是否可以通过函数本身访问函数的原型对象

Question

例如

//creating constructor function

function Something(){}

//adding name property to Something prototype


Something.prototype.name='javascript';

是否可以从对象本身访问 name 属性？

//喜欢

Something.name

// but here output will be 'undefined'

// 代码

function Fan(){}

Fan.prototype.speed='high';

Fan.speed

//output of (Fan.speed) undefined

Answer 1

不，如果您将其设置为Fan.prototype.speed，则该值位于Fan.prototype.speed 而不是Fan.speed。当您使用Fan 作为对象的构造函数时，即为实例化对象设置一个查找链以查找它在构造函数的prototype 上没有的属性：

const f = new Fan();
console.log(f.speed);  // lookup to Fan.prototype.speed

Answer 2

你可以创建一个对象的实例然后访问它：

function Something() {}
Something.prototype.name = 'javascript';
// create an instance for that
var x = new Something();
console.log(x.name);

//Another set of examples, note the prototype on the objects function has no instance so the getName.protytppe.bucket
const another = {
  secret: "skim",
  name: "cream",
  getName: function() {
    return this.name;
  },
  getSecret: function() {
    return this.secret;
  },
  makeButter: function(n) {
    this.name = n;
    return `We made {this.name} from cream`;
  }
};
another.getName.prototype.bucket = "lot-o-milk"

var n = Object.create(another);
console.log(n.name);
console.log(n.getName());
// make cream into butter
n.name = "butter";
console.log(n.name);
console.log(n.getName());
// turn butter back to cream
delete n.name;
console.log(n.name);
console.log(n.getName());
n.makeButter('yellow');
console.log(n.getName());
console.log(n.getSecret(), n.getName.prototype.bucket);