ES6使用`super`出类定义

Question

I'm trying to add extra methods to class, and these extra methods should use the super methods.

If I add them in the model definition, it works.

class A {
    doSomething() {
        console.log('logSomething');
    }

}

class B extends A {
    doSomething() {
        super.doSomething();
        console.log('logSomethingElse');
    }
}

If I try to add the extra method to B.prototype , I'll get SyntaxError: 'super' keyword unexpected here .

class A {
    doSomething() {
        console.log('logSomething');
    }

}

class B extends A {
}

B.prototype.doSomething = function doSomething() {
    super.doSomething();
    console.log('logSomethingElse');
}

It is quite clear, why I get this error. This is a function and not a class method.

Let's try to define the method as a class method, and copy it to the original B class:

class A {
    doSomething() {
        console.log('logSomething');
    }

}

class B extends A {}

class X {
    doSomething() {
        super.doSomething();
        console.log('2 logSomethingElse');
    }
}

B.prototype.doSomething = X.prototype.doSomething;

In this case I'll get TypeError: (intermediate value).doSomething is not a function .

Is there any way to define methods (that refer to super ) outside from the original class definition, and add these methods later to the original class?

Answer 1

super refers to ancestor of a class where the method was defined, it isn't dynamic. As Babel output illustrates this, super is hard-coded to Object.getPrototypeOf(X.prototype) , and thus orphan class like this one doesn't make sense because it doesn't have super :

class X {
    doSomething() {
        super.doSomething();
        ...
    }
}

But super can be substituted with dynamic counterpart:

doSomething() {
    const dynamicSuper = Object.getPrototypeOf(this.constructor.prototype);
    // or
    // const dynamicSuper = Object.getPrototypeOf(Object.getPrototypeOf(this));
    dynamicSuper.doSomething();
    ...
}

class B extends A {}
B.prototype.doSomething = doSomething;

In this case it will refer to ancestor class of class instance where doSomething was assigned as prototype method.

Answer 2

While I think this could be assumed as anti-pattern , you shouldn't use super outside from a class .

You can achieve that using Object Literals .

Refer to Object.setPrototypeOf

 const A = { sayHello() { console.log("I am A"); }, Factory() { return Object.create(this); } } const B = { sayHello() { super.sayHello(); } } Object.setPrototypeOf(B, A); const c = B.Factory(); c.sayHello(); 

Answer 3

如果你没有让类X继承自B类或A类，那么调用该方法的唯一方法是A.prototype.doSomething()或更普遍的A.prototype.doSomething.call(this_substitute, ...args) 。