es6-promise Promise.all error

When I use the es6-promise type definitions, given this code:

Promise.all([
    Promise.resolve('foo'),
    Promise.resolve(['bar'])
]);

I get the following error:

The type argument for type parameter 'R' cannot be inferred from the usage. Consider specifying the type arguments explicitly.
Type argument candidate 'string' is not a valid type argument because it is not a supertype of candidate 'string[]'.

Can you explain why this is? I don't understand the error.

If you're using https://github.com/DefinitelyTyped/DefinitelyTyped/blob/master/es6-promise/es6-promise.d.ts#L59, I found it's because those two items in the array resolve to different promise types. Try specifying it explicitly, usually works for me:

Promise.all<[string, string[]]>([
    Promise.resolve('foo'),
    Promise.resolve(['bar'])
])

Edit: Wrong syntax, but Promise.all<Promise<string> | Promise<string[]>> works.

Can you help me understand why this can't be inferred?

On Fri, 4 Dec 2015, 18:01 Blake Embrey notifications@github.com wrote:

If you're using
https://github.com/DefinitelyTyped/DefinitelyTyped/blob/master/es6-promise/es6-promise.d.ts#L59,
I found it's because those two items in the array resolve to different
promise types. Try specifying it explicitly, usually works for me:

Promise.all<[string, string[]]>([
Promise.resolve('foo'),
Promise.resolve(['bar'])
])

—
Reply to this email directly or view it on GitHub
#5935 (comment)
.

Can you help me understand why this can't be inferred?

I'm not the best person for that, I'll leave it for the TypeScript team to answer.

Blake Embrey (@blakeembrey)

Blake Embrey (@blakeembrey) I believe this is the correct syntax:

Promise.all<string | string[]>([
    Promise.resolve('foo'),
    Promise.resolve(['bar'])
])

If you try adding .then((x) => x) to the end of your assertion, you get errors.

Not causing issues for me. (link) What version (of TypeScript) are you using?

Yes, that works. Sorry, I was trying to say the type assertion you provided didn't work (Promise.all<Promise<string> | Promise<string[]>>).

This is what I have ended up with:

Promise.all<string | number>([
    Promise.resolve('1'),
    Promise.resolve(1)
])
    .then(([x, y]: [string, number]) => {
        x
        y
    })

I provide the type of the tuple in the then callback function so I can continue to use the type checker. If I did the following:

Promise.all<string | number>([
    Promise.resolve('1'),
    Promise.resolve(1)
])
    .then(([x, y]) => {
        x
        y
    })

… then the compiler wouldn't know if x is a string or number.

This is a bit of a PITA. I'm trying to understand if it's possible to write a better type definition for Promise.all so assertions are not needed?

Right, I understand. I don't think it's currently possible to do that since there's no way to express the generic variadically, but this discussion might relate to the proposal in #5453. The TypeScript team should chime in if there's a better approach at the moment, but I've done the same thing as you ended up with in the past.

Nathan Shively-Sanders (@sandersn)

untill #5453 is in, we can add a punch of overloads (say 2-10) to make Promise.all more user friendly. something like:

interface PromiseConstructor {
    all<T1, T2>(values: [T1 | PromiseLike<T1>, T2 | PromiseLike<T2>]): Promise<[T1, T2]>;
    all<T1, T2, T3>(values: [T1 | PromiseLike<T1>, T2 | PromiseLike<T2>, T3 | PromiseLike<T3>]): Promise<[T1, T2, T3]>;
    all<T1, T2, T3, T4>(values: [T1 | PromiseLike<T1>, T2 | PromiseLike<T2>, T3 | PromiseLike<T3>, T4 | PromiseLike<T4>]): Promise<[T1, T2, T3, T4]>;
    ..............
}

Nathan Shively-Sanders (@sandersn) thoughts?

Yes, multiple overloads is the accepted workaround until we have variadic generics from #5453.

I put variadic kinds behind this-types functions because this-types don't have a simple (albeit verbose) workaround. So I think it's worthwhile to add the overloads now until variadic kinds are in. I'll link the PR when it's done.

Here's the PR: #5973.

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es6-promise Promise.all error #5935

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es6-promise Promise.all error #5935

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blakeembrey commented on Dec 4, 2015

OliverJAsh commented on Dec 4, 2015

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sandersn commented on Dec 7, 2015

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