Question

我刚刚注意到以下代码，在one之后没有立即引用let [one, two] = [1, 2]，尝试[one, two] = [two, one]会崩溃：

＆＃13;

let [one, two, three] = [1, 2, 3]
[one, two] = [two, one] // CRASH! one is not defined
console.log(one, two)

＆＃13;

但是，只需在声明和交换之间添加一个未使用的one突然允许代码，但不正确：

＆＃13;

let [one, two, three] = [1, 2, 3]
one // inserted here
[one, two] = [two, one] // no longer crashes! but not actually swapping
console.log(one, two) // should be '2 1', but shows '1 2' instead

＆＃13;

以下代码给出了预期的交换效果

＆＃13;

var a = 1;
var b = 3;

[a, b] = [b, a];
console.log(a); // 3
console.log(b); // 1

＆＃13;

有人可以解释为什么存在这种行为吗？谢谢！

Answer 1

在第1行添加分号，因为解释器假定第1行和第2行声明一起发生，one（和two）尚未定义：

＆＃13;

let [one, two, three] = [1, 2, 3];
[one, two] = [two, one]
console.log(one, two)

＆＃13;

Answer 2

因为它解析如下：

 let [one, two, three] = [1, 2, 3][one, two] = [one, two]

为了使它成功，总是用parens来解释分配任务：

 let [one, two, three] = [1, 2, 3];
 ([one, two] = [two, one]);
 console.log(one, two);

永远不要相信ASI，有些情况会出现问题。

关于ES6数组解构和交换的有趣行为

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