Re: [swift-users] Overload by return type optionality?

[Hooman Mehr via swift-users]

I know, but a simple 

let x = 2/3

becomes ambiguous which I don’t like.

> On Oct 13, 2016, at 8:00 PM, Mark Lacey  wrote:
> 
> 
>> On Oct 13, 2016, at 5:37 PM, Hooman Mehr via swift-users 
>> > wrote:
>> 
>> 
>>> On Oct 13, 2016, at 3:31 PM, Rick Mann via swift-users 
>>> > wrote:
>>> 
>>> Would it make sense to be able to specify priority for a set of overloaded 
>>> methods to help resolve ambiguity?
> 
> I don’t think we want to head down that route, partially because using a 
> contextual type as mentioned below removes the ambiguity.
> 
>> This might be pretty useful in some situations, but I am not sure if the 
>> semantic complexity that it introduces is worth it.
>> 
>> Another example of how this could be useful: 
>> 
>> I made a bare-bones rational number type 
>>  for Swift 
>> a while ago. I would love to be able to overload “/“ operator to create 
>> fractions (rational numbers) instead of dividing two integers. 
>> 
>> If I overloaded “/“ to return rational (Int / Int -> Rational), the result 
>> type of the operator would become ambiguous for every use of it with integer 
>> operands.
> 
> That isn’t the way the type checker works. If you use an explicit type to 
> contextualize the expression, there is no ambiguity. For example this works 
> without any ambiguity.
> 
> struct Rational {}
> func / (lhs: Int, rhs: Int) -> Rational { return Rational() }
> func + (lhs: Rational, rhs: Rational) -> Rational { return Rational() }
> 
> func use(r: Rational) {}
> 
> let x: Rational = (1 / 2) + (2 / 3)  // Rational result type, no ambiguity
> use(r: (1 / 2) + (2 / 3))  // Rational argument type, no ambiguity
> let y = (1 / 2) as Rational// Calls func/(Int,Int)->Rational
> 
> Mark
> 
>> It would be nice if I could prioritize my overload of “/“ over stdlib 
>> definition to resolve the ambiguity. 
>> 
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>> https://lists.swift.org/mailman/listinfo/swift-users
> 

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Re: [swift-users] Overload by return type optionality?

[Joe Groff via swift-users]

> On Oct 13, 2016, at 3:27 PM, Rick Mann  wrote:
> 
>> 
>> On Oct 13, 2016, at 14:47 , Joe Groff  wrote:
>> 
>> 
>>> On Oct 13, 2016, at 2:36 PM, Rick Mann via swift-users 
>>>  wrote:
>>> 
>>> It seems I can write this:
>>> 
>>> extension String
>>> {
>>> public func deleting(prefix inPrefix: String) -> String
>>> public func deleting(prefix inPrefix: String) -> String?
>>> }
>>> 
>>> But I was hoping it would do the right thing:
>>> 
>>> let a = s.deleting(prefix: "foo")
>>> if let b = s.deleting(prefix: "foo") { }
>>> 
>>> But it finds these ambiguous, and I'm not sure how to specify which I want.
>> 
>> The first one is truly ambiguous since either overload works. If you specify 
>> the type of 'a', you should be able to choose one or the other:
>> 
>> let a: String = s.deleting(prefix: "foo")
>> let b: String? = s.deleting(prefix: "foo")
>> 
>> The `if let` should not be ambiguous, since only the Optional-returning 
>> overload is a valid candidate. Got time to file a bug?
> 
> Here you go: https://bugs.swift.org/browse/SR-2942

Thanks!

-Joe
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Re: [swift-users] Overload by return type optionality?

[Rick Mann via swift-users]

> On Oct 13, 2016, at 14:51 , Greg Parker  wrote:
> 
>> 
>> On Oct 13, 2016, at 2:36 PM, Rick Mann via swift-users 
>>  wrote:
>> 
>> It seems I can write this:
>> 
>> extension String
>> {
>>  public func deleting(prefix inPrefix: String) -> String
>>  public func deleting(prefix inPrefix: String) -> String?
>> }
>> 
>> But I was hoping it would do the right thing:
>> 
>> let a = s.deleting(prefix: "foo")
>> if let b = s.deleting(prefix: "foo") { }
>> 
>> But it finds these ambiguous, and I'm not sure how to specify which I want.
> 
> You can specify which you want by explicitly naming the type:
> let a = s.deleting(prefix: "foo") as String
> or
> let a : String = s.deleting(prefix: "foo")
> 
> …but presumably that doesn't give you the usability that you were hoping for.

Eh, it's close, especially if Joe Groff is right that the if-let case shouldn't 
be ambiguous.

A better question would be to ask, is this a reasonable approach, to overload a 
method like this for this kind of behavior. There are two ways to think about 
stripping the prefix off a string: a string minus a non-existent prefix is just 
the string, or you want to know that the prefix didn't exist. This gives you 
both, but might make for subtly unexpected behavior.

Would it make sense to be able to specify priority for a set of overloaded 
methods to help resolve ambiguity?


-- 
Rick Mann
rm...@latencyzero.com


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Re: [swift-users] Overload by return type optionality?

[Greg Parker via swift-users]

> On Oct 13, 2016, at 2:36 PM, Rick Mann via swift-users 
>  wrote:
> 
> It seems I can write this:
> 
> extension String
> {
>  public func deleting(prefix inPrefix: String) -> String
>  public func deleting(prefix inPrefix: String) -> String?
> }
> 
> But I was hoping it would do the right thing:
> 
> let a = s.deleting(prefix: "foo")
> if let b = s.deleting(prefix: "foo") { }
> 
> But it finds these ambiguous, and I'm not sure how to specify which I want.

You can specify which you want by explicitly naming the type:
let a = s.deleting(prefix: "foo") as String
or
let a : String = s.deleting(prefix: "foo")

…but presumably that doesn't give you the usability that you were hoping for.


-- 
Greg Parker gpar...@apple.com  Runtime 
Wrangler


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Re: [swift-users] Overload by return type optionality?

[Nevin Brackett-Rozinsky via swift-users]

It works if you specify the types of the variables:

let a: String = …
if let b: String = …

Nevin


On Thu, Oct 13, 2016 at 5:36 PM, Rick Mann via swift-users <
swift-users@swift.org> wrote:

> It seems I can write this:
>
> extension String
> {
>   public func deleting(prefix inPrefix: String) -> String
>   public func deleting(prefix inPrefix: String) -> String?
> }
>
> But I was hoping it would do the right thing:
>
> let a = s.deleting(prefix: "foo")
> if let b = s.deleting(prefix: "foo") { }
>
> But it finds these ambiguous, and I'm not sure how to specify which I want.
>
> I'm having trouble googling for an answer on this. Some answers around
> generics, but that doesn't apply in this case.
>
> --
> Rick Mann
> rm...@latencyzero.com
>
>
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[swift-users] Overload by return type optionality?

[Rick Mann via swift-users]

It seems I can write this:

extension String
{
  public func deleting(prefix inPrefix: String) -> String
  public func deleting(prefix inPrefix: String) -> String?
}

But I was hoping it would do the right thing:

let a = s.deleting(prefix: "foo")
if let b = s.deleting(prefix: "foo") { }

But it finds these ambiguous, and I'm not sure how to specify which I want.

I'm having trouble googling for an answer on this. Some answers around 
generics, but that doesn't apply in this case.

-- 
Rick Mann
rm...@latencyzero.com


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