关于JS单例模式的一个问题

Question

function Demo(name) {
    this.name = name;
}
var getSingle = function(fn) {
    var result;
    return function() {
        console.log(arguments.length);
        return result || (result = fn.apply(this, arguments));
    }
};
const aa = new getSingle(Demo)('a');
const bb = new getSingle(Demo)('b');
console.log(aa === bb); //true
console.log(aa.name); // TypeError: aa is undefined
console.log(bb); // undefined

为什么aa.name和bb未定义呢?

Answer 1

这个是在终端打出的结果，中文翻译，你认为的name并不能够读取到，

Answer 2

const aa = new (getSingle(Demo))('a'); 加括号可以解决undefined的问题，
但是单例不是这么写的：
result = fn.apply(this, arguments)：result永远为undefined，这明显是写错了；
而且每次执行getSingle 都会var result,光这一点就可以看出来这绝对不是单例写法了

Answer 3

function Demo(name) {
        this.name = name;
    }
    var getSingle = function(fn) {
        var result;
        return function() {
            console.log(arguments.length);
            return result || (result = fn.apply(this, arguments));
        }
    };
    const ss=getSingle(Demo);
    const aa = new ss('a');
    const bb = new ss('b');
    console.log(aa === bb);
    console.log(aa.name);
    console.log(bb);
    const m=new (getSingle(Demo))('m');
    console.log(m.name);
    const m2=new (getSingle(Demo))('m2');
    console.log(m==m2);

1.运算符优先级的问题
2.每次走getSingle 都会重新定义result 返回新function 所以你那样调即便成功了也不会相等

Answer 4

--------更新
需用借用ES6的语法,ES5的话我实在想不出来

function Demo(name) {
    this.name = name;
}

let getSingleBuilder = function (fn) {
    let result;
    return function () {
        return result || (result = new fn(...arguments));
    }
};

let SingleDemo = getSingleBuilder(Demo);

const aa = new SingleDemo('a');
const bb = new SingleDemo('b');
console.log(aa === bb); // true

-------------原答案

运算符优先级问题，加上两个括号就好了，否则会把getSingle当作了构造函数。
运算符优先级

const aa = new (getSingle(Demo))('a');
const bb = new (getSingle(Demo))('b');