As a follow-up, here is code that you can test now. It demonstrates the
same, but with normal function result.
func createArbitrary<T>(usingGenerator f: () -> (T)) -> T {
let any: Any = 42 as Any
if Int.self is T.Type {
let t = any as! T
return t
}
return f()
}
print(createArbitrary(usingGenerator: { return 5 })) //=> 42
And yes, nobody should do these tricks in production code :P
2017-02-22 21:15 GMT+03:00 Anton Zhilin <[email protected]>:
I understand how parametric polymorphism works *in Haskell*. But we talk
> about Swift, and there *is* a way to get an instance of E. I’ll explain
> it another way:
>
> func bypassRethrows<E: Error>(_ f: () throws(E) -> ()) throws(E) {
> let error: Error = MyError() // create an instance of `MyError`
> if MyError.self is E.Type { // in case `E` happens to be `MyError`
> let e: E = error as! E // then we've actually created an instance
> of `E`, and we can downcast safely
> throw e // voila, acquired an instance of `E`
> }
> }
> let f: () throws MyError -> () = { }
> try bypassRethrows(f) // actually throws `MyError`, without ever
> calling `f`
>
> What line here seems impossible?
>
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