更改原型后，Javascript对象丢失了构造函数 [英] Javascript object lost constructor when the prototype was changed

问题描述

在下面的代码中

var A = function() {};
var a = new A();
var b = new A();

A.prototype = {};

var c = new A();
console.log(a.constructor === b.constructor);
console.log(a.constructor === c.constructor);

输出为 true 和 false .

我对错误的输出感兴趣. a 和 c 是使用相同构造函数(即 A )创建的-为什么它们的构造函数属性不同?看来我很想念东西.

I am interested in the false output. a and c were created using same constructor function which is A - why is their constructor property different? It seems I miss something.

PS.如果我删除要更改A输出原型的行，则为: true true .

PS. If I remove the line where I am changing prototype of A output is: true true.

推荐答案

构造对象时，构造函数的 prototype 属性被复制到新对象的__proto __ ，因此，虽然 a 和 b 保留了旧的原型(还包含 .constructor 属性)，但 c使用新的空白原型(没有 constructor 作为自己的属性).这是一个图:

When you construct an object, the constructor's prototype property is copied to the new object's __proto__, so while a and b retain the old prototype (which also contains the .constructor property), c uses the new blank prototype (which doesn't have constructor as own property). Here's a diagram:

在 a之后=新的A;b =新A :

在 A.prototype = {}之后；c =新的A()

如您所见，直接分配给原型会破坏JS对象系统，并可能导致令人惊讶的结果(您的问题就说明了这一点).这就是为什么它通常不被接受的原因.

As you can see, direct assignment to a prototype breaks the JS object system and may lead to surprising results (which your very question demonstrates). That's why it's generally frowned upon.

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