This is a contrived example, but it illustrates my problem much more concisely then the code I'm using - and I've tested this and it exhibits the problem:

$image = imagecreatefromjpeg('test.jpg');    
$copy_of_image = $image;

// The important bit
imagedestroy($image);

header('Content-type: image/jpeg');
imagejpeg($copy_of_image);

Now, my expectation is that $copy_of_image is exactly that, but when I run this, it fails, printing out the URL of the script of all things. Comment out the imagedestroy() and it works just fine.

a var_dump of $image provides:

resource(3) of type (gd) 

So why can't I copy this? Apparently the assignment $copy_of_image = $image is creating a reference rather then a copy - is there a way to prevent that?