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Inferred type symbol in conditional type should be visible-but-banned in falsy arm #22610
Description
Currently if you write
type C<T> = T extends infer U ? U : U;
// ~ Cannot find name "U"We give you an error because U is not in scope at all in the false arm.
This is a) confusing (U is right there!) and b) a landmine, because you might have an outer type of the same name
// Elsewhere
type U = string;
// ...
/// Not what I meant!
type C<T> = T extends infer U ? number : U;We should keep U in scope in both cases and disallow its use with an explicit error message, e.g. (please bikeshed):
Inferred type parameter "U" is unmatched in this case and cannot be used
This should not block nesting of lexical scopes; i.e. this should still be legal because each U would shadow the outer expression's:
type C<T> = T extends Array<infer U> ? U :
T extends Promise<infer U> ? U :
never;
Activity
RyanCavanaugh commentedon Mar 15, 2018
cc Anders Hejlsberg (@ahejlsberg) Daniel Rosenwasser (@DanielRosenwasser) since we should 👍 / 👎 before release
krryan commentedon Mar 15, 2018
Why keep
Uin scope in the falsy case at all? Why not allow the use of the outerU? Just for readability/least-surprise concerns? I can buy that but I'm wondering if I'm missing something more.RyanCavanaugh commentedon Aug 13, 2018
This one missed the boat