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