splice正在影响以前复制的变量 [英] splice is affecting previously copied variables

问题描述

可能重复：

在javascript中按值复制数组

我有一个有趣的JavaScript问题。我复制一个数组变量只对副本进行修改，然后拼接副本删除一个元素。但原始数组变量受拼接影响 - 好像副本是'按引用复制'：

i have a funny problem with javascript. i copy an array variable to make modifications on the copy only, then splice the copy to delete an element. however the original array variable is affected by the splice - as if the copy was a 'copy by reference':

window.onload = function() {
  var initial_variable = ['first', 'second', 'third'];
  var copy_initial_variable = initial_variable;
  copy_initial_variable.splice(0, 1);
  alert('initial variable - ' + initial_variable);
};
//output: initial variable - second,third

首先，这是javascript的故意行为还是一个bug？

firstly, is this intentional behaviour for javascript or is it a bug?

其次，如何制作数组副本并删除副本中的一个元素但不是原始元素？

and secondly, how can i make a copy of an array and delete an element in the copy but not in the original?

有一件事让我觉得上面可能是一个javascript错误，这个行为只发生在数组而不是用整数。例如：

one thing which makes me think that the above may be a javascript bug is that this behaviour only happens with arrays and not with integers. for example:

window.onload = function() {
  var initial_variable = 1;
  var copy_initial_variable = initial_variable;
  copy_initial_variable = 2;
  alert('initial variable - ' + initial_variable);
};
//output: initial variable - 1

如果行为一致那么这应该是输出 2 因为作业大概是通过参考？

if the behaviour were consistent then this ought to output 2 since the assignment would presumably be by reference?

推荐答案

这个绝不是一个错误，而是一个非常普遍的误解。让我们看看当我说

This is in no way a bug, but a very common misunderstanding. Let's see what happens when I say

var a = b;

整数和其他javascript原语（如浮点数和布尔值）是按值分配。
这意味着 b 所具有的任何值都将被复制到 a 。对于计算机，这意味着将 b 引用的内存部分复制到 a 引用的内存中。这就是你期望的行为。

Integers and other javascript primitives, like floats and booleans, are "assigned by value". Which means that whatever value b has is going to be copied to a. To the computer, it means having the part of memory that b references copied to the memory that a references. That's the behavior you were expecting.

当数组和其他对象（以及 new Object（）的后代 call）就像这样使用，有一个参考副本。这意味着 a 的值现在引用了 b 的值， b 引用的内存不会被复制或修改。因此，在写作时

When arrays and other objects (and "descendants" of a new Object() call) are used like that, there is a copy by reference. Meaning that the value of a now references the value of b, the memory that b references isn't copied or modified. Thus, when writing

a = [1,2,3];
b = a;

b 和 a 可以互换。他们引用相同的内存地址。要实现您的目标，请使用

b and a become interchangeable. They're referencing the same memory address. To achieve what you're trying to do, use

var copy_initial_variable = initial_variable.slice(0);

阅读 JavaScript是否通过引用传递？以获取更多信息。

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