imagedestroy

(PHP 4, PHP 5, PHP 7, PHP 8)

imagedestroy — 销毁一图像

说明

imagedestroy ( resource $image ) : bool

imagedestroy() 释放与 image 关联的内存。image 是由图像创建函数返回的图像标识符，例如 imagecreatetruecolor()。

User Contributed Notes

devel at kijote dot com dot ar 03-Jan-2020 01:16


Important note: When the imagedestroy is called, the resource is freed but the printed resource output is the same as before calling the function:



<?php



  $img = imagecreate(1, 1);



  print_r([$img, $img ? 'TRUE': 'FALSE', is_resource($img) ? 'TRUE' : 'FALSE', get_resource_type($img) ?: 'FALSE']);



  /*

  Result:

  Array

  (

    [0] => Resource id #1

    [1] => TRUE

    [2] => TRUE

    [3] => gd



  )

  */



  imagedestroy($img);



  print_r([$img, $img ? 'TRUE': 'FALSE', is_resource($img) ? 'TRUE' : 'FALSE', get_resource_type($img) ?: 'FALSE']);



  /*

  Result:

  Array

  (

    [0] => Resource id #1

    [1] => TRUE

    [2] => FALSE

    [3] => Unknown

  )

  */

?>



As you can see in the above example, the first index in array is TRUE in both cases. So, despite the common thinking you can't trust in something like:



<?php



  if ($img)  {  // it's casted as boolean and returns true even after imagedestroy is called

     // do something

  }



?>



If you need to ensure the availability of a certain resource, you must to use is_resource and get_resource_type functions.

Claude D. 19-Nov-2015 02:35


Caution should be observed with imagedestroy(); copying your reference variable over to another variable will cause imagedestroy to destroy both at once.



Eg.: 



$a = imagecreate(...);

$b = $a;

imagedestroy($a);



While you'd think $b still contains your image, it doesn't. Both $a and $b are destroyed.

corpus-deus at softhome dot net 12-Feb-2010 06:33


In theory creating an image object and calling imagedestroy in your destructor should be a good way of doing it; something like:





<?php


final class My_Image() {





  private $img;





  public function __construct() {


    $this->img = imagecreatetruecolor();


    // ... other stuff ...


  }





  public function __destruct() {


    if(is_resource($this->img)) {


      imagedestroy($this->img);


    }


  }





  // ... other methods...





}


?>





I check that $this->img is a resource in case imagecreatetruecolor() fails or you null the value somewhere down the line from a method call meaning that $this->img is NOT a resource, in which case imagedestroy is an unecessary function call that just fails with a warning message.

webmaster at codefisher dot org 05-Oct-2007 10:26


And to continue what Docey said, if php did not destroy all resources when the script stopped it would be a huge memory leak and everyone would be crying out for it to be fixed right away.



I have been using this function during a script that was breaking an image made of many little icons into the little parts, which could mean 400+ images in the one script, which was using a lot of memory so I needed to destroy them.

Docey 20-Oct-2006 10:59


When the script stop PHP will automatic destory ANY

resources, this goes for aswell for images, thus in the

case the use clicks the stop button, php will automatic

clear the resource.



thus imagedestroy is used to clear the memory BEFORE

the script ends. this is usefull to keep memory usage

DURING the script to an acceptable level.



hope this clear somethings up.

Andrew Hoffmann - ahoffmann at wisc dot edu 23-Jun-2005 07:25


When working with a lot of high-resolution images, it's IMPERATIVE that you use the imagedestroy() function.



In my scenario, I was taking two high resolution desktop wallpapers and shrinking them down into successively smaller ones (preventing the user from having to upload a dozen files).



At first, my script would run, then just stop.  I realized later that I had not destroyed the source image and the newly resized image in memory after I had finished writing the file out to disk.  Thus, I quickly reached the memory limit that my hosting provider placed in their php.ini file.



Reusing an image variable does NOT clear the old data out of memory!  You must use imagedestroy() to clear the data out.  (I don't know if unset() works as well).



Also note that the image data in memory is raw, so don't base how much memory you are using based on the original filesize of the compressed image (such as jpeg or png).

dan at mlodecki dot net 04-Mar-2004 04:42


I have noticed that gd drawing functions can behave oddly if there is a previous image which has not been imagedestroy()'ed.  You should always imagedestroy when you are done with an image object.