Function type union works unexpectedly

[ethanresnick]

TypeScript Version: nightly (2.2.0-dev.20161121)

Code

declare const fn: (((...args: any[]) => any) | (() => boolean));

fn(1, 2, 3);

Expected behavior:
The call to fn(1, 2, 3) should succeed, as it should match the (...args: any[]) => any part of the union.

Actual behavior:

I get the error: "Supplied parameters do not match any signature of call target."

[alitaheri]

This is Working as Intended. The operator => takes precedence over | so:

type A = (...args: any[]) => any | (() => boolean);

Is equivalent to:

type A = (...args: any[]) => (any | (() => boolean));

Not:

type A = ((...args: any[]) => any) | (() => boolean);

Try:

type AnyFn = (...args: any[]) => any;
declare const fn: (((...args: any[]) => any) | (() => boolean));
declare const fn2: (AnyFn | (() => boolean));

fn(1, 2, 3);
fn2(1, 2, 3);

I tested it, works fine.

Be more careful with operator precedence, it bites bad 🙀

[ethanresnick]

@alitaheri Thanks. Good call on the precedence, and you're right that with your revised code the two declarations behave equivalently. However, now both function calls give an error! So I guess the issue is with how the two types are being unioned? I'll update the OP accordingly.

[thorn0]

Duplicate of #7294 ?

[ethanresnick]

Seems like it might be, though I'll wait for @RyanCavanaugh to confirm before closing this.

[ethanresnick]

Looking at #7294 more closely, they're definitely the same problem conceptually. I'm gonna give this a close.