Specify only first type argument

Question

function f<T, U>(foo: T, bar: U) {
}

f(1, "x"); // OK, inferes <number, string>
f<number>(1, "x"); // ERROR: Expected 2 type arguments, but got 1.

How can I pass only the first type argument and let TypeScript infer the other?

Answer 1

As @Joe-Clay says, you can set default parameters to make generics optional, but the compiler will not infer them the way you want.

A workaround I've used is to split functions up into multiple ones (via currying), each with one generic parameter. For example:

function f<T,U>(foo:T, bar: U): [T, U] {
  return [foo, bar];
}

becomes

function curriedF<T>(foo: T) {
    return function <U>(bar: U): [T, U] {
        return [foo, bar]
    }
}

Which allows this:

var z = curriedF(1)("x"); // [number, string]
var z = curriedF<number>(1)("x"); // also [number, string]

Hope that helps; good luck.

---

UPDATE 2018-10-19

There is some progress on partial inference (using the * sigil to mark "please infer this for me") and it might be present in TypeScript 3.2 or thereabouts. See this pull request for more information.