Question

此代码示例是从Mozilla

派生和调整的

如果您看到代码，则B.prototype的doSomething()方法在应用B＆＃39; s doSomething() obj时调用A.protoype＆＃39; s this 。那么为什么这段代码的输出从B.prototype而不是A.prototype返回？可能是我在这里遗漏了一些东西，但无法弄清楚原因。

function A(a) {
    this.varA = a;
}

// What is the purpose of including varA in the prototype when A.prototype.varA will always be shadowed by
// this.varA, given the definition of function A above?
A.prototype = {
    varA: null, // Shouldn't we strike varA from the prototype as doing nothing?
    // perhaps intended as an optimization to allocate space in hidden classes?
    // https://developers.google.com/speed/articles/optimizing-javascript#Initializing instance variables
    // would be valid if varA wasn't being initialized uniquely for each instance
    doSomething: function(a) {
        return "DoSomething From A"
    }
}

function B(a, b) {
    A.call(this, a);
    this.varB = b;
}
B.prototype = Object.create(A.prototype, {
    varB: {
        value: null,
        enumerable: true,
        configurable: true,
        writable: true
    },
    doSomething: {
        value: function() { // override
            A.prototype.doSomething.apply(this, arguments);
            return "DoSomething From B" // call super
                // ...
        },
        enumerable: true,
        configurable: true,
        writable: true
    }
});
B.prototype.constructor = B;

var b = new B();
b.doSomething(7)

Answer 1

调用A.prototype.doSomething，您只是对A原型上的方法返回的值没有做任何事情。

A.prototype.doSomething.apply(this, arguments);返回“来自A的DoSomething”并且该值被丢弃
return "DoSomething from B"将返回b.doSomething(7)。

将您的return更改为console.log，您会看到它的效果与您预期的一样，您会收到A消息，然后会收到B消息

原型继承返回值

1 个答案: