TypeScript type definition for promise.reject

Question

The following code is correct in terms of the type that is returned, because then always return the promise array.

Promise.resolve(['one', 'two'])
.then( arr =>
{
  if( arr.indexOf('three') === -1 )
    return Promise.reject( new Error('Where is three?') );

  return Promise.resolve(arr);
})
.catch( err =>
{
  console.log(err); // Error: where is three?
})

TypeScript throw error:

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

But in reality, then never will return void.

I can explicitly specify type .then<Promise<any>>, but it's more like a workaround, not the right solution.

How to write this right?

basarat · Accepted Answer · 2016-09-08 03:14:19Z

12

You should not return Promise.resolve and Promise.reject inside a promise chain. The resolve should be driven by simple return and reject should be driven by an explicit throw new Error.

Promise.resolve(['one', 'two'])
.then( arr =>
{
  if( arr.indexOf('three') === -1 )
    throw new Error('Where is three?');

  return arr;
})
.catch( err =>
{
  console.log(err); // Error: where is three?
})

More

More on promise chaining https://basarat.gitbooks.io/typescript/content/docs/promise.html

answered Sep 8, 2016 at 3:14

basarat

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2 Comments

user663031 Over a year ago

As a best practice, of course you are right, but at the same time there is no reason why you could not return Promise.resolve() and/or Promise.reject() from within a then handler, and the OP has a legitimate question as to why TypeScript is unhappy about that.

DanielM Over a year ago

What is the reason why "you should not return Promise.resolve and Promise.reject"? I don't understand why anyone would consider that "a best practice". Chainability is at the core of promises, and it's a great principle. Why would you throw an exception that will be converted into a rejected promise instead of creating the rejected promise yourself?

Ben Southgate · Accepted Answer · 2016-09-08 03:17:15Z

3

Typescript is complaining about the difference in return type between your Promise.reject return value (Promise<void>) and your Promise.resolve value (Promise<string[]>).

Casting your then call as .then<Promise<void | string[]>> will let the compiler know of the union return type.

as @basarat notes, you should just throw an error instead of using Promise.reject (which will be passed to whatever catch handler is provided).

edited Sep 8, 2016 at 3:17

answered Sep 8, 2016 at 3:10

Ben Southgate

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