value expression. This allows for recursive functions while still
being “live”.

The consequence here is that *one* top-level binding after a series of
functions may be visible. This is obviously not optimal.

To fix this, we should:

- Distinguish between script-mode top-level locals and module-scope
variables that happen to be declared. My personal preference is to treat
anything with explicit access control as a normal lazy global and anything
without access as a top-level local.

- Consider parsing everything up front, even if we don’t type-check it,
so that we can say “use of ‘b’ before it’s initialized” instead of
“undeclared name ‘b’”

Note that we *do* need to be conservative here. This code should
continue to be rejected, even though ‘f’ doesn’t refer to ‘local’ directly,
because *calling* ‘f' would be dangerous before the initialization of
‘local':

internal func f() -> Int {
  return g()
}
*// more code here*

let local = 42
private func g() -> Int {
  return local
}

Thanks for bringing this up, if only so I have an opportunity to write
out the issue. :slight_smile:
Jordan

On Sep 21, 2016, at 23:04, Jens Persson <jens@bitcycle.com> wrote:

Did you see the other code examples that came up in that twitter
conversations?
For example:

This worrying little program compiles:
func f() -> Int {
    return a
}
let a = f()

It also compiles if you print(a) at the end, and it will print 0.

If we replace Int with [Int] it will still compile but crash when run.

And also this:

AnotherFile.swift containing:
func f() -> Int {
    return a
}
let a = f()

main.swift containing
print(a)

Compile, run (for eternity, at 0% CPU).

/Jens

On Thu, Sep 22, 2016 at 3:13 AM, Joe Groff <jgroff@apple.com> wrote:

> On Sep 21, 2016, at 2:22 PM, Jens Persson via swift-users < >>> swift-users@swift.org> wrote:
>
> // This little Swift program compiles (and runs) fine:
>
> func foo() -> Int { return a }
> let a = 1
> let b = 2
> print(foo())
>
> But if `foo()` returns `b` instead of `a`, I get this compile time
error:
> "Use of unresolved identifier `b`"

This looks like a bug to me (cc-ing Jordan, who's thought about global
scoping issues more than me). In "script mode", it shouldn't be possible to
refer to a variable before its initialization is executed. However, the way
this is currently modeled is…problematic, to say the least, among other
reasons because script globals are still visible to "library" files in the
same module.

-Joe


(Joe Groff) #11

Oh, but how can the following (earlier mentioned) example have anything to do with Script-mode top-level locals being treated as globals?

Create "AnotherFile.swift" containing:
func f() -> Int { return a }
let a = f()

In this case, you have a deadlock, since the initialization of `a` depends on its own initialization.

-Joe

···

On Sep 22, 2016, at 11:23 AM, Jens Persson <jens@bitcycle.com> wrote:

Create "main.swift" containing:
print(a)

Compile. Run. For ever. At zero % CPU.

/Jens

On Thu, Sep 22, 2016 at 8:03 PM, Jens Persson <jens@bitcycle.com> wrote:
Thank you for the thorough explanation!
/Jens

On Thu, Sep 22, 2016 at 7:28 PM, Jordan Rose <jordan_rose@apple.com> wrote:
Yep, it really is a long-standing bug. Script-mode top-level locals are treated as globals (module-scope bindings) by the compiler, but their initial bindings are evaluated eagerly instead of lazily (as you’d want in a script). Taken together, this means that you can get this completely unsafe behavior.

So, why is ‘a’ accepted but ‘b’ not in your original example?

func foo() -> Int { return b }
let a = 1
let b = 2
print(foo())

The secret to the current behavior is that script mode is executed interactively, instead of parsing it all up front. To make things a little better, it actually parses any number of declarations until it sees something it actually needs to execute—a statement or a declaration with an initial value expression. This allows for recursive functions while still being “live”.

The consequence here is that one top-level binding after a series of functions may be visible. This is obviously not optimal.

To fix this, we should:

- Distinguish between script-mode top-level locals and module-scope variables that happen to be declared. My personal preference is to treat anything with explicit access control as a normal lazy global and anything without access as a top-level local.

