PHP unsetting a class object in an array of class objects calls destruct for ALL class objects?

Question

I have an array of Client class objects stored in a PHP array. Sometimes I need to unset() one of those objects in the array and I don't want to have to worry about explicitly closing the socket connection. I want __destruct() to do that for me.

class A{

    private $id;

    public function __construct( $id ){
        $this->id = $id;
    }

    public function __destruct(){
        echo "Calling destruct on " . $this->id . ".\n";
    }

}

$array = array();

for($i=0;$i<5;$i++){
    $array[$i] = new A($i);
}

unset($array[3]);

print_r($array);

The destructor is fired as it should for the element we're destroying. But then every other destructor in the array is called even though the elements are not being destroyed. Why?

Calling destruct on 3.
Array
(
    [0] => A Object
        (
            [id:A:private] => 0
        )

    [1] => A Object
        (
            [id:A:private] => 1
        )

    [2] => A Object
        (
            [id:A:private] => 2
        )

    [4] => A Object
        (
            [id:A:private] => 4
        )

)
Calling destruct on 0.
Calling destruct on 1.
Calling destruct on 2.
Calling destruct on 4.

Why is this happening and what alternatives do I have?

jimp · Accepted Answer · 2012-09-19 01:11:58Z

2

If $array goes out of scope, then the rest of the objects will be destroyed, too.

It looks like that is happening. You see your destruct trace message, then your print_r() output, then the rest of the objects destruct.

answered Sep 19, 2012 at 1:11

jimp

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5 Comments

nick Over a year ago

the destruct is being called because the script ended.. got it

jimp Over a year ago

That will do it, too. :) I wasn't sure if that was the end of the script or not.

nick Over a year ago

yeah, but this is for a daemon.. which never ends :)

nick Over a year ago

so far, it's not. and i don't think it will. why?

jimp Over a year ago

I misunderstood your comment about a daemon never ending to mean your code wasn't exiting. In that case, I was going to say if your daemon was forking processes to handle its tasks in parallel, then the child processes exiting would have the same effect in the output. But you aren't doing that, so nevermind. :]

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