如何改变$数组2值，而不参照$ ARRAY1？ [英] How to changing value in $array2 without referring $array1?

问题描述

考虑下面的PHP code段。

  $ ARRAY1 =阵列（1,20）;
$ X =安培; $数组1 [1];
$数组2 = $数组1;
$数组2 [1] = 22;
的print_r（$数组1 [1]）; //输出为22
 

在这里， $数组2 未引用到 $数组1 ，但如何在<$ C $改变价值C> $数组2 $数组1 ？>

解决方案

如果你想 $数组2 是 $数组1 那么你做同样的事情与 $ X 。

  $数组2 =＆放大器; $数组1;
 

现在任何你在改变任何 $数组1 或 $数组2 是在两个数组可见，因为 $数组2 仅仅是一个参考 $数组1 。

---

更新

想着它，你可能会看到的是改变价值的方式，但仍然有数组的完整副本。这是可行的与对象

  $ OBJ =新stdClass的（）;
$ ARRAY1 =阵列（1，20）;
$数组1 [1] = $ OBJ;
$数组1 [1]  - ＆GT;彩色= 22;$数组2 = $数组1;
$数组2 [1]  - ＆GT;彩色= 33;回声$数组1 [1]  - ＆GT;颜色; //输出33
 

这是因为对象总是以引用复制，而数字和字符串原样复制。

Consider the following PHP code segment.

$array1 = array(1,20);
$x = &$array1[1];
$array2 = $array1;
$array2[1] = 22;
print_r($array1[1]); // Output is 22

Here, $array2 is not referencing to $array1, but how to change value in $array2 by changing value of $array1?

解决方案

If you want $array2 to be a reference of $array1 then you do the same thing as with $x.

$array2 = &$array1;

Now anything you change in either $array1 or $array2 is visible in both arrays since $array2 is just a reference to $array1.

---

Update

Thinking about it, what you may be looking at is a way to change a value, but still have a full copy of the arrays. This is doable with an object.

$obj = new stdClass();
$array1 = array(1, 20);
$array1[1] = $obj;
$array1[1]->color = 22;

$array2 = $array1;
$array2[1]->color = 33;

echo $array1[1]->color;  // prints 33

This is because objects are always copied by reference, whereas numbers and strings are copied as is.

这篇关于如何改变$数组2值，而不参照$ ARRAY1？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！