[swift-users] overloading methods where only difference is optional vs. non-optional type

[Howard Lovatt]

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 at 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 at 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
>> * https://leanpub.com/wepntk <- buy my book?
>> 
>> 
>>> On Oct 10, 2017, at 9:06 AM, Phil Kirby via swift-users <swift-users at swift.org> wrote:
>>> 
>>> 2 down vote favorite
>>> Original StackOverflow post:
>>> 
>>> https://stackoverflow.com/questions/46620311/overloading-methods-where-only-difference-is-optional-vs-non-optional-type
>>> 
>>> 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 at swift.org
>>> https://lists.swift.org/mailman/listinfo/swift-users
>> 
> 
> _______________________________________________
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> swift-users at swift.org
> https://lists.swift.org/mailman/listinfo/swift-users
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