简体   繁体   English

JavaScript中的原型链

[英]Prototype chaining in JavaScript

I'm reading a book called JavaScript patterns but there's one part where I think the guy is confusing. 我正在读一本书,叫做《 JavaScript模式》,但有一部分我觉得这家伙很困惑。

The guy actually led up in the book to the klass design pattern, where he developed it piece by piece. 这个家伙实际上是在书中引领了玻璃设计的模式,在那里他逐步地开发了它。 He first he presents the problem: 他首先提出问题:

function inherit(C, P) {
C.prototype = P.prototype;
}

He says: 他说:

" This gives you short and fast prototype chain lookups because all objects actually share the same prototype. But that's also a DRAWBACK because if one child or grandchild somewhere down the inheritance chain MODIFIES the prototype, it AFFECTS all parents and grandparents. " 这为您提供了快速而又简短的原型链查找,因为所有对象实际上共享同一个原型。但这也是一个缺点,因为如果继承链中某个子级或子级对原型进行修改,则会影响所有父母和祖父母。

However, I actually tried to modify the prototype say() in Child and it had no affect on Parent and in fact Child still pointed to Parent and completely ignored its own prototype of same name, which makes sense since it's pointing to a different memory position. 但是,我实际上尝试修改Child中的原型say(),它对Parent没有影响,实际上Child仍然指向Parent并完全忽略了自己的同名原型,这很有意义,因为它指向的是不同的内存位置。 So how can the guy say something like that? 那家伙怎么能这样说呢? Below proves my point: 下面证明了我的观点:

function Parent(){}

Parent.prototype.say = function () {
return 20;
};

function Child(){
}

Child.prototype.say = function () {
return 10;
};

inherit(Child, Parent);

function inherit(C, P) {
C.prototype = P.prototype;
 } 

 var parent = new Parent();
var child = new Child();


var child2 = new Child()
alert(child.say(); //20
alert(parent.say()); //20
alert(child2.say()); //20

It's impossible for any child or grandchild to modify the prototype! 任何子孙都无法修改原型!

This leads to my second point. 这引出我的第二点。 He says the solution to the problem of the possibility of accidentially modifying parent prototypes down inheritance chain (which I can't reproduce) is to break the direct link between parent's and child's prototype while at the same time benefiting from the prototype chain. 他说,意外地沿继承链向下修改父原型的可能性(我无法复制)的解决方案是打破父子原型之间的直接联系,同时从原型链中受益。 He offers the following as a solution: 他提供以下解决方案:

function inherit(C, P) {
var F = function () {};
F.prototype = P.prototype;
C.prototype = new F();
}

The problem is this outputs the same exact values as the other pattern: 问题在于,它输出与其他模式相同的确切值:

function Parent(){}

Parent.prototype.say = function () {
return 20;
};

function Child(){
}

 Child.prototype.say = function () {
return 10;
};

inherit(Child, Parent);

function inherit(C, P) {
var F = function () {};
F.prototype = P.prototype;
C.prototype = new F();
}

var parent = new Parent();
var child = new Child();


var child2 = new Child()
alert(child.say(); //20
alert(parent.say()); //20
alert(child2.say()); //20

It doesn't make sense that an empty function somehow breaks a link. 空函数以某种方式断开链接没有意义。 In fact, the Child points to F and F in turn points to the Parent's prototype. 实际上,子级指向F,而F依次指向父级的原型。 So they are ALL still pointing to the same memory position. 因此,它们仍然都指向相同的内存位置。 This is demonstrated above, where it outputs the same exact values as the first example. 上面已经说明了这一点,在这里输出的值与第一个示例相同。 I have no clue what this author is trying to demonstrate and why he makes claims that don't gel for me and that I can't reproduce. 我不知道作者试图证明什么,以及为什么他提出的主张对我不利,并且我无法复制。

Thanks for response. 感谢您的回复。

For your first point: 对于您的第一点:

What this guy is trying to say, is that the methods for both child and parent will change if you modify the prototype after you created instances. 这个家伙想说的是,如果您创建实例修改原型,则针对父级和子级的方法都会更改。

For example: 例如:

function inherit(C, P) {
  C.prototype = P.prototype;
} 

function Parent(){}
function Child(){}

inherit(Child, Parent);

Parent.prototype.say = function () {
  return 20;
};

var parent = new Parent();
var child = new Child();


// will alert 20, while the method was set on the parent.
alert( child.say() );

The same thing happens when you change the child's constructor (which is shared with the parent). 当您更改子代的构造函数(与父代共享)时,会发生相同的事情。

// same thing happens here, 
Child.prototype.speak = function() {
  return 40;
};