- Consider parsing everything up front, even if we don’t type-check it, so that we can say “use of ‘b’ before it’s initialized” instead of “undeclared name ‘b’”

Note that we do need to be conservative here. This code should continue to be rejected, even though ‘f’ doesn’t refer to ‘local’ directly, because calling ‘f' would be dangerous before the initialization of ‘local':

internal func f() -> Int {
  return g()
}
// more code here

let local = 42
private func g() -> Int {
  return local
}

Thanks for bringing this up, if only so I have an opportunity to write out the issue. :slight_smile:
Jordan

On Sep 21, 2016, at 23:04, Jens Persson <jens@bitcycle.com> wrote:

Did you see the other code examples that came up in that twitter conversations?
For example:

This worrying little program compiles:
func f() -> Int {
    return a
}
let a = f()

It also compiles if you print(a) at the end, and it will print 0.

If we replace Int with [Int] it will still compile but crash when run.

And also this:

AnotherFile.swift containing:
func f() -> Int {
    return a
}
let a = f()

main.swift containing
print(a)

Compile, run (for eternity, at 0% CPU).

/Jens

On Thu, Sep 22, 2016 at 3:13 AM, Joe Groff <jgroff@apple.com> wrote:

> On Sep 21, 2016, at 2:22 PM, Jens Persson via swift-users <swift-users@swift.org> wrote:
>
> // This little Swift program compiles (and runs) fine:
>
> func foo() -> Int { return a }
> let a = 1
> let b = 2
> print(foo())
>
> But if `foo()` returns `b` instead of `a`, I get this compile time error:
> "Use of unresolved identifier `b`"

This looks like a bug to me (cc-ing Jordan, who's thought about global scoping issues more than me). In "script mode", it shouldn't be possible to refer to a variable before its initialization is executed. However, the way this is currently modeled is…problematic, to say the least, among other reasons because script globals are still visible to "library" files in the same module.

-Joe


(Jens Persson) #12

Yes, but should the compiler silently accept that? And is this issue really
related to the other issue?
/Jens

···

On Thu, Sep 22, 2016 at 8:23 PM, Joe Groff <jgroff@apple.com> wrote:

> On Sep 22, 2016, at 11:23 AM, Jens Persson <jens@bitcycle.com> wrote:
>
> Oh, but how can the following (earlier mentioned) example have anything
to do with Script-mode top-level locals being treated as globals?
>
> Create "AnotherFile.swift" containing:
> func f() -> Int { return a }
> let a = f()

In this case, you have a deadlock, since the initialization of `a` depends
on its own initialization.

