Super constructor spread

[anvish]

TypeScript Version: 2.0.3

Code

class A {
    constructor(a: number, b: string) { }
}

class B extends A {
    constructor(...args) {
        super(...args) // Supplied parameters do not match any signature of call target.
    }
}

Expected behavior:
It's possible to pass arguments to super constructor without specifying them explicitly.
E.g. when extending classes of external framework, like React, which calls constructors.

Actual behavior:
Error: Supplied parameters do not match any signature of call target

[yortus]

I think the compiler is correct to flag the example as non-type-safe. You could instantiate B with zero arguments, one argument, ten arguments, and each argument can be any type. Those same arguments will then be passed through to A's constructor. But A expects exactly two arguments of specific types, so this pattern is not type-safe at all.

You'd get the same error if you wrote new A(42) or new A(1, 'foo', true), even though B's constructor would allow any of those.

[remojansen]

If the constructor of A can take an undetermined number of arguments, then it would make sense to use ...:

class A {
   values: any[];
    constructor(...args: any[]) {
        this.values = args;
    }
}

class B extends A {
    constructor(...args: any[]) {
        super(args);
    }
}

However, in this case it looks like the number of arguments of A is determined. So it would be better to do the following:

class A {
    constructor(a: number, b: string) {}
}

class B extends A {
    constructor(...args: any[]) {
        if (args.length < 2) throw new Error();
        let a: number = args[0];
        let b: string = args[1];
        super(a, b);
    }
}

If the number of constructor arguments in B can be determined, then I would recommend to avoid using ... on both A and B.

class A {
    constructor(a: number, b: string) {}
}

class B extends A {
    constructor(a: number, b: string) {
        super(a, b);
    }
}

The ... operator will save you writing some code but you are losing certain level of security provided by the type system.

[DanielRosenwasser]

For the record, if you just didn't write a constructor at all, TypeScript would just "inherit" the constructor. See here.