Type query for a result of a function call
Strate opened this issue · 11 comments
Is it possible to use typeof type query operator to get function's return type?
Seems that no. It would be great to do something like:
declare var do: (arg: string) => number
let a: typeof do // a now has (arg: string) => number type
let b: typeof do() // b now has number type
This comment is related: #2710 (comment)
I have also encountered this problem many times.
What would be the inferred type in the following situation?
function fn<a>(value: a) : { value: a; } { return { value: value }; }
let a: typeof fn;If the answer is {value: {}} (assuming the current state of affairs) then such feature is useless.
@DanielRosenwasser, how feasible is it to allow variables to hold unresolved its own declared (not from the context) type parameters?
Might be related #5959
I think answer should be an error about required type argument, and developer should write
let a: typeof fn<number>()to get {value: number}
@Strate
this is pure syntax sugar over
let b = fn<number>();
let a : typeof b;if this is what you are looking for then it might be better phrased as "Type query for a result of a function call"
You are right, I'm gonna to change the caption :)
I'm wondering how this would handle overloaded functions? I guess one must pass in the type of the arguments to let it overload correctly the return type.
function f(a: string): string;
function f(a: string, b: number): boolean;
function f(a: string, b?: string | number, c?: string): boolean | string {
return true;
}
typeof f(string, number) // booleanIsn't ellipses ... a better syntax for not overloaded functions? Because only f() (no arguments) doesn't corresponds to the function signature f(a: string) (with one argument).
function f(a: string): boolean;
typeof f(...) // boolean