Preserving method overloading signatures in TypeScript while carrying a function

Question

Preserving method overloading signatures in TypeScript while carrying a function

Let's dive into an example that involves strict null checking. Imagine I have the following function declaration:

declare function f(value: null, multiplier: number): null;
declare function f(value: string, multiplier: number): number;
declare function f(value: string | null, multiplier: number): number | null;

Now, I am looking to utilize it in this way:

const f2 = (value) => f(value, 2);

It appears that the inferred parameter type is "any" and the inferred return type is "number | null".

Is there a method to have TypeScript correctly infer the new function type so that it retains overloads without needing to redefine all signatures?

The desired call signature should match exactly with this:

function f2(value: null): null;
function f2(value: string): number;

typescript

Answer 1

Answer №1

In TypeScript, there is no direct way to partially specify a function's call signature from another function without explicitly defining the types. The language does not make assumptions about the type of parameters.

Consider the following example for better understanding:

function someOtherFunction(x: number) {}
function test(x) {
   // we cannot assume that `x` is a number just because it is used as one in the input parameter for another function.
   someOtherFunction(x, 5);
}

A possible refactor could be like this:

type ValueType = null | string;
declare function f(value: ValueType, multiplier: number): number | null;

const f2 = (value: ValueType) => f(value, 2);

The nearest alternative approach could resemble something similar to the following implementation:

declare function f<T = string | null>(
  value: T,
  multiplier: number
): T extends string ? number : null;

const f2 = (value: string | null) => f<typeof value>(value, 5);

const x = f2(null); // x: number | null
const x2 = f2("5"); // x2: number | null

Essentially, this method achieves function overloading through conditional types in TypeScript automatically.

Answer 2

In TypeScript, there is no direct way to partially specify a function's call signature from another function without explicitly defining the types. The language does not make assumptions about the type of parameters.

Consider the following example for better understanding:

function someOtherFunction(x: number) {}
function test(x) {
   // we cannot assume that `x` is a number just because it is used as one in the input parameter for another function.
   someOtherFunction(x, 5);
}

A possible refactor could be like this:

type ValueType = null | string;
declare function f(value: ValueType, multiplier: number): number | null;

const f2 = (value: ValueType) => f(value, 2);

The nearest alternative approach could resemble something similar to the following implementation:

declare function f<T = string | null>(
  value: T,
  multiplier: number
): T extends string ? number : null;

const f2 = (value: string | null) => f<typeof value>(value, 5);

const x = f2(null); // x: number | null
const x2 = f2("5"); // x2: number | null

Essentially, this method achieves function overloading through conditional types in TypeScript automatically.

Preserving method overloading signatures in TypeScript while carrying a function

Answer №1

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