Another possibility, other than generics, would be to drop rethrows all
together and have the compiler infer if a throw is possible or not,
including:
struct FStore {
let f: () throws -> Void
func call() throws { try f() }
}
The compiler can make two versions, one if f can throw and one if it
definitely doesn't.
Just a thought.
On Tue, 10 Jan 2017 at 4:29 pm, Jacob Bandes-Storch <jtban...@gmail.com>
wrote:
> Moving to swift-users list.
>
> No, there's no way to do this today. The point of rethrows is that within
> one call site, "f(block)" can be treated as throwing if the block throws,
> or not throwing if the block doesn't throw. In your example, once the
> FStore object is constructed, the information about the original passed-in
> function is lost, so the caller has no way to know whether call() can throw
> or not.
>
> If this *were* possible, the information would somehow need to be encoded
> in the type system when creating FStore(f: block). That would require
> something like dependent typing, or generic-param-based-rethrows, e.g.
>
> struct FStore<T: () throws -> Void> { // made-up syntax
> let f: T
> func call() rethrows(T) { try f() } // throws-ness of this function
> depends on throws-ness of T
> }
>
>
>
> On Mon, Jan 9, 2017 at 9:21 PM, Howard Lovatt via swift-evolution <
> swift-evolut...@swift.org> wrote:
>
> Hi,
>
> If I have an escaping function that I store and then call, I need to
> declare the calling function as throwing, not rethrowing. EG:
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
> struct FStore {
> let f: () throws -> Void
> init(f: @escaping () throws -> Void) { self.f = f }
> func call() throws { try f() } // Can't put rethrows here - have
> to use throws
> }
> Is there a better solution?
>
> Thanks for any suggestions,
>
> -- Howard.
>
>
>
>
>
> _______________________________________________
>
>
> swift-evolution mailing list
>
>
> swift-evolut...@swift.org
>
>
> https://lists.swift.org/mailman/listinfo/swift-evolution
>
>
>
>
>
>
> --
-- Howard.
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