overloading methods where only difference is optional vs. non-optional type

Phil_Kirby · 2017-10-10T10:51:52.000Z

2 down vote favorite
Original StackOverflow post:

overloading methods where only difference is optional vs. non-optional type

swift

asked by user6902806 on 12:49PM - 07 Oct 17 UTC

I was under the impression that swift can have overloaded methods that differ only in the type of object that the methods return. I would think that I could have two funcs with the same signature yet they differ in return type.
import Foundation

// ambiguous use of 'IsTextEmpty(text:)'
func IsTextEmpty(text : String?) -> Bool? {
return text?.isEmpty
}

func IsTextEmpty(text : String?) -> Bool {
  guard let text = text else {
    return true
  }

return text.isEmpty
}

let text: String? = nil

if let empty = IsTextEmpty(text:"text") {
print("Not Empty")
}

if IsTextEmpty(text: text) {
print("Empty")
}
Here, both functions have the same input parameters but one func returns an optional Bool? and the other returns a Bool. In this case I get an error:
ambiguous use of 'IsTextEmpty(text:)'
If I change the name of one of the input parameters I no longer get the ambiguous error:
// Works
func IsTextEmpty(foo : String?) -> Bool? {
return foo?.isEmpty
}

func IsTextEmpty(text : String?) -> Bool {
  guard let text = text else {
    return true
  }

return text.isEmpty
}

let text: String? = nil

if let empty = IsTextEmpty(foo:"text") {
print("Not Empty")
}

if IsTextEmpty(text: text) {
print("Empty")
}
Shouldn't the compiler detect that they are two distinct methods even though their return types are different, since an optional Bool? is a different type from a non-optional Bool?

ckeithray · 2017-10-10T17:27:31.000Z

nope. it's the same as this example:

func funny() -> Int {
return 1
}

func funny() -> String {
return "2"
}

print(funny()) // the compiler doesn't know which one you want.

// the above doesn't compile.
// error: forloop.playground:8:1: error: ambiguous use of 'funny()'

You have to have some difference visible in the caller:

func funny(useInt: Bool) -> Int {
return 1
}

func funny(useString: Bool) -> String {
return "2"
}

print(funny(useInt: true), funny(useString: true))
// prints "1 2\n"

···

--
C. Keith Ray
Senior Software Engineer / Trainer / Agile Coach
* http://www.thirdfoundationsw.com/keith_ray_resume_2014_long.pdf
* What Every Programmer Needs To… by C. Keith Ray [PDF/iPad/Kindle] <- buy my book?

On Oct 10, 2017, at 9:06 AM, Phil Kirby via swift-users <swift-users@swift.org> wrote:

<>2 down vote <> favorite <swift - overloading methods where only difference is optional vs. non-optional type - Stack Overflow;
Original StackOverflow post:

swift - overloading methods where only difference is optional vs. non-optional type - Stack Overflow
I was under the impression that swift can have overloaded methods that differ only in the type of object that the methods return. I would think that I could have two funcs with the same signature yet they differ in return type.

import Foundation

// ambiguous use of 'IsTextEmpty(text:)'
func IsTextEmpty(text : String?) -> Bool? {
  return text?.isEmpty
}

func IsTextEmpty(text : String?) -> Bool {
   guard let text = text else {
     return true
   }

   return text.isEmpty
}

let text: String? = nil

if let empty = IsTextEmpty(text:"text") {
   print("Not Empty")
}

if IsTextEmpty(text: text) {
   print("Empty")
}
Here, both functions have the same input parameters but one func returns an optional Bool? and the other returns a Bool. In this case I get an error:

ambiguous use of 'IsTextEmpty(text:)'
If I change the name of one of the input parameters I no longer get the ambiguous error:

// Works
func IsTextEmpty(foo : String?) -> Bool? {
  return foo?.isEmpty
}

func IsTextEmpty(text : String?) -> Bool {
   guard let text = text else {
     return true
   }

   return text.isEmpty
}

let text: String? = nil

if let empty = IsTextEmpty(foo:"text") {
   print("Not Empty")
}

if IsTextEmpty(text: text) {
   print("Empty")
}
Shouldn't the compiler detect that they are two distinct methods even though their return types are different, since an optional Bool? is a different type from a non-optional Bool?

