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 *
/*
以下输入了局是我(夜猫子)私自编纂添加的,次要是为后来人检查便利加在这里,越肉代办,向longnetpro道歉
输入了局:
============ 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
*/