In what ways can one determine the function parameter of a union type?

I am working with a union type of functions:

function Function1(arg0: string, arg1: any[], name: "hello" | "bye") {
  return name;
}

function Function2(arg0: string, arg1: any[], name: "foo" | "bar") {
  return name;
}


type Functions = typeof Function1 | typeof Function2;

Currently, I am trying to define a new type that can take a tuple where the first element is the actual function and the second element is the corresponding "name" argument.


const baz = [Function1, "foo"];

I attempted the following approach:

type FooBar<T = Functions> = [T, Parameters<T>[2]];

Initially, it throws an error at Parameters<T>[2] stating "Type 'T' does not satisfy the constraint '(...args: any) => any' "

While it almost works, it retrieves all arguments from both functions instead.

Answer №1

To distribute your operation of distributing your [T, Parameters<T>[0]] over unions in type T, a distributive conditional type can be written for this purpose. Let's introduce the utility type:

type FuncThirdArgTuple<T> = 
  T extends (a1: any, a2: any, a3: infer A) => any ? [T, A] : never

This type eliminates the usage of Parameters<T>[2] and directly applies infer to acquire the third parameter type.

When applied to Functions, we obtain:

type FuncThirdArgTupleForFunctions = FuncThirdArgTuple<Functions>
/* type FuncThirdArgTupleForFunctions = 
    [
      (arg0: string, arg1: any[], name: "hello" | "bye") => "hello" | "bye", 
      "hello" | "bye"
    ] | [
      (arg0: string, arg1: any[], name: "foo" | "bar") => "foo" | "bar", 
      "foo" | "bar"
    ]
*/

This result aligns with the desired type. Let's test it out:

let baz: FuncThirdArgTupleForFunctions;
baz = [function1, "hello"]; // valid
baz = [function1, "foo"]; // error

All looks good.

Link to Playground for Code Visualization

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