引用传递

你可以将一个变量通过引用传递给函数，这样该函数就可以修改其参数的值。语法如下：


<?php

function foo (&$var)

{

    $var++;

}



$a=5;

foo ($a);

// $a is 6 here

?>

注意在函数调用时没有引用符号－只有函数定义中有。光是函数定义就足够使参数通过引用来正确传递了。

以下内容可以通过引用传递：

变量，例如 foo($a)
New 语句，例如 foo(new foobar())
从函数中返回的引用，例如：

<?php function &bar() { $a = 5; return $a; } foo(bar()); ?>
详细解释见引用返回。

任何其它表达式都不能通过引用传递，结果未定义。例如下面引用传递的例子是无效的：


<?php

function bar() // Note the missing &

{

    $a = 5;

    return $a;

}

foo(bar());



foo($a = 5) // 表达式，不是变量

foo(5) // 常量，不是变量

?>

这些条件是 PHP 4.0.4 以及以后版本有的。

add a note User Contributed Notes
引用传递

blistwon-php at designfridge dot com
18-Jan-2004 10:33

One thing to note about passing by reference. If you plan on assigning the
reference to a new variable in your function, you must use the reference
operator in the function declaration as well as in the assignment. (The
same holds true for
classes.)

-------------------------------------------------------
CODE
-------------------------------------------------------
function
f1(&$num) {
	$num++;
}
function f2(&$num) {
	$num1 =
$num;
	$num1++;
}
function f3(&$num) {
	$num1 =
&$num;
	$num1++;
}

$myNum = 0;
print("Declare
myNum: " . $myNum . "<br
/>\n");
f1($myNum);
print("Pass myNum as ref 1: " .
$myNum . "<br />\n");
f2($myNum);
print("Pass
myNum as ref 2: " . $myNum . "<br
/>\n");
f3($myNum);
print("Pass myNum as ref 3: " .
$myNum . "<br
/>\n");
-------------------------------------------------------

-------------------------------------------------------
OUTPUT
-------------------------------------------------------

Declare myNum: 0
 Pass myNum as ref 1: 1
 Pass myNum as ref 2: 1

Pass myNum as ref 3:
2
-------------------------------------------------------

Hope
this helps people trying to detangle any problems with pass-by-ref.

add a note