关于 ES6 数组解构和交换的有趣行为

Question

我刚刚注意到下面的代码，在 let [one, two] = [1, 2] 之后没有立即引用 one，尝试 [one, two] = [two, one] 会崩溃：

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

然而，简单地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：

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

而下面的代码给出了预期的交换效果

var a = 1;
var b = 3;

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

有人可以解释为什么存在这种行为吗？谢谢！

Answer 1

在第 1 行添加一个分号，因为解释器假定第 1 行和第 2 行声明一起发生，并且 one（和 two）尚未定义：

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

Answer 2

因为它是这样解析的：

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

为了让它工作，总是用括号包围解构赋值：

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

永远不要相信 ASI，如果它出错了，也会有这样的情况。

关于 ES6 数组解构和交换的有趣行为

Interesting behavior about ES6 array destructuring and swapping

javascript

destructuring

ecmascript-6