PHP 删除“引用"从一个变量.
问题描述：
我有下面的代码.我想更改 $b 以再次使用它的值.如果我这样做，它也会改变 $a .之前将值分配为 $a 的引用后，如何再次为 $b 分配值?
I have the code below. I want to change $b to use it again with values. If I do so it changes $a as well. How can I assign a value to $b again after previously assigning it as a reference to $a?
$a = 1;
$b = &$a;
// later
$b = null;
答
查看内嵌说明
$a = 1; // Initialize it
$b = &$a; // Now $b and $a becomes same variable with just 2 different names
unset($b); // $b name is gone, vanished from the context But $a is still available
$b = 2; // Now $b is just like a new variable with a new value. Starting new life.