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 <keith...@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@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@swift.org
>>> https://lists.swift.org/mailman/listinfo/swift-users
>>
>
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