Javascript数组反向功能意外行为 [英] Javascript array reverse function unexpected behavior
本文介绍了Javascript数组反向功能意外行为的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
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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