Can't use type aliases or conditional types resolving to Promise for async/await return types

[tao-cumplido]

TypeScript Version: 3.1

Search Terms:

• async promise alias
• async promise conditional

Code

type P<T> = Promise<T>

async function foo<T>(x: T): P<T> {
    return x;
}

async function bar<T>(x: T): T extends number ? Promise<number> : Promise<any> {
    return x;
}

Expected behavior:
No error.

Actual behavior:
Type 'P' is not a valid async function return type in ES5/ES3 because it does not refer to a Promise-compatible constructor value.

Type 'T extends number ? Promise : Promise' is not a valid async function return type in ES5/ES3 because it does not refer to a Promise-compatible constructor value.

Playground Link:
Link

Related Issues:

[weswigham]

As a workaround, can you write Promise<T extends number ? number : any> instead?

[tao-cumplido]

Yes, that works for my use case. Should have thought of it myself, thanks!

[swar30]

just had same issue, this looks to be regression in latest TS version (3.1.3) in prev versions this results in compilation error (as far as I can see).

Work around indeed works, but it would be better if this will break compilation, so will know immediately this is not supported

[trusktr]

I just had this problem, and I did not think to place the whole conditional Promise<inside of here>.

Here's my playground example.

Thanks for the work around, I'll use that. A fix for this would be nice, so it works for people trying it for the first time (f.e. me).

[rbuckton]

This is behaving by design. ES5 and earlier runtimes did not have a native Promise implementation, and ES5 down-level transpiled async functions use a "bring-your-own promise" solution that uses the return type position to determine which Promise implementation to use.

@swar30: Are you suggesting that T extends number ? Promise<number> : Promise<any> should also be an error in ES2015 and later?