How to implement method pointers in PHP - php

Question

Is there a way to implement method pointers in PHP?
I keep getting the following error:
Fatal error: Call to undefined function create_jpeg() in /Users/sky/Documents/images.php on line 175
This is line 175:
if ($this->ImageType_f[$pImageType]($pPath) != 0)
class CImage extends CImageProperties
{
private $Image;
private $ImagePath;
private $ImageType;
private function create_jpeg($pFilename)
{
if (($this->Image = imagecreatefromjepeg($pFilename)) == false)
{
echo "TEST CREATION JPEG\n";
echo "Error: ".$pFilename.". Creation from (JPEG) failed\n";
return (-1);
}
return (0);
}
private function create_gif($pFilename)
{
if (($this->Image = imagecreatefromgif($pFilename)) == false)
{
echo "Error: ".$pFilename.". Creation from (GIF) failed\n";
return (-1);
}
return (0);
}
private function create_png($pFilename)
{
if (($this->Image = imagecreatefrompng($pFilename)) == false)
{
echo "Error: ".$pFilename.". Creation from (PNG) failed\n";
return (-1);
}
return (0);
}
function __construct($pPath = NULL)
{
echo "Went through here\n";
$this->Image = NULL;
$this->ImagePath = $pPath;
$this->ImageType_f['JPEG'] = 'create_jpeg';
$this->ImageType_f['GIF'] = 'create_gif';
$this->ImageType_f['PNG'] = 'create_png';
}
function __destruct()
{
if ($this->Image != NULL)
{
if (imagedestroy($this->Image) != true)
echo "Failed to destroy image...";
}
}
public function InitImage($pPath = NULL, $pImageType = NULL)
{
echo "pPath: ".$pPath."\n";
echo "pImgType: ".$pImageType."\n";
if (isset($pImageType) != false)
{
if ($this->ImageType_f[$pImageType]($pPath) != 0)
return (-1);
return (0);
}
echo "Could not create image\n";
return (0);
}
}

Answer 1

Just call the method you need with $this->$method_name() where $method_name is a variable containing the method you need.
Also it is possible using call_user_func or call_user_func_array
What is callable is described here: http://php.net/manual/ru/language.types.callable.php
So assuming $this->ImageType_f['jpeg']' must be callable: array($this, 'create_jpeg').
Alltogether: call_user_func($this->ImageType_f[$pImageType], $pPath) is the way to do it.
Or if $this->ImageType_f['jpeg'] = 'create_jpeg':
$this->{$this->ImageType_f['jpeg']]($pPath);
Some documentation on functions I mentioned here:
http://us2.php.net/call_user_func
http://us2.php.net/call_user_func_array

Answer 2

Your problem is is line:
if ($this->ImageType_f[$pImageType]($pPath) != 0)
-since $this->ImageType_f[$pImageType] will result in some string value, your call will be equal to call of global function, which does not exists. You should do:
if ($this->{$this->ImageType_f[$pImageType]}($pPath) != 0)
-but that looks tricky, so may be another good idea is to use call_user_func_array():
if (call_user_func_array([$this, $this->ImageType_f[$pImageType]], [$pPath]) != 0)

Answer 3

I think you need to use this function call_user_func
In your case call will looks like
call_user_func(array((get_class($this), ImageType_f[$pImageType]), array($pPath));