TypeScript: function expression overloads - function

Question

I have this TypeScript code which uses overloads on a function declaration. This code works as expected.
function identity(x: string): string;
function identity(x: number): number;
function identity(x: string | number): string | number {
return x;
}
const a = identity('foo') // string
const b = identity(1) // number
const c = identity({}) // type error (expected)
I am trying to achieve the equivalent of this using function expressions instead of function declarations, however I get a type error:
/* Type '(x: string | number) => string | number' is not assignable to type '{ (x: string): string; (x: number): number; }'.
Type 'string | number' is not assignable to type 'string'.
Type 'number' is not assignable to type 'string' */
const identity: {
(x: string): string;
(x: number): number;
} = (x: string | number): string | number => x;
I want to know how I can achieve the same effect of overloading the function but with function expressions.

Answer 1

You can use a type assertion on the function implementation instead. In an assignment the checks are stricter for compatibility, with an assertion they are weaker. Despite this, we still get a decent amount of type safety (I am not sure it is equivalent to the overloads to implementation signature checks but it seems pretty close):
//OK
const identity = ((x: string | number): string | number => x) as {
(x: string): string;
(x: number): number;
};
// Error argument is incompatible
const identity2 = ((x: boolean): string | number => x) as {
(x: string): string;
(x: number): number;
};
// Error return type is incompatible
const identity3 = ((x: string | number) => false) as {
(x: string): string;
(x: number): number;
};