Typescript: Surprising `Not all code paths return a value`

[wycats]

TypeScript Version: 2.5.2

Code

type AsyncFunction<T> = () => Promise<T>;

const f: AsyncFunction<number | void> = async () => {
  if (Math.random()) {
    return 1;
  }
};

Expected behavior:

No tsc errors.

Actual behavior:

const f: AsyncFunction<number | void> = async () => {
//                                      ~~~~~~~~~~~~~ [ts] Not all code paths return a value.

  if (Math.random()) {
    return 1;
  }
};

Notably, when using return-type annotation, this error does not occur:

async function a(): Promise<number | void> {
  if (Math.random()) {
    return 1;
  }
}

const b = async (): Promise<number | void> => {
  if (Math.random()) {
    return 1;
  }
};

In our real-world scenario, it would not be sufficient to look for this syntactic pattern. Rather, the fact that the async function is assigned to a union containing void should be the trigger that silences this error.

[RyanCavanaugh]

Doesn't require async

type fn = () => (void | number);

var x: fn = () => {
  if (Math.random()) {
    return 1;
  }
}

I'm not really sure if this error is "unexpected" or not. It's controlled by the "No implicit returns" flag and this is definitely an implicit return.

[chancancode]

Just wanted to point out that this generally works without errors with the "No implicit returns" flag:

class Foo {
  maybeReturn(): string | void {
    if (Math.random() > 0.5) {
      return 'hello';
    }
  }
}

Playground link

[mhegazy]

PRs welcome

[bcherny]

Related bug in v3.7.2. noImplicitReturns doesn't affect the behavior.

// Error TS2366: Function lacks ending return statement and return
// type does not include 'undefined'.
function f(x: number): number {
  if (typeof x === 'number') {
    return 1
  }
}

Playground Link

Notably, this prevents you from expressing React props as unions:

import * as React from 'react'

type Props =
   | { a: number }
   | { b: string }

// Error TS2366: Function lacks ending return statement and return
// type does not include 'undefined'.
function MyComponent(props: Props): JSX.Element {
  if ('a' in props) {
    return <div>a is {props.a}</div>
  }
  if ('b' in props) {
    return <div>b is {props.b}</div>
  }
}

Playground Link

For this example to type check, we need to add a catch-all return statement at the end of MyComponent's body, which is not actually needed since at that point, props has already been refined to never.

[matthieusieben]

Not sure if related to this but I get the same behavior when returning void:

const aa = (a: any) => { // No error
  if (a) return
}

const bb = (b: any) => { // Not all code paths return a value.
  if (b) return void console.log("foo")
}

return void should be considered as a path that does not return a value, right?

[wycats]

@mhegazy since some time has gone by: is this still "PRs welcome"?