属性涉及对象和原型的继承

Question

我是 JS 的新手，在我的一项测试中，我一直试图弄清楚以下代码在属性继承方面的工作原理。

function doSomething(){}

doSomething.prototype.foo = "bar"; //Add "foo" property to doSomething's prototype
let anInstance = new doSomething();
anInstance.prop = "value"; //Add "prop" property to object "anInstance"
doSomething.prop = "another value"; //Add "prop" property to "doSomething"
let anotherInstance = new doSomething();

console.log(doSomething.prop);
console.log(anInstance);
console.log(anotherInstance.prop);

这是上述脚本在控制台中的输出：

another value

doSomething{prop: "value"}
  prop: "value"
  __proto__:
    foo: "bar"
    constructor: ƒ doSomething()
    __proto__: Object

undefined

如您所见，在添加 returns 后打印 doSomething 的 prop 属性 another value，但访问 [=16] =]的propreturnsundefined.

难道 anotherInstance 不应该 "inherit" 这样 prop 属性因为它是在创建它的函数中定义的吗？

提前致谢。

Answer 1

向函数添加属性与向函数的原型对象添加属性不同。函数的实例继承了函数的 prototype 属性，而不是函数自己的属性：

function doSomething(){}

doSomething.prototype.foo = "bar"; //Add "foo" property to doSomething's prototype

doSomething.prop = "another value"; //Add "prop" property to "doSomething" but no the prototype

let anotherInstance = new doSomething();

// only has foo:
console.log(doSomething.prototype)

// this is the object it will inherit from
// only has foo
console.log(Object.getPrototypeOf(anotherInstance))

//they are the same object:
console.log(doSomething.prototype === Object.getPrototypeOf(anotherInstance))

上面的代码中doSomething.prop只是一个属性的函数，对原型继承没有任何作用

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