ES6 deconstruction assignment

ES6 Allow to follow a certain pattern , Extract values from arrays and objects , Assign values to variables , This is called deconstruction
var [a,b,c] = [1,2,3] //a=1.b=2.c=3
This is the way it's written ‘ Pattern matching ’, As long as the patterns on both sides of the equal sign are the same , The variables on the left will be assigned corresponding values
let [aa, [[bb], cc] = [1, [[2], 3]] //aa=1, bb=2, cc=3
let [x, …y] = [1,2,3,4] //x=1,y=[2,3,4]
let [aaa,bbb] = [11] //aaa=11, bbb=undefined

Deconstruction values allow default values to be specified
let [aaa = true] = [] //aaa = true
let [aa, bb=1] = [11] //aa=11, bb=1
let [xa, xb = 22] = [11, undefined] //xa = 11, xb = 22
es6 The strict equality operator is used internally to determine whether a position has a value , Only if a member is strictly equal to undefined The default value can only be used when

The default value can refer to other variables assigned by deconstruction , But the variable must have been declared
let [d=1, e=d] = [] //d=1 e=1
let [g=2, h=g] = [2] //g=2, h=2
let [i=3, j=i] = [11,22] //i=11, j =22
let [k=m, m =1] = [] // Report errors ,m Undefined

let obj2 = {
    
            ob:{
    
                st:{
    
                    fir:1,
                    las:5
                }
            }
        }
        let {
    ob, ob:{
    st},ob:{
    st:{
    fir}}} = obj2 //ob={st: {…}} st={fir: 1, las: 5} fir=1