有一段时间一直迷惑于PHP中引用的传递,后来查手册及C源程序,并反复测试,大致对引用传递在内存中的模式有了一定的了解,后来为了加深印象,写了个总结,应该不会有大的问题——当然这是在PHP4中,在以后的版本中可能会有变化。当时写总结的时候,想锻炼一下英语,因此就凑合了一篇。不过本人英语不好,也懒得翻译,反正当时想自己看得懂就行了。今天心血来潮,突然觉得还蛮有用的,于是在这里现丑了,请大家指正。那位看得懂的帮忙翻译一下吧,我没空了。 
/* 
filename: SetGetTest.php 
comment on assignment by value and referrence 
assuming $var is a varialbe, its handle(pointer) is named as *var, 
and its content is named as @var 
the memory area of @var is referred by *var, if the *var is the same, 
then the memory areas are the same, so *var is just like a pointer. 
1. when $var = $var1 
@var copied from @var1, but in the different memory area, 
new *var assigned by the system, pointing to the memory area holding @var 
*var and *var1 are different 
2. when $var =& $var1, 
*var assigned by *var1, and the @var is not assigned or copied, 
it is absolutely the same as @var1, and in the same memory area 
both *var and *var1 pointing to the memory area, that means they are the same. 
passing by referrence 
3. 
function set1(&$s){ 
$var =& $s; 
} 
set1($var1) results: 
*var1 passing to the function, and *s is the same as *var1, 
then *var is the same as *s, the result is that *var is the same as *var1 
and all the contents are the same obviously. 
4. 
function set2(&$s){ 
$var = $s; 
} 
set2($var1) results: 
*var1 passing to the function, and *s is the same as *var1, 
but when $var = $s executes, from 1. we can see @var is the same as @s, 
but *var is different from *s, so @var and @s is not in the same memory area, 
while @s and @var1 is sharing the same memory area, also *var1 and *s are the same. 
5. 
normal function return: 
function get(){ return $var1; } 
assuming the result is referred by a variable $result. 
then @result is copied from @var1 but *result is not the same as *var1 
when $var = get(); 
first you get a variable $result, as I said above, @result is the same as @var1, but *result 
is different from *var1, and next $var = $result executes. 
As I said in 1., you can find, @var is the same as @result and the same as @var1, 
but *var is different from *result AND *var1; 
while $var =& get() just means: 
*var is the same as *result, so @var and @result are in the same memory area, 
but they are still different from those of $var1, 
both the memory area of @var1 and *var1, 
6. 
returning by referrence 
function &get(){ return $var1; } 
there are two ways to get the result 
$var = get(); and $var =& get(); now I will tell the difference 
I. $var = get(); 
the *result is the same as *var1 and so @result and @var1 are the same. 
and then $var = $result executes, 
*var is not the same as *result, and also different from *var1, 
but their contents are the same. 
I. $var =& get(); 
the *result is the same as *var1 and so @result and @var1 are the same. 
and then $var =& $result executes, 
this means $var and $result are the same, both of @ and * 
*/ 
// the test is the following 
function println($s = ""){ 
print "$s
\n"; 
} 
class GetSetTest 
{ 
var $var = null; 
function setByRef(&$arg){ 
$this->var =& $arg; 
} 
function passByRef(&$arg){ 
$this->var = $arg; 
} 
function setByVal($arg){ 
$this->var = $arg; 
} 
function &getByRef(){ 
return $this->var; 
} 
function getByVal(){ 
return $this->var; 
} 
} 
$o = new GetSetTest; 
println("============ setByRef getByRef ============="); 
println("-----------------Before change----------------"); 
$in = "before change"; 
$o->setByRef($in); 
$outByVal = $o->getByRef(); 
$outByRef =& $o->getByRef(); 
println("\$in: ".$in); 
println("\$outByVal: ".$outByVal); 
println("\$outByRef: ".$outByRef); 
println("\$this->var: ".$o->var); 
println("-----------------After change-----------------"); 
$in = "after change"; 
println("\$in: ".$in); 
println("\$outByVal: ".$outByVal); 
println("\$outByRef: ".$outByRef); 
println("\$this->var: ".$o->var); 
println(); 
println("============ setByRef getByVal ============="); 
println("-----------------Before change----------------"); 
$in = "before change"; 
$o->setByRef($in); 
$outByVal = $o->getByVal(); 
$outByRef =& $o->getByVal(); 
println("\$in: ".$in); 
println("\$outByVal: ".$outByVal); 
println("\$outByRef: ".$outByRef); 
println("\$this->var: ".$o->var); 
println("-----------------After change-----------------"); 
$in = "after change"; 
println("\$in: ".$in); 
println("\$outByVal: ".$outByVal); 
println("\$outByRef: ".$outByRef); 
println("\$this->var: ".$o->var); 
println(); 
println("============ passByRef getByVal ============="); 
println("-----------------Before change----------------"); 
$in = "before change"; 
$o->passByRef($in); 
$outByVal = $o->getByVal(); 
$outByRef =& $o->getByVal(); 
println("\$in: ".$in); 
println("\$outByVal: ".$outByVal); 
println("\$outByRef: ".$outByRef); 
println("\$this->var: ".$o->var); 
println("-----------------After change-----------------"); 
$in = "after change"; 
println("\$in: ".$in); 
println("\$outByVal: ".$outByVal); 
println("\$outByRef: ".$outByRef); 
println("\$this->var: ".$o->var); 
println(); 
/* 
以下输出结果是我(夜猫子)擅自编辑添加的,主要是为后来人查看方便加在这里,越肉代庖,向longnetpro致歉 
输出结果: 
============ setByRef getByRef ============= 
-----------------Before change---------------- 
$in: before change 
$outByVal: before change 
$outByRef: before change 
$this->var: before change 
-----------------After change----------------- 
$in: after change 
$outByVal: before change 
$outByRef: after change 
$this->var: after change 
============ setByRef getByVal ============= 
-----------------Before change---------------- 
$in: before change 
$outByVal: before change 
$outByRef: before change 
$this->var: before change 
-----------------After change----------------- 
$in: after change 
$outByVal: before change 
$outByRef: before change 
$this->var: after change 
============ passByRef getByVal ============= 
-----------------Before change---------------- 
$in: before change 
$outByVal: before change 
$outByRef: before change 
$this->var: before change 
-----------------After change----------------- 
$in: after change 
$outByVal: before change 
$outByRef: before change 
$this->var: after change 
*/ 
?>