> On 9 Feb 2017, at 19:08, Hooman Mehr via swift-evolution
> <swift-evolution@swift.org> wrote:
>
> I think the best solution is overloading the existing ?? operator. It is very
> easy to do:
>
> func ??<T,U: Error>(lhs: T?, rhs: U) throws -> T {
>
> if let lhs = lhs { return lhs } else { throw rhs }
> }
>
> then you can say:
>
> do {
>
> let y = try x ?? myError
>
> } catch ...
>
> It might even make sense to add to the standard library.
Interesting idea, but what if you actually want that error assigned to y?
I wonder if an alternative to the original proposal might be to allow throw on
the right hand side? So you could do:
let y = x ?? throw myError
That'd be a lot more explicit about what's going on than a new operator would
be. It might require defining throw as being some kind of special type though
that can be used to satisfy only function signatures that call for it
explicitly like:
func ??<T,U:_Throw>(lhs:T?, rhs:U) throws -> T {
guard let lhs = lhs else { throw rhs.error }
return lhs
}
Same idea really, but the difference being that this wouldn't be a type that
satisfies generics, but can be used for any operator that could allow a throw
to occur within its operands._______________________________________________
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