javascript for loop order

Question

var x = ["a", "b", "c"]; 
for(var i = 0; i < x.length; i++){ 

x[i] = x[2 - i]; 

} 

My approach: 
for i = 0 => x[0] = x[2] (which is "c", so replace "a" with "c") 
for i = 1 => x[1] = x[1] (which is "b", so replace "b" with "b") 
for i = 2 => x[2] = x[0] (which is "a" so replace "c" with "a") 
for i = 3 test failed, stop. 
so x = ["c", "b", "a"] 

为什么控制台返回x为[“c”，“b”，“c”]？ 有人可以告诉我，我是否完全误解了循环逻辑？谢谢！

Answer 1

编辑：整洁！不知道Array.reverse()，这绝对比下面更容易！

到第三次迭代发生时，第一次迭代中第一个元素已经被设置为“c”。

做是简单地使对输出的第二阵列的最简单方法：

var x = ["a", "b", "c"]; 
var y = new Array(x.length); 
for(var i = 0; i < x.length; i++){ 
    y[i] = x[2 - i]; 
} 
console.log(y) 

Answer 2

在i = 2，X [0]是 “c”，则替换它，当 “I” 为0所以它会是[“c”，“b”，“c”]。

Answer 3

var x = ["a", "b", "c"]; 
for(var i = 0; i < x.length; i++){ 

x[i] = x[2 - i]; 

} 

让我们写这样的代码进行手写：

var x = ['a', 'b', 'c']; 

x[0] = x[2]; // ['c', 'b', 'c'] 
x[1] = x[1]; // ['c', 'b', 'c'] 
x[2] = x[0]; // ['c', 'b', 'c'] 

的问题是，到时候你到i = 2你已经修改x[0] = x[2]，所以x[2] = x[0]不出所料没有得到结果。

可以使用Array#reverse方法，我认为：

var x = ['a', 'b', 'c']; 
x.reverse(); // ['c', 'b', 'a']