为您详解PHP开发工具的使用与分析

网络整理 - 08-18

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。

以下为引用的内容：
<?php
/*

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<br>\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);