TypeScript - Check if Class implements an Interface

Question

I'm using an interface in TypeScript to define a function that is only available on some of the classes that extend the base class. This is a simplified version of the code I have so far:

class Animal {
}

interface IWalkingAnimal {
    walk(): void;
}

class Dog extends Animal implements IWalkingAnimal {
}

class Cat extends Animal implements IWalkingAnimal {
}

class Snake extends Animal {
}

private moveAnimal(animal: Animal) {
    if (animal instanceof Cat || animal instanceof Dog) {
        animal.walk();
    }
}

Now, the trouble is I'll be adding more 'walking' animals so the moveAnimal functional will grow too large to be manageable. What I would like to do is something like this:

private moveAnimal(animal: Animal) {
    if (animal implements IWalkingAnimal ) {
        animal.walk();
    }
}

However the 'implements' check does not work, and I cannot find an equivalent to 'instanceof' when using interfaces. In Java it seems that the use of 'instanceof' would work here, but TypeScript will not allow this.

Does such a thing exist in TypeScript, or is there a better approach here? I am using the TypeScript 1.8.9.

Answer 1

Unlike classes, interfaces exist only at compile-time, they are not included into the resulting JavaScript, so you cannot do an instanceof check.

You could make IWalkingAnimal a subclass of Animal (and use instanceof ), or you could check if the object in question has a walk method:

if (animal['walk']) {}

You can wrap this in a user defined type guard (so that the compiler can narrow the type when used in an if statement, just like with instanceof ).

/**
* User Defined Type Guard!
*/
function canWalk(arg: Animal): arg is IWalkingAnimal {
   return (arg as IWalkingAnimal).walk !== undefined;
}


private moveAnimal(animal: Animal) {
    if (canWalk(animal)) {
        animal.walk();  // compiler knows it can walk now
    }
}