How do I write the signature for a function where the argument types are dependent?

Question

Is there a type-safe way of expressing this function? In this example, I'm adding numbers, but the question is about the general pattern.

type signature = 
    ((a: number, b: number) => number) & 
    ((a: bigint, b: bigint) => bigint);

// Operator '+' cannot be applied to types 'number | bigint' and 'number | bigint'.
const add : signature = (a, b) => a + b;

The two arguments can be two different types, but should never be both at the same time. Is there a way to write this function type so it works?

related, possibly near enough for a dupe - https://stackoverflow.com/q/59745527 I haven't found a good way — Snow, Dec 04 '21 at 02:51
That's unfortunate. This *might* be different as no type-guard is involved. — recursive, Dec 04 '21 at 02:56
See [this](https://stackoverflow.com/questions/65280785/is-it-possible-to-declare-a-typescript-function-which-works-on-both-numbers-and#answer-65281351) — captain-yossarian from Ukraine, Dec 04 '21 at 09:18

Ahmad Dehnavi · Answer 1 · 2021-12-04T08:29:53.050

you can define all method types and then define a method to handle all defined types.


function add(a: {option1:string}, b: number): number
function add(a: number, options: { option2: number }): number
function add(a: string, b: string, options: { options3: boolean }): number
function add(
    a: string | number|{option1:string},
    b: string | number | { option2: number }, 
    c: { options3: boolean } = {options3: false}
) {
    // todo
    return 0;
}

add({option1:'a1'},2)
add(1,{option2:2})
add('a','b',{options3:false})

function add(a: number, b: number): number
function add(a: bigint, b: bigint): bigint
function add(a: any, b: any): number | bigint {
    return a + b;
}

or use generic types


function add<T extends (string|number|bigint)>(a: T, b: T): T {
    return a as any + b;
}
add('a','b');
add(1,2);
add(1,'b'); // error

How do I write the signature for a function where the argument types are dependent?

1 Answers1