Get argument types for function / class constructor

Question

I am trying to do something I am not sure is possible in TypeScript: inferring the argument types/return types from a function.

For example:

function foo(a: string, b: number) {   return `${a}, ${b}`; }  type typeA = <insert magic here> foo; // Somehow, typeA should be string; type typeB = <insert magic here> foo; // Somehow, typeB should be number; 

My use case is to try to create a config object that contains constructors and parameters.

For example:

interface IConfigObject<T> {     // Need a way to compute type U based off of T.     TypeConstructor: new(a: U): T;     constructorOptions: U; }  // In an ideal world, could infer all of this from TypeConstructor  class fizz {     constructor(a: number) {} }  const configObj : IConfigObj = {     TypeConstructor: fizz;     constructorOptions: 13; // This should be fine }  const configObj2 : IConfigObj = {     TypeConstructor: fizz;     constructorOptions: 'buzz'; // Should be a type error, since fizz takes in a number } 

Answer 1

Typescript now has the ConstructorParameters builtin, similar to the Parameters builtin. Make sure you pass the class type, not the instance:

ConstructorParameters<typeof SomeClass> 

ConstructorParameter Official Doc

Parameters Official Doc

Answer 2

With TypeScript 2.8 you can use the new extends keyword:

type FirstArgument<T> = T extends (arg1: infer U, ...args: any[]) => any ? U : any; type SecondArgument<T> = T extends (arg1: any, arg2: infer U, ...args: any[]) => any ? U : any;  let arg1: FirstArgument<typeof foo>; // string; let arg2: SecondArgument<typeof foo>; // number; let ret: ReturnType<typeof foo>; // string;