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关于php数组引用的问题

时间:2021-07-01 10:21:17 帮助过:18人阅读

php$a =  array(1); 
$b =& $a[0];   //注释这条语句最后
输出2,1 $c = $a; $c[0]++; echo $c[0].$a[0]; // 输出 2,2 注释第二条语句,输出 2,1 。

回复内容:

php$a =  array(1); 
$b =& $a[0];   //注释这条语句最后
输出2,1 $c = $a; $c[0]++; echo $c[0].$a[0]; // 输出 2,2 注释第二条语句,输出 2,1 。

http://php.com/manual/en/language.references.whatdo.php

Note, however, that references inside arrays are potentially dangerous. Doing a normal (not by reference) assignment with a reference on the right side does not turn the left side into a reference, but references inside arrays are preserved in these normal assignments. This also applies to function calls where the array is passed by value.

在赋值数组的时候,如果=右边的数据存在引用,那边赋值的新数组对应的元素也是引用,所以改变$c[0]的值 也会同时改变$a[0]的值。

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