How to type class method arguments based on function arguments passed to class constructor

Question

I have class example:

export class Test{
    private func:  (...args: any[]) => void;
    constructor(func: (...args: any[]) => void) {
       this.func = func;
    }
    method(...args: any[]) {
        return this.func(...args);
    }
}

I want to not see TS warning that the arguments not in constructor, and i want to see TS warning that the method require "testFunction" function arguments.

Use example:

const testFunction = (a: number, b: string) => {}

const test = new Test(testFunction); // I want to not see TS warning that the arguments were not passed

test.method(1, 'e') // I want to see TS warning that the method require "testFunction" function arguments

Xavier Pedraza · Accepted Answer · 2025-07-31 18:36:14Z

You should be able to accomplish this with generics:

class Test<P extends any[], R> {

    constructor(
        private func: (...params: P) => R
    ) {
        // Defensive check
        if (typeof this.func !== 'function') {
            throw new Error("Initializer must be a function");
        }
    }

    method(...params: P): R {
        return this.func(...params);
    }

}

When creating a new Test instance, the generics will be inferrred:

// "test" is a Test<[number, number], number>
const test = new Test((a: number, b: number) => a + b);

And will inform the Typescript compiler:

test.method(5, 6);                 // ✅ Compiles
test.method(5, '6');               // ❌ Does not compile
test.method(42);                   // ❌ Does not compile
test.method(1, 2, 3);              // ❌ Does not compile

let n: number = test.method(5, 6); // ✅ Compiles
let s: string = test.method(5, 6); // ❌ Does not compile

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