Quoting Abu Warez <[EMAIL PROTECTED]>:
Hi,
I'm using php 5.2.1 on an Ubuntu machine.
I have the following class diagram (observer pattern):
+---------+ +---------+ +-------------+
| class A |<#>--------->| class B |< >--------->| interface C |
| | +---------+ | |
| |-----------------------------------|>| |
+---------+ +-------------+
in my case class A, the creator of class B, is also the class to observe
for class B. The following code implements the above diagram:
<?php
interface C
{
}
class B
{
var $m_ObsC;
public function __construct( C $p_ObsC )
{
$this->m_ObsC = $p_ObsC;
}
public function __destruct()
{
echo "B::destruct called <br>";
}
}
class A implements C
{
var $m_b;
public function __construct()
{
$this->m_b = new B( $this );
}
public function __destruct()
{
echo "A::destruct called <br>";
}
}
$a = new A();
unset( $a );
echo "<br>-----end-of-script----<br>";
?>
the output of the above code is:
-----end-of-script----
A::destruct called
B::destruct called
So actually the memory used by object $a is freed only after the script
ends and NOT when unset was called. Now this is a big problem for me,
because i need to free the memory when I call unset and not after the
script ends else php runs out of memory in my real application. Object $a
is not referenced by any other object. I also tried with:
$this->m_b = new B( &$this );
but the result is the same.
Any clue how to determine php tho free the memory when unset is called and
not after the script ends?
Thanks,
Abu
I think this happens because there is still a reference to the B object.
According to the manual :
The destructor method will be called as soon as all references to a
particular object are removed or when the object is explicitly
destroyed or in any order in shutdown sequence.
So as long as A has a reference to B the __destructor will not be called.
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