为什么 Javascript 中用于交换数组中的 2 个元素的这段代码不起作用？

Question

我想创建一个在 Javascript 中交换数组的 2 个元素的函数，所以我创建了这段代码：let arrayOne = ["elementA","elementB","elementC","elementD","elementE","elementF","elementG","elementH","elementI"];function swapThatFails(element1,element2) {    arrayOne[arrayOne.indexOf(element1)] = element2;    arrayOne[arrayOne.indexOf(element2)] = element1;    console.log("arrayOne = ",arrayOne);}swapThatFails ("elementA", "elementC");但是，它对原始数组没有任何改变。我设法通过获取要交换的数组的索引来使其工作：let arrayTwo = ["elementA","elementB","elementC","elementD","elementE","elementF","elementG","elementH","elementI"];function swapThatWorks(element1,element2) {    let index1 = arrayTwo.indexOf(element1);    let index2 = arrayTwo.indexOf(element2);    arrayTwo[index1] = element2;    arrayTwo[index2] = element1;    console.log("arrayTwo = ",arrayTwo);}swapThatWorks ("elementA", "elementC");对我来说，它们都是相同的，除了第二个是分两步完成的，所以看起来更容易理解。为什么第一个不起作用，而第二个起作用？let arrayOne = ["elementA","elementB","elementC","elementD","elementE","elementF","elementG","elementH","elementI"];let arrayTwo = ["elementA","elementB","elementC","elementD","elementE","elementF","elementG","elementH","elementI"];function swapThatFails(element1,element2) {    arrayOne[arrayOne.indexOf(element1)] = element2;    arrayOne[arrayOne.indexOf(element2)] = element1;    console.log("arrayOne = ",arrayOne);  }function swapThatWorks(element1,element2) {    let index1 = arrayTwo.indexOf(element1);    let index2 = arrayTwo.indexOf(element2);    arrayTwo[index1] = element2;    arrayTwo[index2] = element1;    console.log("arrayTwo = ",arrayTwo);  }  swapThatFails ("elementA", "elementC");swapThatWorks ("elementA", "elementC");

Answer 1

让我们考虑一下

function swapThatFails(element1,element2) {

  arrayOne[arrayOne.indexOf(element1)] = element2;

  arrayOne[arrayOne.indexOf(element2)] = element1;

  console.log("arrayOne = ",arrayOne);

}

假设数组是['a','b']并且你用swapThatFails('a', 'b').

首先你搜索element1并放在element2那里

arrayOne[arrayOne.indexOf(element1)] = element2;

你的数组变成['b','b']

然后你搜索element2并放在element1那里。但是数组已经被修改了！第一次出现element2的不是原来的位置，而是新的位置，也就是上一步刚刚被替换的位置。

因此这

  arrayOne[arrayOne.indexOf(element2)] = element1;

查找b为第一个元素并将其替换为a. 你得到了['a','b']。这不是你所期望的！

另一方面，如果您首先搜索元素，获取正确的索引然后替换元素，它会按预期工作。