Typescript type T or function () => T usage

Question

You can see a demo in this playground.

I've made a simple generic type which can represent either a variable or a function that returns a variable. But, unfortunately, it doesn't work with a typical typeof arg === 'function' check. It produces the following error:

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

Is there a way to make it work without using type guard function?

type Initializer<T> = T | (() => T)

function correct(arg: Initializer<string>) {
    return typeof arg === 'function' ? arg() : arg
}

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

const isFunction = (arg: any): arg is Function => typeof arg === 'function'

function correct_2<T>(arg: Initializer<T>) {
    return isFunction(arg) ? arg() : arg
}

Answer 1

You can write:

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

function correct<T>(arg: Initializer<T>): T {
    return typeof arg === 'function' ? arg() : arg // works
    // arg is Initializer<T> & Function in the true branch
}

const r1 = correct(2) // const r1: 2
const r2 = correct(() => 2) // const r2: number

In the original version, arg is resolved to (() => T) | (T & Function) in the true branch. TS apparently can't recognize for this union function type, that both constituents are callable. At least in above version, it is clear for the compiler that you can invoke arg after a function check.

Might also be worth to create a github issue for this case in the TypeScript repository - in my opinion T & Function should represent some (wide) type of function.