// will alert 40, while the method was set on the child
alert( parent.speak() );

And about your second point: 关于第二点:

function inherit(C, P) {
  var F = function () {};
  F.prototype = P.prototype;
  C.prototype = new F();
}

The new inheriting function will actually separate the constructor of the parent from the child, because it's not pointing to the same object anymore, but now it's pointing to a prototype of a newly created function that has nothing to do with the parent. 新的继承函数实际上会将父代的构造函数与子代分开,因为它不再指向同一对象,但是现在它指向的是与父代无关的新创建函数的原型。 So, you actually create a local copy of the parent's constructor, and then create a new instance of the copy, which returns all constructor methods an properties. 因此,您实际上是创建父级构造函数的本地副本,然后创建副本的新实例,该实例将所有构造函数方法返回一个属性。 By changing the constructor of the child now, it will not affect the parent. 通过现在更改子代的构造函数,它不会影响父代。

function inherit(C, P) {
  var F = function () {};
  F.prototype = P.prototype;
  C.prototype = new F();
}

function Parent(){}
function Child(){}

inherit(Child, Parent);

// same thing happens here, 
Child.prototype.speak = function() {
  return 40;
};

var parent = new Parent();

// will throw an error, because speak is undefined
alert( parent.speak() );

The fact that you can change prototype object by pointing another prototype to it is normal JavaScript behavior. 您可以通过将另一个原型指向原型对象来更改原型对象,这是正常的JavaScript行为。 JavaScript's primitive values are immutable but objects and arrays aren't. JavaScript的原始值是不可变的,但对象和数组不是。 I'll explain it with simple example: 我将用一个简单的例子来解释它:

var person = {name: 'greg', age: 20};

>>>person.name; //prints 'greg'

>>>person.age; //prints 20

var a = person;

>>>a.name; //prints 'greg'

a.name = 'peter';

>>>a.name; //prints 'peter'

>>>person.name; //prints 'peter'

//I've changed person.name through a.name. That's why objects in JavaScript are called mutable

Arrays have the same behavior: 数组具有相同的行为:

var arr = ['first', 'second', 'third'],
    newArr = arr;

newArr.pop();

>>>newArr; //prints ['first', 'second']

>>>arr; //prints ['first', 'second']

//Frist array was also changed

Let's look at strings numbers and booleans(primitive data types): 让我们看一下字符串数字和布尔值(原始数据类型):

var str = 'hello world',

    newStr = str;

>>>str; //prints 'hello world'

>>>newStr; //prints 'hello world'

>>>newStr.toUpperCase(); //prints 'HELLO WORLD'

>>>str; //prints 'hello world'

>>newStr; //prints 'hello world'

//Numbers and booleans have similiar behavior

I had the same issue but i fixed it. 我有同样的问题,但我已解决。 Look, i've commented your code, some hints in it should help you: 看,我已经评论了您的代码,其中的一些提示应该可以帮助您:

function Parent(){}

Parent.prototype.say = function () {
return 20;
};

function Child(){
}



/**
*
* The area you should examine i've selected below.
*
*/

//Start BUG

//new say method will not affect the Parent.prototype beacuse it wasn't assigned yet
Child.prototype.say = function () {
return 10;
};

//rewrite Child.prototype and all it's methods with Parent.prototype
inherit(Child, Parent);

//End BUG



function inherit(C, P) {
C.prototype = P.prototype;
 } 

var parent = new Parent();
var child = new Child();


var child2 = new Child()
alert(child.say(); //20
alert(parent.say()); //20
alert(child2.say()); //20

The problem here is, that instead of copying and changing the Parent.prototype you creating new Child.prototype.say method and right after it you rewriting the whole Child.prototype object through Parent.prototype assignment. 这里的问题是,不是复制和更改Parent.prototype,而是创建新的Child.prototype.say方法,然后紧接着通过Parent.prototype赋值重写整个Child.prototype对象。 Just change their order and it should work fine. 只需更改其顺序,它就可以正常工作。

If you change the prototype after inheriting it, you see it change the prototype for the parent also: 如果您在继承原型后更改了原型,则会看到它也更改了父级的原型:

function Parent(){}

Parent.prototype.say = function () {
return 20;
};

function Child(){
}

inherit(Child, Parent);

Child.prototype.say = function () {
return 10;
};

function inherit(C, P) {
C.prototype = P.prototype;
 } 

 var parent = new Parent();
var child = new Child();


var child2 = new Child()
alert(child.say()); //10
alert(parent.say()); //10
alert(child2.say()); //10

If you use the modifed version of the inherit function, the Child and Parent prototypes remain separate. 如果您使用的版本体改inherit功能, ChildParent的原型保持独立。

声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM