If you have an object ({}) as a map then the keys must be strings (or numbers which are converted to strings automatically).

In this case you can use the toString() method:

class A { }

console.log(A.toString());

Will print:

function A() {

    }

You can also use the name property:

console.log(A.name); // A

You can also override the toString() method to return something of your own:

class A {

    static toString() {

        return class A;

    }

}

console.log(A.toString()); // class A

And then:

let m = {} as { [name: string]: string };

m[A.toString()] = something;



// or

m[A.name] = something;

If you are using a Map then all of the above still work, but you are not restricted to have string keys therefor you can use the class itself:

let m = new Map<{ new (): A }, string>();

m.set(A, A.toString());



console.log(m.get(A)); // class A

Edit

If you have an instance of a class, you can get the class using the constructor property:

let a = new A();

...

m.set(a.constructor, SOME_VALUE);

The constructor property comes from object and it looks like so:

interface Object {

    /** The initial value of Object.prototype.constructor is the standard built-in Object constructor. */

    constructor: Function;



    ...

}

So you always get a Function and you'll need to cast:

m.set(a.constructor as typeof A, SOME_VALUE);

As you probably don't know the type of the class, you'll need to do something else.

You can cast to any, but that's not very pretty.

Here's a working solution that should work well for you:

interface Base { }

type BaseContructor = { new (): Base };



class A implements Base { }

class B implements Base { }



let m = new Map<BaseContructor, string>();

let a = new A();

let b = new B();



m.set(a.constructor as BaseContructor, value);

m.set(b.constructor as BaseContructor, value);

(code in playground)