84669 person learning
152542 person learning
20005 person learning
5487 person learning
7821 person learning
359900 person learning
3350 person learning
180660 person learning
48569 person learning
18603 person learning
40936 person learning
1549 person learning
1183 person learning
32909 person learning
想要用一个值来保存另一个值，如下：
var source=[1,2,3] var des=source; des[0]=5;
要怎么样，才能在修改des的时候，不改变source那？
光阴似箭催人老，日月如移越少年。
var source=[1,2,3] var des=source.slice(); des[0]=5; des [5, 2, 3] source [1, 2, 3] var obj = {a: 25, b: 50, c: 75}; var A = Object.create(obj); A.a = 30; console.log(obj.a + " " + A.a); // 25 30 if (!Object.create){ Object.create = function(proto) { function F(){} F.prototype = proto; return new F; } }
var des = JSON.parse(JSON.stringify(source))
深拷贝即可避免浅拷贝 :)
拷贝个副本即可：
var des = source.concat();
var des = source.slice(0);
![[Web front-end] Node.js quick start](https://img.php.cn/upload/course/000/000/067/662b5d34ba7c0227.png)
var des = JSON.parse(JSON.stringify(source))深拷贝即可避免浅拷贝 :)
拷贝个副本即可：
var des = source.slice(0);