Javascript数组反向功能意外行为 [英] Javascript array reverse function unexpected behavior

问题描述

我想了解为什么情况1和情况2无法返回相同的结果.

I would like to understand why situation 1 and situation 2 don't return the same results.

情况1:

var array1 = ["1", "2", "3"];
var array2 = array1.reverse();

console.log(array1); // ["3", "2", "1"]
console.log(array2); // ["3", "2", "1"] Why this doesn't work ?

情况2:

var array1 = ["1", "2", "3"];
var array2 = array1;

console.log(array1);           // ["1", "2", "3"]
console.log(array2.reverse()); // ["3", "2", "1"] Why this works ?

推荐答案

在情况1

reverse方法将调用数组对象的元素转置到位，变异该数组，并返回对该数组的引用.

The reverse method transposes the elements of the calling array object in place, mutating the array, and returning a reference to the array.

在情形2 中，您具有相同参考.您要在之前反转 array1 数组.

In Situation 2 you have the same reference. You are printing array1 array before reverse it.

var array1 = ["1", "2", "3"];
var array2 = array1;

console.log(array1);           // [ '1', '2', '3' ]
console.log(array2.reverse()); // [ '3', '2', '1' ]
console.log(array1);           // [ '3', '2', '1' ]

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