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

ckeithray · 2017-10-10T17:36:15.000Z

You need to know that the NAME of a method or function isn't just outside the parenthesis.

The name of

func IsTextEmpty(foo : String?) -> Bool?

is

"IsTextEmpty(foo:)"

and the name of

func IsTextEmpty(text : String?) -> Bool

is

"IsTextEmpty(text:)"

The argument labels are important.

func IsTextEmpty(foo : String?) -> Bool? {
return foo?.isEmpty
}

func IsTextEmpty(text : String?) -> Bool {
    guard let text = text else {
        return true
    }

    return text.isEmpty
}

print(IsTextEmpty(foo: "1"), IsTextEmpty(text: "2"))
// prints Optional(false) false

···

--
C. Keith Ray
Senior Software Engineer / Trainer / Agile Coach
* http://www.thirdfoundationsw.com/keith_ray_resume_2014_long.pdf

On Oct 10, 2017, at 10:27 AM, C. Keith Ray <keithray@mac.com> wrote:

nope. it's the same as this example:

func funny() -> Int {
return 1
}

func funny() -> String {
return "2"
}

print(funny()) // the compiler doesn't know which one you want.

// the above doesn't compile.
// error: forloop.playground:8:1: error: ambiguous use of 'funny()'

You have to have some difference visible in the caller:

func funny(useInt: Bool) -> Int {
return 1
}

func funny(useString: Bool) -> String {
return "2"
}

print(funny(useInt: true), funny(useString: true))
// prints "1 2\n"

--
C. Keith Ray
Senior Software Engineer / Trainer / Agile Coach
* http://www.thirdfoundationsw.com/keith_ray_resume_2014_long.pdf
* What Every Programmer Needs To… by C. Keith Ray [PDF/iPad/Kindle] <- buy my book?

On Oct 10, 2017, at 9:06 AM, Phil Kirby via swift-users <swift-users@swift.org <mailto:swift-users@swift.org>> wrote:

<>2 down vote <> favorite <swift - overloading methods where only difference is optional vs. non-optional type - Stack Overflow;
Original StackOverflow post:

swift - overloading methods where only difference is optional vs. non-optional type - Stack Overflow
I was under the impression that swift can have overloaded methods that differ only in the type of object that the methods return. I would think that I could have two funcs with the same signature yet they differ in return type.

import Foundation

// ambiguous use of 'IsTextEmpty(text:)'
func IsTextEmpty(text : String?) -> Bool? {
  return text?.isEmpty
}

func IsTextEmpty(text : String?) -> Bool {
   guard let text = text else {
     return true
   }

   return text.isEmpty
}

let text: String? = nil

if let empty = IsTextEmpty(text:"text") {
   print("Not Empty")
}

if IsTextEmpty(text: text) {
   print("Empty")
}
Here, both functions have the same input parameters but one func returns an optional Bool? and the other returns a Bool. In this case I get an error:

ambiguous use of 'IsTextEmpty(text:)'
If I change the name of one of the input parameters I no longer get the ambiguous error:

// Works
func IsTextEmpty(foo : String?) -> Bool? {
  return foo?.isEmpty
}

func IsTextEmpty(text : String?) -> Bool {
   guard let text = text else {
     return true
   }

   return text.isEmpty
}

let text: String? = nil

if let empty = IsTextEmpty(foo:"text") {
   print("Not Empty")
}

if IsTextEmpty(text: text) {
   print("Empty")
}
Shouldn't the compiler detect that they are two distinct methods even though their return types are different, since an optional Bool? is a different type from a non-optional Bool?

