Php exit function question

I have noticed a behavior in PHP that makes sense, but I am unsure how to get around it.

I have a long script, something like this

<?php 
 if ( file_exists("custom_version_of_this_file.php") ) {
  require_once "custom_version_of_this_file.php";
  exit;
 }
 // a bunch of code etc
 function x () {
  // does something
 }
?>

Interestingly, the function x() will get registered with the script BEFORE the require_once() and exit are called, and therefore, firing the exit; statement does not prevent functions in the page from registering. Therefore, if I have a function x() in the require_once() file, the script will crash.

Because of the scenario I am attempting (which is, use the custom file if it exists instead of the original file, which will likely be nearly identical but slightly different), I would like to have the functions in the original (calling) file NOT get registered so that they may exist in the custom file.

Anyone know ho开发者_开发技巧w to accomplish this?

Thanks

---

You could use the function_exists function. http://us.php.net/manual/en/function.function-exists.php

if (!function_exists("x")) {
    function x()
    {
        //function contents
    }
}

---

You can wrap your functions in a class and extend it to add custom versions.

this_file.php

class DefaultScripts {
    function x () {
        echo "base x";
    }

    function y() {
        echo "base y";
    }
}

if(file_exists('custom_version_of_this_file.php'))
    require_once('custom_version_of_this_file.php');
else
    $scripts = new DefaultScripts();

custom_version_of_this_file.php

class CustomScripts extends DefaultScripts {
    function x () {
        echo "custom x";
    }
}

$scripts = new CustomScripts();

Results if file exists

$scripts->x(); // 'custom x'
$scripts->y(); // 'base y'

and if it doesn't

$scripts->x(); // 'base x'
$scripts->y(); // 'base y'