-Joe

> Create "main.swift" containing:
> print(a)
>
> Compile. Run. For ever. At zero % CPU.
>
> /Jens
>
>
>
> On Thu, Sep 22, 2016 at 8:03 PM, Jens Persson <jens@bitcycle.com> wrote:
> Thank you for the thorough explanation!
> /Jens
>
> On Thu, Sep 22, 2016 at 7:28 PM, Jordan Rose <jordan_rose@apple.com> > wrote:
> Yep, it really is a long-standing bug. Script-mode top-level locals are
treated as globals (module-scope bindings) by the compiler, but their
initial bindings are evaluated eagerly instead of lazily (as you’d want in
a script). Taken together, this means that you can get this completely
unsafe behavior.
>
> So, why is ‘a’ accepted but ‘b’ not in your original example?
>
>> func foo() -> Int { return b }
>> let a = 1
>> let b = 2
>> print(foo())
>
> The secret to the current behavior is that script mode is executed
interactively, instead of parsing it all up front. To make things a little
better, it actually parses any number of declarations until it sees
something it actually needs to execute—a statement or a declaration with an
initial value expression. This allows for recursive functions while still
being “live”.
>
> The consequence here is that one top-level binding after a series of
functions may be visible. This is obviously not optimal.
>
> To fix this, we should:
>
> - Distinguish between script-mode top-level locals and module-scope
variables that happen to be declared. My personal preference is to treat
anything with explicit access control as a normal lazy global and anything
without access as a top-level local.
>
> - Consider parsing everything up front, even if we don’t type-check it,
so that we can say “use of ‘b’ before it’s initialized” instead of
“undeclared name ‘b’”
>
> Note that we do need to be conservative here. This code should continue
to be rejected, even though ‘f’ doesn’t refer to ‘local’ directly, because
calling ‘f' would be dangerous before the initialization of ‘local':
>
> internal func f() -> Int {
> return g()
> }
> // more code here
>
> let local = 42
> private func g() -> Int {
> return local
> }
>
> Thanks for bringing this up, if only so I have an opportunity to write
out the issue. :slight_smile:
> Jordan
>
>
>> On Sep 21, 2016, at 23:04, Jens Persson <jens@bitcycle.com> wrote:
>>
>> Did you see the other code examples that came up in that twitter
conversations?
>> For example:
>>
>> This worrying little program compiles:
>> func f() -> Int {
>> return a
>> }
>> let a = f()
>>
>>
>> It also compiles if you print(a) at the end, and it will print 0.
>>
>> If we replace Int with [Int] it will still compile but crash when run.
>>
>> And also this:
>>
>> AnotherFile.swift containing:
>> func f() -> Int {
>> return a
>> }
>> let a = f()
>>
>> main.swift containing
>> print(a)
>>
>> Compile, run (for eternity, at 0% CPU).
>>
>> /Jens
>>
>>
>> On Thu, Sep 22, 2016 at 3:13 AM, Joe Groff <jgroff@apple.com> wrote:
>>
>> > On Sep 21, 2016, at 2:22 PM, Jens Persson via swift-users < > swift-users@swift.org> wrote:
>> >
>> > // This little Swift program compiles (and runs) fine:
>> >
>> > func foo() -> Int { return a }
>> > let a = 1
>> > let b = 2
>> > print(foo())
>> >
>> > But if `foo()` returns `b` instead of `a`, I get this compile time
error:
>> > "Use of unresolved identifier `b`"
>>
>> This looks like a bug to me (cc-ing Jordan, who's thought about global
scoping issues more than me). In "script mode", it shouldn't be possible to
refer to a variable before its initialization is executed. However, the way
this is currently modeled is…problematic, to say the least, among other
reasons because script globals are still visible to "library" files in the
same module.
>>
>> -Joe
>>
>
>
>


(Joe Groff) #13

Yes, but should the compiler silently accept that? And is this issue really related to the other issue?

No, this is a separate issue. The compiler might be able to catch some obvious cases, but it'd be impossible to statically prevent all circularities. The runtime could probably do a better job detecting this situation, though, and give a runtime error instead of just letting the deadlock happen.

-Joe

···

On Sep 22, 2016, at 11:28 AM, Jens Persson <jens@bitcycle.com> wrote:

/Jens

On Thu, Sep 22, 2016 at 8:23 PM, Joe Groff <jgroff@apple.com> wrote:

> On Sep 22, 2016, at 11:23 AM, Jens Persson <jens@bitcycle.com> wrote:
>
> Oh, but how can the following (earlier mentioned) example have anything to do with Script-mode top-level locals being treated as globals?
>
> Create "AnotherFile.swift" containing:
> func f() -> Int { return a }
> let a = f()

In this case, you have a deadlock, since the initialization of `a` depends on its own initialization.

