javascript 类型错误：调用 Function.prototype.method() 时 this.prototype 未定义
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TypeError: this.prototype is undefined when calling Function.prototype.method()
提问by Volodymyr Bezuglyy
I am reading the book "Javascript: The good parts".
Now I am reading chapter about Augmenting Types:
我正在阅读“Javascript：好的部分”一书。
现在我正在阅读关于增强类型的章节：
Function.prototype.method = function (name, func) {
this.prototype[name] = func;
return this;
};
UPDATE:
Why following code does not work?
更新：
为什么下面的代码不起作用？
js> Function.prototype.method("test", function(){print("TEST")});
typein:2: TypeError: this.prototype is undefined
But following code works without problems:
但是以下代码可以正常工作：
js> Function.method("test", function(){print("TEST")});
function Function() {[native code]}
Why this code works?
为什么这段代码有效？
js> var obj = {"status" : "ST"};
js> typeof obj;
"object"
js> obj.method = function(){print(this.status)};
(function () {print(this.status);})
js> obj.method();
ST
"obj" is object.
But I can call method "method" on it.
What is the difference between Function.prototype.method and obj.method?
“obj”是对象。
但是我可以在它上面调用方法“方法”。
Function.prototype.method 和 obj.method 有什么区别？
采纳答案by pimvdb
thisrefers to Function.prototypebecause you called .methodon that. So, you're using Function.prototype.prototypewhich does not exist.
this指的是Function.prototype因为你调用.method了那个。所以，你使用的Function.prototype.prototype是不存在的。
Either use Function.method(...)or this[name] = ...to eliminate one of the .prototypes.
使用Function.method(...)或this[name] = ...消除.prototypes 之一。
回答by Rob W
Because:
因为：
Function instanceof Function // <--- is true
Function.prototype instanceof Function // <-- is false
Function.prototypeis an Object, and does not inherit anything from the Function contructor.Functionis a constructor, but also a function, so it inherits methods fromFunction.prototype.- When calling
Function.method, you're calling themethodmethod of an instance ofFunction. So,thispoints to the created instance ofFunction. - When calling
Function.prototype.method, you're invoking an ordinary method of an object.thispoints toFunction.prototype.
Function.prototype是一个对象，不从函数构造函数继承任何东西。Function是一个构造函数，也是一个函数，所以它从Function.prototype.- 调用 时
Function.method，您正在调用 的method实例的方法Function。因此，this指向创建的Function. - 调用 时
Function.prototype.method，您正在调用对象的普通方法。this指向Function.prototype.
To clarify, here's an example:
为了澄清，这里有一个例子：
Function.method() // is equivalent to
(function Function(){}).method()
(new Function).method() // Because Function is also a function
Function.prototype.method // No instance, just a plain function call
回答by Kory Hodgson
prototype is only used when you are declaring the function, but not when you are calling it. Prototype makes the function a member of the object that gets created with every instance of that object.
原型仅在您声明函数时使用，而不是在调用时使用。Prototype 使函数成为使用该对象的每个实例创建的对象的成员。
