为什么我不能设置使用“绑定"创建的函数的“原型"? [英] Why can't I set the 'prototype' of a function created using 'bind'?

问题描述

考虑这个代码:

function foo(something) {
  this.a = something;
}

var obj1 = {};

var bar = foo.bind(obj1);

现在不执行以下语句:

bar.prototype.newprop = "new"; // Cannot execute this

据我所知，每个函数都有一个原型对象.那为什么不能执行上面的语句呢?

As I understood, every function has a prototype object. Then why can't we execute the above statement?

bar 确实是一个我们可以称之为函数的函数:

And bar is indeed a function as we can call it:

bar(2);
console.log(obj1.a); // 2

推荐答案

据我所知，每个函数都有一个原型对象.

As I understood, every function has a prototype object.

嗯，每个规则都有例外:-) 你发现了一个:绑定函数没有 .prototype 属性，因为它们不需要它.当您使用 new 调用绑定函数时，它会调用原始函数作为构造函数，使用原始的 .prototype 对象作为新实例的原型.

Well, there are exceptions to every rule :-) You found one: bound functions don't have a .prototype property because they don't need it. When you call a bound function with new, it calls the original function as a constructor, using the original's .prototype object as the prototype of the new instance.

事实上，由于 ECMAScript 6 很多函数没有带有对象的 .prototype 属性，因为它们不是构造函数 - 它们不能用 new 调用所以他们不需要它.其中有

In fact, since ECMAScript 6 many functions don't have a .prototype property with an object, because they are not constructors - they cannot be called with new so they don't need it. Among those are

• 箭头函数 (() => {…})
• 方法(method() { ... } 在对象字面量和类中)
• 内置非构造函数(如Math.sin)

• arrow functions (() => {…})
• methods (method() { … } in object literals and classes)
• builtin non-constructor functions (like Math.sin)

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