Using a Typescript Generic with simple addition isn't working

Question

This seems so trivial, yet I am not understanding why this isn't working. Since I am just concatinating/adding depending on a number or string, the return type should match.

function M<T>(arg: T): T {
  return arg + arg;
}


M(3) // 6
M('bill') // billbill

What did you want? const a = M<number>(3); // a will be a number, const b = M<string>('bill');// b will be a string — hoangdv
– hoangdv, Commented Aug 21, 2019 at 6:20
See also this issue on the topic github.com/Microsoft/TypeScript/issues/12410 — DAG
– DAG, Commented Aug 21, 2019 at 8:53

Colliot · Accepted Answer · 2019-08-21 06:37:49Z

1

There is no type constraint in TypeScript saying a certain operator can be applied to a certain type. So it is not possible to do this.

Something close and valid is:

interface Addable<T> {
    add(that: T): T;
}

function M<T extends Addable<T>>(arg: T): T {
  return arg.add(arg);
}

But there is just no way to express it for an operator, + in this case.

answered Aug 21, 2019 at 6:37

Colliot

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2 Comments

james emanon Over a year ago

typescriptlang.org/play/index.html , doesn't seem to work when used with: M(3) or M(test)

Colliot Over a year ago

@jamesemanon Certainly it doesn’t. There’s just no way to do that.

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