关于 php 数组引用的问题

Question

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 。

Answer 1

http://php.net/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。