Method inferred as property for inheritance and method overrides

[panthus]

TypeScript Version: 2.0.3

Code

class Animal {
    public getA(a: Animal) { }
}

class Cat extends Animal {
    public x: Cat[];
    public getA(a: Cat) { }
}

class Dog extends Animal {
    public getA(a: Dog) {}
}

let x: Cat | Dog;
// Inferred type: (property) getA: ((a: Cat) => void) | ((a: Dog) => void)
x.getA(new Animal()); // Error

Expected behavior:
Matching method overload is selected.

Actual behavior:
The method is inferred to be a property if subclasses have a different signature.

[dead-claudia]

Seems working as intended. It's unsafe to elevate those, because ((a: Cat) => void) | ((a: Dog) => void) are two incompatible types. If x is in fact a Cat, you would want x.getA(dog) to be a compile-time error, and similarly, if x is a Dog, you'd expect the same with x.getA(cat). The function incompatibility is why you need to narrow it first (there's no overload here because the types don't intersect).

[RyanCavanaugh]

Duplicate #7294