Call a member of an object with arguments

Question

function whatever(object, methodName, args) {   return object[methodName](...args); } 

Can the above be typed so the the following is enforced:

• methodName is a key of object.
• object[methodName] is callable and its args are ...args.
• The return type of whatever(object, methodName, args) is the return type of object[methodName](...args).

The closest thing I could find is the definition of function.apply, but it isn't quite the same as the above.

Answer 1

I think this does the trick:

function callMethodWithArgs<   M extends keyof T,   T extends { [m in M]: (...args: Array<any>) => any },   F extends T[M] >(obj: T, methodName: M, args: Parameters<F>) {   return obj[methodName](...args) as ReturnType<F>; } 

Requires TS 3 though!

Answer 2

type Dictionary = { [key: string]: any }  type MethodNames<T extends Dictionary> = T extends ReadonlyArray<any>   ? Exclude<keyof [], number>   : { [P in keyof T]: T[P] extends Function ? P : never }[keyof T]  function apply<T extends Dictionary, P extends MethodNames<T>>(   obj: T,   methodName: P,   args: Parameters<T[P]> ): ReturnType<T[P]> {   return obj[methodName](...args); }  // Testing object types: const obj = { add: (...args: number[]) => {} } apply(obj, 'add', [1, 2, 3, 4, 5])  // Testing array types: apply([1, 2, 3], 'push', [4])  // Testing the return type: let result: number = apply(new Map<number, number>(), 'get', [1]) 

Playground link

The Dictionary type allows T[P] to be used.

The Parameters and ReturnType types are baked into TypeScript.

The MethodNames type extracts any keys whose value is assignable to the Function type. Array types require a special case.

Checklist

• Method name is validated? ✅
• Arguments are type-checked? ✅
• Return type is correct? ✅