- 只要等号两边的模式相同，左边的变量就会被赋予对应的值
let [ a, b, c] = [ 1 , 2 , 3 ] ; console. log ( "a=" + a) ; console. log ( "b=" + b) ; console. log ( "c=" + c) ; 效果 let [ , , c] = [ 1 , 2 , 3 ] ; console. log ( "c=" + c) ;
效果 示例二(前者部分赋值) let [ a, b] = [ 1 , 2 , 3 ] ; console. log ( "a=" + a) ; console. log ( "b=" + b) ;
效果 如果解构不成功，一般变量的值就等于undefined let [ a, b, c, d] = [ 1 , 2 , 3 ] ; console. log ( "a=" + a) ; console. log ( "b=" + b) ; console. log ( "c=" + c) ; console. log ( "d=" + d) ;
效果 let [ a, [ [ b] , c] ] = [ 1 , [ [ 2 ] , 3 ] ] ; console. log ( "a=" + a) ; console. log ( "b=" + b) ; console. log ( "c=" + c) ;
效果 let [ first, ... other] = [ 1 , 2 , 3 , 4 ] ; console. log ( "first=" + first) ; console. log ( "other=" + other) ;
效果 - 解构赋值允许指定默认值
let [ a, b = 1 ] = [ 1 , 2 , 3 ] ; console. log ( "a=" + a + ",b=" + b) ; let [ a1, b1 = 1 ] = [ 1 ] ; console. log ( "a1=" + a1 + ",b1=" + b1) ; let [ a2 = 1 ] = [ null ] ; console. log ( "a2=" + a2) ; let [ a3 = 1 ] = [ undefined ] ; console. log ( "a3=" + a3) ;
效果 问题 - 大家可能有个疑惑
- 问题：为什么语句let [ a3 = 1 ] = [ undefined] ; 中a3对应的值为undefined，为什么运行结果a3= 1 ?
- 答案：ES6 内部使用严格相等运算符（== = ），判断一个位置是否有值，当一个数组成员严格等于undefined，默认值就会生效
- 默认值可以引用解构赋值的其他变量，但该变量必须已经声明
let [ x = 1 , y = x] = [ ] ;
let [ x = 1 , y = x] = [ 2 ] ;
let [ x = 1 , y = x] = [ 1 , 2 ] ;
let [ x = y, y = 1 ] = [ ] ;