-Joe

> Create "main.swift" containing:
> print(a)
>
> Compile. Run. For ever. At zero % CPU.
>
> /Jens
>
>
>
> On Thu, Sep 22, 2016 at 8:03 PM, Jens Persson <jens@bitcycle.com> wrote:
> Thank you for the thorough explanation!
> /Jens
>
> On Thu, Sep 22, 2016 at 7:28 PM, Jordan Rose <jordan_rose@apple.com> wrote:
> Yep, it really is a long-standing bug. Script-mode top-level locals are treated as globals (module-scope bindings) by the compiler, but their initial bindings are evaluated eagerly instead of lazily (as you’d want in a script). Taken together, this means that you can get this completely unsafe behavior.
>
> So, why is ‘a’ accepted but ‘b’ not in your original example?
>
>> func foo() -> Int { return b }
>> let a = 1
>> let b = 2
>> print(foo())
>
> The secret to the current behavior is that script mode is executed interactively, instead of parsing it all up front. To make things a little better, it actually parses any number of declarations until it sees something it actually needs to execute—a statement or a declaration with an initial value expression. This allows for recursive functions while still being “live”.
>
> The consequence here is that one top-level binding after a series of functions may be visible. This is obviously not optimal.
>
> To fix this, we should:
>
> - Distinguish between script-mode top-level locals and module-scope variables that happen to be declared. My personal preference is to treat anything with explicit access control as a normal lazy global and anything without access as a top-level local.
>
> - Consider parsing everything up front, even if we don’t type-check it, so that we can say “use of ‘b’ before it’s initialized” instead of “undeclared name ‘b’”
>
> Note that we do need to be conservative here. This code should continue to be rejected, even though ‘f’ doesn’t refer to ‘local’ directly, because calling ‘f' would be dangerous before the initialization of ‘local':
>
> internal func f() -> Int {
> return g()
> }
> // more code here
>
> let local = 42
> private func g() -> Int {
> return local
> }
>
> Thanks for bringing this up, if only so I have an opportunity to write out the issue. :slight_smile:
> Jordan
>
>
>> On Sep 21, 2016, at 23:04, Jens Persson <jens@bitcycle.com> wrote:
>>
>> Did you see the other code examples that came up in that twitter conversations?
>> For example:
>>
>> This worrying little program compiles:
>> func f() -> Int {
>> return a
>> }
>> let a = f()
>>
>>
>> It also compiles if you print(a) at the end, and it will print 0.
>>
>> If we replace Int with [Int] it will still compile but crash when run.
>>
>> And also this:
>>
>> AnotherFile.swift containing:
>> func f() -> Int {
>> return a
>> }
>> let a = f()
>>
>> main.swift containing
>> print(a)
>>
>> Compile, run (for eternity, at 0% CPU).
>>
>> /Jens
>>
>>
>> On Thu, Sep 22, 2016 at 3:13 AM, Joe Groff <jgroff@apple.com> wrote:
>>
>> > On Sep 21, 2016, at 2:22 PM, Jens Persson via swift-users <swift-users@swift.org> wrote:
>> >
>> > // This little Swift program compiles (and runs) fine:
>> >
>> > func foo() -> Int { return a }
>> > let a = 1
>> > let b = 2
>> > print(foo())
>> >
>> > But if `foo()` returns `b` instead of `a`, I get this compile time error:
>> > "Use of unresolved identifier `b`"
>>
>> This looks like a bug to me (cc-ing Jordan, who's thought about global scoping issues more than me). In "script mode", it shouldn't be possible to refer to a variable before its initialization is executed. However, the way this is currently modeled is…problematic, to say the least, among other reasons because script globals are still visible to "library" files in the same module.
>>
>> -Joe
>>
>
>
>


(Jens Persson) #14

Ah, yes. Thanks.

···

On Thu, Sep 22, 2016 at 8:29 PM, Joe Groff <jgroff@apple.com> wrote:

> On Sep 22, 2016, at 11:28 AM, Jens Persson <jens@bitcycle.com> wrote:
>
> Yes, but should the compiler silently accept that? And is this issue
really related to the other issue?

No, this is a separate issue. The compiler might be able to catch some
obvious cases, but it'd be impossible to statically prevent all
circularities. The runtime could probably do a better job detecting this
situation, though, and give a runtime error instead of just letting the
deadlock happen.

