typescript override generic method

Question

How can I overload a generic method with concrete type? It seems to work if I specify R at class level ie Foo but not at method's level.

class Foo {
    find<R>(arg: R[]): R[] {
        return arg;
    }
}

class Bar extends Foo {
    find(arg: string[]) {
        return super.find(arg);
    }
}

Gives the following error

Property 'find' in type 'Bar' is not assignable to the same property in base type 'Foo'. Type '(arg: string[]) => string[]' is not assignable to type '(arg: R[]) => R[]'. Types of parameters 'arg' and 'arg' are incompatible. Type 'R[]' is not assignable to type 'string[]'. Type 'R' is not assignable to type 'string'.(2416)

cyco130 · Accepted Answer · 2021-08-26 06:18:53Z

2

The compiler does not allow it because it would create a hole in the type system. Consider the following:

function findNumbers(obj: Foo) {
    return obj.find([1, 2, 3]);
}

findNumbers(new Bar());

findNumbers should accept an instance of Bar (since Bar extends Foo) but Bar.prototype.find can only accept string[] whereas Foo.prototype.find accepts any array. In TypeScript's type system an instance of a subclass should be able to do everything an instance of a superclass can. In your example Bar.prototype.find wouldn't be able to accept a number[] whereas Foo.prototype.find could. Therefore your Bar would not be a valid subclass of Foo.

edited Aug 26, 2021 at 6:18

answered Feb 10, 2020 at 21:22

cyco130

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