Obtaining the return type of a function

Use built-in ReturnType:

type SomeType = ReturnType<typeof SomeFunc>

ReturnType expands to:

type ReturnType<T extends (...args: any) => any> = T extends (...args: any) => infer R ? R : any;

The code below works without executing the function. It's from the react-redux-typescript library (https://github.com/alexzywiak/react-redux-typescript/blob/master/utils/redux/typeUtils.ts)

interface Func<T> {
    ([...args]: any, args2?: any): T;
}
export function returnType<T>(func: Func<T>) {
    return {} as T;
}


function mapDispatchToProps(dispatch: RootDispatch, props:OwnProps) {
  return {
    onFinished() {
      dispatch(action(props.id));
    }
  }
}

const dispatchGeneric = returnType(mapDispatchToProps);
type DispatchProps = typeof dispatchGeneric;

EDIT

As of TypeScript 2.8 this is officially possible with ReturnType<T>.

type T10 = ReturnType<() => string>;  // string
type T11 = ReturnType<(s: string) => void>;  // void
type T12 = ReturnType<(<T>() => T)>;  // {}
type T13 = ReturnType<(<T extends U, U extends number[]>() => T)>;  // number[]

See this pull request to Microsoft/TypeScript for details.

TypeScript is awesome!

---

Old-school hack

Ryan's answer doesn't work anymore, unfortunately. But I have modified it with a hack which I am unreasonably happy about. Behold:

const fnReturnType = (false as true) && fn();

It works by casting false to the literal value of true, so that the type system thinks the return value is the type of the function, but when you actually run the code, it short circuits on false.

The easiest way in the TypeScript 2.8:

const foo = (): FooReturnType => {
}

type returnType = ReturnType<typeof foo>;
// returnType = FooReturnType