How to implement method pointers in 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);
  }
}

Community · Accepted Answer · 2023-11-17 20:38:18Z

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:

https://www.php.net/call_user_func https://www.php.net/call_user_func_array

edited Nov 17, 2023 at 20:38

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answered Nov 15, 2013 at 12:45

baldrs

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1 Comment

cernunnos Over a year ago

You could also save the function name in a variable then call it: $funcName = $this->ImageType_f[$pImageType]; $this->$funcName($pPath);

Community · Accepted Answer · 2023-11-17 20:33:26Z

1

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)

edited Nov 17, 2023 at 20:33

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answered Nov 15, 2013 at 12:48

Alma Do

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Comments

Community · Accepted Answer · 2023-11-20 22:45:59Z

0

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));

edited Nov 20, 2023 at 22:45

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answered Nov 15, 2013 at 12:49

GremniX

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