PHP 的 =& 运算符
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PHP's =& operator
提问by Kev
Are both these PHP statements doing the same thing?:
这两个 PHP 语句都在做同样的事情吗？：
$o =& $thing;
$o = &$thing;
采纳答案by judda
Yes, they are both the exact same thing. They just take the reference of the object and reference it within the variable $o. Please note, thingshould be variables.
是的，它们是完全相同的东西。他们只是获取对象的引用并在变量中引用它$o。请注意，thing应该是变量。
回答by Scott Lahteine
They're not the same thing, syntactically speaking. The operator is the atomic =&and this actually matters. For instance you can't use the =&operator in a ternary expression. Neither of the following are valid syntax:
从语法上讲，它们不是一回事。运算符是原子=&，这实际上很重要。例如，您不能在三元表达式中使用=&运算符。以下都不是有效的语法：
$f = isset($field[0]) ? &$field[0] : &$field;
$f =& isset($field[0]) ? $field[0] : $field;
So instead you would use this:
所以你会使用这个：
isset($field[0]) ? $f =& $field[0] : $f =& $field;
回答by Rich Bradshaw
They both give an expected T_PAAMAYIM_NEKUDOTAYIM error.
它们都给出了预期的 T_PAAMAYIM_NEKUDOTAYIM 错误。
If you meant $o = &$thing;then that assigns the reference of thing to o. Here's an example:
如果您的意思$o = &$thing;是将事物的引用分配给 o。下面是一个例子：
$thing = "foo";
$o = &$thing;
echo $o; // echos foo
$thing = "bar";
echo $o; // echos bar
回答by bbe
The difference is very important:
区别非常重要：
<?php
$a = "exists";
$b = $a;
$c =& $a;
echo "a=".$a.", b=".$b.", c=".$c."<br/>"; //a=exists b=exists c=exists
$a = null;
echo "a=".$a.", b=".$b.", c=".$c; //a= b=exists c=
?>
Variable $c dies as $a becomes NULL, but variable $b keeps its value.
变量 $c 在 $a 变为 NULL 时消失，但变量 $b 保持其值。
回答by Emil Vikstr?m
回答by Karl Laurentius Roos
If you meant thingwith a $before them, then yes, both are assigning by reference. You can learn more about references in PHP here: http://www.php.net/manual/en/language.references.whatdo.php
如果您的意思thing是$在它们之前加上 a ，那么是的，两者都是通过引用分配的。您可以在此处了解有关 PHP 引用的更多信息：http: //www.php.net/manual/en/language.references.whatdo.php
