PHP 从变量中删除 "reference"。

Question

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 。在之前将其作为对 $a 的引用赋值后，如何再次将值赋给 $b？

$a = 1;
$b = &$a;

// later
$b = null;

Answer 1

查看内联解释

$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 a new life.