Union 'this' type with original 'this' type?

Question

In typescript, I can specific this type like this:

let data = {
   prop1: "prop1"
}

let obj = {
    fun1(this: typeof data){
        this.prop1;  // works
        this.fun2(); // doesn't work
    },
    fun2(){}
}

But how can I declare 'this' type that I can make union with original 'this' type so I can write like this.fun2(), maybe something like fun1(this: typeof _this | typeof data).

Answer 1

I think you're looking for something like this:

let data = {
    prop1: "prop1",
};

interface ObjType {
    fun1(this: (typeof data) | ObjType): void;
    fun2(): void;
}

let obj: ObjType = {
    fun1(this: (typeof data) | ObjType) {
        if ("prop1" in this) {
            this.prop1;  // works
        }
        if ("fun2" in this) {
            this.fun2(); // works
        }
    },
    fun2() {
    }
}

Playground link

(We need the type declaration because we can't use typeof obj in obj's constructor if we don't have a type on it. Well, we can, but we get any.)

Note the need for type guards, because this could be typeof data (and so have prop1) or ObjType (and so have fun2). So we have to figure out which it is before using them.