...运算符可用于浅拷贝

• 深拷贝

let obj1={name:'Jacl'}
let obj2=obj1
obj1["age"]=12
obj2//{name: "Jacl", age: 12}

用Object.assign(obj)实现深拷贝

let obj1={name:'Jacl'}
let obj2=Object.assign( obj1)
obj1["age"]=12
obj2//{name: "Jacl", age: 12}

• 浅拷贝

let obj1={name:'Jacl'}
let obj2={...obj1}
obj1["age"]=12
obj2//{name: "Jacl"}

当然，我们也可以在浅拷贝的同时修改所获对象的属性

let obj1={name:'Jacl'}
let obj2={...obj1,name:'Zha'}
obj1["age"]=12
obj2//{name: "Zha"}

用Object.assign({}, obj)实现浅拷贝

let obj1={name:'Jacl'}
let obj2=Object.assign({}, obj1)
obj1["age"]=12
obj2//{name: "Jacl"}

数组同理

[...arr1]

• 那么如果想对[{},{},{}...]形式的数据对象进行浅拷贝呢?

一层浅拷贝(不符合要求)

二层浅拷贝(符合要求)

...运算符也有其它用处

• 类型转换
  set转arr(数组去重)

let s1=new Set([2,1,3,4,3,1])
let arr=[...s1]
arr //[2, 1, 3, 4]

• 获取对象属性/方法

let obj={name:'Li',age:2}
let {name}={...obj}
name//"Li"

• 扩充对象属性(增加数组元素)

let a=[1,2,3]
let b=[4,...a]
b//[4, 1, 2, 3]

let x={name:'Li'};
let y={age:3,...x};
y//{age: 3, name: "Li"}