I don’t think that explains it (or perhaps I did not understand your response
correctly).
Here is the same issue with a custom type (which does not have a failable
initializer):
struct A { }
if let a = A() { }
// error: initializer for conditional binding must have Optional type, not
'A'
if let a: A = A() { }
// warning: non-optional expression of type 'A' used in a check for optionals
> Am 02.06.2017 um 15:49 schrieb Zhao Xin <[email protected]>:
>
> I think it did an unnecessary implicitly casting. For example,
>
> let y = Int(exactly: 5)
> print(type(of:y)) // Optional<Int>
>
> You code equals to
>
> if let y:Int = Int(exactly: 5) { }
>
> However, I don't know why it did that. Maybe because of type inferring?
>
> Zhaoxin
>
> On Fri, Jun 2, 2017 at 8:40 PM, Martin R via swift-users
> <[email protected]> wrote:
> This following code fails to compile (which is correct, as far as I can judge
> that):
>
> if let x = 5 { }
> // error: initializer for conditional binding must have Optional type, not
> 'Int'
>
> But why is does it compile (with a warning) if an explicit type annotation is
> added?
>
> if let y: Int = 5 { }
> // warning: non-optional expression of type 'Int' used in a check for
> optionals
>
> Tested with Xcode 8.3.2 and both the build-in Swift 3.1 toolchain and the
> Swift 4.0 snapshot from May 25, 2017.
>
> I am just curious and would like to understand if there is fundamental
> difference between those statements.
>
> Regards, Martin
>
>
>
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>
>
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