TypeScript: conditional types and using a boolean parameter to control the return type

Question

How do I rewrite this without overload signatures, using conditional types instead?

function foo(returnString: true): string;
function foo(returnString: false): number;
function foo(returnString: boolean) {
  return returnString ? String(Math.random()) : Math.random();
}

I tried the following code, but it doesn't compile without as any:

function foo<T extends boolean>(returnString: T): T extends true ? string : number {
  return (returnString ? String(Math.random()) : Math.random()) as any;
}

How can I get rid of as any?

The error message is super-unhelpful:

Type 'string | number' is not assignable to type 'T extends true ? string : number'.
  Type 'string' is not assignable to type 'T extends true ? string : number'.

Answer 1

I'm not exactly sure why the compiler can't accept this as is (not extremely familiar with TypeScript), but here's what you could do:

function foo<T extends boolean>(returnString: T): T extends true ? string : number;
function foo<T extends boolean>(returnString: T): string | number {
  return returnString ? String(Math.random()) : Math.random();
}

Basically you separate the declaration (public signature) and the implementation, giving the more accurate signature to the declaration and the broader one to the implementation.