TalkPHP - View Single Post - Question about 'return' statement in UDF...

Salathe · 04-18-2008, 11:42 AM

Quote:

Originally Posted by Wildhoney

...
We could thus refactor the above to something like the following:
...

php Code:

$szFilename = 'image.jpg';
$aParts = explode('.', $szFilename);

if(count($aParts) == 1)
{
    printf('Unable to find the extension for %s', $szFilename);
    return;
}

$szExtension = end($aParts);
$szExtension = strtolower($szExtension);
printf('The extension is: %s', $szExtension);

Quote:

Originally Posted by Wildhoney

Ach! Not necessarily. Challenge me

!

I'll challenge you. In terms of readability, does that code snippet say to you "find the extension for this file name"? How about:

PHP Code:

$path = 'image.jpg';
$ext  = pathinfo($path, PATHINFO_EXTENSION);
// return ($ext === '') ? FALSE : $ext;

if ($ext === '')
{
    printf('Unable to find extension for %s', $path);
    return FALSE;
}

printf('Extension for %s is %s', $path, $ext);
return $ext;

Quote:

Originally Posted by Wildhoney

If it reads nicer, and also easier to debug, by utilising a couple more variables then I would say take that approach.

It's also often nicer to entirely rethink what you're writing and find a more suitable method of doing it. I could be wrong but is my snippet nicer to read, easier to debug and uses less variables (than your refactoring, the same as your original messy one)?