-Joe

> /Jens
>
> On Thu, Sep 22, 2016 at 8:23 PM, Joe Groff <jgroff@apple.com> wrote:
>
> > On Sep 22, 2016, at 11:23 AM, Jens Persson <jens@bitcycle.com> wrote:
> >
> > Oh, but how can the following (earlier mentioned) example have
anything to do with Script-mode top-level locals being treated as globals?
> >
> > Create "AnotherFile.swift" containing:
> > func f() -> Int { return a }
> > let a = f()
>
> In this case, you have a deadlock, since the initialization of `a`
depends on its own initialization.
>
> -Joe
>
> > Create "main.swift" containing:
> > print(a)
> >
> > Compile. Run. For ever. At zero % CPU.
> >
> > /Jens
> >
> >
> >
> > On Thu, Sep 22, 2016 at 8:03 PM, Jens Persson <jens@bitcycle.com> > wrote:
> > Thank you for the thorough explanation!
> > /Jens
> >
> > On Thu, Sep 22, 2016 at 7:28 PM, Jordan Rose <jordan_rose@apple.com> > wrote:
> > Yep, it really is a long-standing bug. Script-mode top-level locals
are treated as globals (module-scope bindings) by the compiler, but their
initial bindings are evaluated eagerly instead of lazily (as you’d want in
a script). Taken together, this means that you can get this completely
unsafe behavior.
> >
> > So, why is ‘a’ accepted but ‘b’ not in your original example?
> >
> >> func foo() -> Int { return b }
> >> let a = 1
> >> let b = 2
> >> print(foo())
> >
> > The secret to the current behavior is that script mode is executed
interactively, instead of parsing it all up front. To make things a little
better, it actually parses any number of declarations until it sees
something it actually needs to execute—a statement or a declaration with an
initial value expression. This allows for recursive functions while still
being “live”.
> >
> > The consequence here is that one top-level binding after a series of
functions may be visible. This is obviously not optimal.
> >
> > To fix this, we should:
> >
> > - Distinguish between script-mode top-level locals and module-scope
variables that happen to be declared. My personal preference is to treat
anything with explicit access control as a normal lazy global and anything
without access as a top-level local.
> >
> > - Consider parsing everything up front, even if we don’t type-check
it, so that we can say “use of ‘b’ before it’s initialized” instead of
“undeclared name ‘b’”
> >
> > Note that we do need to be conservative here. This code should
continue to be rejected, even though ‘f’ doesn’t refer to ‘local’ directly,
because calling ‘f' would be dangerous before the initialization of ‘local':
> >
> > internal func f() -> Int {
> > return g()
> > }
> > // more code here
> >
> > let local = 42
> > private func g() -> Int {
> > return local
> > }
> >
> > Thanks for bringing this up, if only so I have an opportunity to write
out the issue. :slight_smile:
> > Jordan
> >
> >
> >> On Sep 21, 2016, at 23:04, Jens Persson <jens@bitcycle.com> wrote:
> >>
> >> Did you see the other code examples that came up in that twitter
conversations?
> >> For example:
> >>
> >> This worrying little program compiles:
> >> func f() -> Int {
> >> return a
> >> }
> >> let a = f()
> >>
> >>
> >> It also compiles if you print(a) at the end, and it will print 0.
> >>
> >> If we replace Int with [Int] it will still compile but crash when run.
> >>
> >> And also this:
> >>
> >> AnotherFile.swift containing:
> >> func f() -> Int {
> >> return a
> >> }
> >> let a = f()
> >>
> >> main.swift containing
> >> print(a)
> >>
> >> Compile, run (for eternity, at 0% CPU).
> >>
> >> /Jens
> >>
> >>
> >> On Thu, Sep 22, 2016 at 3:13 AM, Joe Groff <jgroff@apple.com> wrote:
> >>
> >> > On Sep 21, 2016, at 2:22 PM, Jens Persson via swift-users < > swift-users@swift.org> wrote:
> >> >
> >> > // This little Swift program compiles (and runs) fine:
> >> >
> >> > func foo() -> Int { return a }
> >> > let a = 1
> >> > let b = 2
> >> > print(foo())
> >> >
> >> > But if `foo()` returns `b` instead of `a`, I get this compile time
error:
> >> > "Use of unresolved identifier `b`"
> >>
> >> This looks like a bug to me (cc-ing Jordan, who's thought about
global scoping issues more than me). In "script mode", it shouldn't be
possible to refer to a variable before its initialization is executed.
However, the way this is currently modeled is…problematic, to say the
least, among other reasons because script globals are still visible to
"library" files in the same module.
> >>
> >> -Joe
> >>
> >
> >
> >
>
>