有关PHP引用

1. 引用不能被传递

<?php 

$a=1;

$b=&$a;

$c=$b;

$c=99;

echo $a;   // 1

?>

2.如果对一个未定义的变量进行引用赋值、引用参数传递或引用返回，则会自动创建该变量。

<?php

function foo(&$var) { }



foo($a); // $a is "created" and assigned to null



$b = array();

foo($b['b']);

var_dump(array_key_exists('b', $b)); // bool(true)



$c = new StdClass;

foo($c->d);

var_dump(property_exists($c, 'd')); // bool(true) 

?>

3.有关global

<?php
$var1 = "Example variable";
$var2 = "";

function global_references($use_globals)
{
    global $var1, $var2;
    if (!$use_globals) {
        $var2 =& $var1; // visible only inside the function
    } else {
        $GLOBALS["var2"] =& $var1; // visible also in global context
    }
}

global_references(false);
echo "var2 is set to '$var2'\n"; // var2 is set to ''
global_references(true);
echo "var2 is set to '$var2'\n"; // var2 is set to 'Example variable'
?> 

//把 global $var; 当成是 $var =& $GLOBALS['var']; 的简写。从而将其它引用赋给 $var 只改变了本地变量的引用。  

 3.如果在 foreach 语句中给一个具有引用的变量赋值，被引用的对象也被改变。

<?php
$ref = 0;
$row =& $ref;
foreach (array(1, 2, 3) as $row) {
    // do something
}
echo $ref; // 3 - last element of the iterated array
?>