T | (() => T)

TypeScript Version: 3.9.0 (Nightly)

Search Terms: T | (() => T), T & Function

Code

type Initializer<T> = T | (() => T)
// type Initializer<T> = T extends any ? (T | (() => T)) : never

function correct<T>(arg: Initializer<T>) {
    return typeof arg === 'function' ? arg() : arg // error
}

Line 2 provides a workaround for this. More info on stackoverflow.

Expected behavior: no errors

Actual behavior: This expression is not callable. Not all constituents of type '(() => T) | (T & Function)' are callable. Type 'T & Function' has no call signatures.

Playground Link: here.

Related Issues: none

[kentcdodds]

Found that this reveals the issue:

function a<T>(b: T | (() => T)) {
  return typeof b === 'function' ? b() : b
}

But this does not (and compiles file):

function a<T>(b: T | (() => T)) {
  return b instanceof Function ? b() : b
}

Kind of odd, because typeof checks seem to work normally. But this is a reasonable workaround for anyone who bumps into it. HT to @pbteja1998 😃

NOTE: this workaround can have problems in some situations: https://github.com/microsoft/TypeScript/issues/37663#issuecomment-856866935 Check a safer (though more verbose) workaround: https://github.com/microsoft/TypeScript/issues/37663#issuecomment-856961045)

[Svish]

Ran into this while trying to use the SetStateAction<T> type in react:

type SetStateAction<S> = S | ((prevState: S) => S);

If using typeof foobar === 'function' isn’t the correct way to check whether we have a value, or a callable initializer function, what is?