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

hlovatt · 2017-10-10T19:28:53.000Z

func funny() -> Int {
return 1
}

func funny() -> String {
return "2"
}

let i: Int = funny()

let s: String = funny()

Is fine. Since Swift knows the return type required. However, it can’t infer the type of:

let q = funny()

-- Howard.

···

On 11 Oct 2017, at 4:36 am, C. Keith Ray via swift-users <swift-users@swift.org> wrote:

You need to know that the NAME of a method or function isn't just outside the parenthesis.

The name of

    func IsTextEmpty(foo : String?) -> Bool?

is

    "IsTextEmpty(foo:)"

and the name of

    func IsTextEmpty(text : String?) -> Bool

is

    "IsTextEmpty(text:)"

The argument labels are important.

func IsTextEmpty(foo : String?) -> Bool? {
    return foo?.isEmpty
}

func IsTextEmpty(text : String?) -> Bool {
    guard let text = text else {
        return true
    }

    return text.isEmpty
}

print(IsTextEmpty(foo: "1"), IsTextEmpty(text: "2"))
// prints Optional(false) false

--
C. Keith Ray
Senior Software Engineer / Trainer / Agile Coach
* http://www.thirdfoundationsw.com/keith_ray_resume_2014_long.pdf

On Oct 10, 2017, at 10:27 AM, C. Keith Ray <keithray@mac.com> wrote:

nope. it's the same as this example:

func funny() -> Int {
return 1
}

func funny() -> String {
return "2"
}

print(funny()) // the compiler doesn't know which one you want.

// the above doesn't compile.
// error: forloop.playground:8:1: error: ambiguous use of 'funny()'

You have to have some difference visible in the caller:

func funny(useInt: Bool) -> Int {
return 1
}

func funny(useString: Bool) -> String {
return "2"
}

print(funny(useInt: true), funny(useString: true))
// prints "1 2\n"

--
C. Keith Ray
Senior Software Engineer / Trainer / Agile Coach
* http://www.thirdfoundationsw.com/keith_ray_resume_2014_long.pdf
* What Every Programmer Needs To… by C. Keith Ray [PDF/iPad/Kindle] <- buy my book?

On Oct 10, 2017, at 9:06 AM, Phil Kirby via swift-users <swift-users@swift.org> wrote:

2 down vote favorite
Original StackOverflow post:

swift - overloading methods where only difference is optional vs. non-optional type - Stack Overflow

I was under the impression that swift can have overloaded methods that differ only in the type of object that the methods return. I would think that I could have two funcs with the same signature yet they differ in return type.

import Foundation

// ambiguous use of 'IsTextEmpty(text:)'
func IsTextEmpty(text : String?) -> Bool? {
  return text?.isEmpty
}

func IsTextEmpty(text : String?) -> Bool {
   guard let text = text else {
     return true
   }

   return text.isEmpty
}

let text: String? = nil

if let empty = IsTextEmpty(text:"text") {
   print("Not Empty")
}

if IsTextEmpty(text: text) {
   print("Empty")
}
Here, both functions have the same input parameters but one func returns an optional Bool? and the other returns a Bool. In this case I get an error:

ambiguous use of 'IsTextEmpty(text:)'
If I change the name of one of the input parameters I no longer get the ambiguous error:

// Works
func IsTextEmpty(foo : String?) -> Bool? {
  return foo?.isEmpty
}

func IsTextEmpty(text : String?) -> Bool {
   guard let text = text else {
     return true
   }

   return text.isEmpty
}

let text: String? = nil

if let empty = IsTextEmpty(foo:"text") {
   print("Not Empty")
}

if IsTextEmpty(text: text) {
   print("Empty")
}
Shouldn't the compiler detect that they are two distinct methods even though their return types are different, since an optional Bool? is a different type from a non-optional Bool?

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

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