问题
Are both these PHP statements doing the same thing?:
$o =& $thing;
$o = &$thing;
回答1:
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, thing should be variables.
回答2:
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:
$f = isset($field[0]) ? &$field[0] : &$field;
$f =& isset($field[0]) ? $field[0] : $field;
So instead you would use this:
isset($field[0]) ? $f =& $field[0] : $f =& $field;
回答3:
They both give an expected T_PAAMAYIM_NEKUDOTAYIM error.
If you meant $o = &$thing; then that assigns the reference of thing to o. Here's an example:
$thing = "foo";
$o = &$thing;
echo $o; // echos foo
$thing = "bar";
echo $o; // echos bar
回答4:
The difference is very important:
<?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.
回答5:
If you meant thing with 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
回答6:
Yes, they do. $o will become a reference to thing in both cases (I assume that thing is not a constant, but actually something meaningful as a variable).
来源：https://stackoverflow.com/questions/5930177/phps-operator