> On Apr 13, 2016, at 9:26 AM, Sean Heber <[email protected]> wrote:
>
>> On Apr 13, 2016, at 10:08 AM, Erica Sadun <[email protected]> wrote:
>>
>>
>>> On Apr 13, 2016, at 8:25 AM, Sean Heber via swift-evolution
>>> <[email protected]> wrote:
>>>
>>> This pair works pretty well, too, if you don’t mind free functions:
>>>
>>> func with<T>(inout this: T, @noescape using: inout T->Void) { using(&this) }
>>> func with<T>(this: T, @noescape using: T->Void) { using(this) }
>>>
>>> It works either with classes or mutable structs if you call it correctly
>>> and the type doesn’t matter.
>>>
>>> l8r
>>> Sean
>>
>> My current version is this:
>>
>> func with<T>(item: T, @noescape update: (inout T) -> Void) -> T {
>> var this = item; update(&this); return this
>> }
>>
>> Used, for example:
>>
>> struct Foo { var (a, b, c) = ("a", "b", "c") }
>> class Bar: DefaultReflectable { var (a, b, c) = ("a", "b", 2.5) }
>> var f = with(Foo()) { $0.a = "X" }
>> var b = with(Bar()) { $0.a = "X" }
>> print(f) // Foo(a: "X", b: "b", c: "c")
>> print(b) // Bar(a=X, b=b, c=c)
>>
>> Is there an advantage to having a pair of functions?
>
> I don’t know if it’s a huge advantage or not, but with warnings on unused
> results, using a with() that has a return with a class instance would mean
> you’d have to discard the return result explicitly or pointlessly reassign
> the results to your instance (thus meaning not using a let) just to avoid the
> warning. If you annotated the with() to allow discarding the result, then
> it’d be error-prone for structs. It seemed “safer” to me to have a pair.
>
> l8r
> Sean
>
If you want to call
>> let f = with(Foo()) { $0.a = "X" }
>> let b = with(Bar()) { $0.a = "X" }
with constant assignment and an initializer, I'm not sure how yours would work.
Could they?
-- E, trying right now in the playground
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