[RESOLVED] Functions with same name

Piranha

Is there a possibility to use two functions with the same name but different parameters in PHP? Something like this:

function test()
{
	echo "function test() <br />";
}

function test($abc)
{
	echo "function test($abc) <br />";
}

test();
test("abc");

I know that the code above does not work, it returns the message Fatal error: Cannot redeclare test() (previously declared in C:\php\akvarium\html\test\test1.php:12). I tried to put it in a class instead, but it doesn't work either. Is there another way to solve this problem? I am currently using PHP 4, but am thinking of upgrading to PHP 5 anyway.

Edit: It is possible to do like this:

function test($abc)
{
	echo "function test($abc) <br />";
}

test();
test("abc");

But if I do that I get an ugly error message: Warning: Missing argument 1 for test(). But some built-in functions can take different arguments, for example [man]echo[/man].

Mythotical

As far as I know you can't have the same function as another. You would have to make the second function different than the first. Don't hold me to that though as lately I've been seeing that I can be wrong.

Houdini

You can only have a function that has one unique name however you can add parameters to the function here is a good explaination of functions at this Link so just add parameters that go into the () part of the function or if you have functions that do a certain thing then name them accordingly for the purpose you designed them for.

NogDog

There is no overloading of function names in PHP. About the best you can do is to have logic in the single function definition that branches based on the number (and possibly type) of parameters entered, using the optional parameter definition option:

<?php
function example($arg1, $arg2=0)
{
   if(func_num_args() == 2)
   {
      printf("Two args were supplied: '%s' and '%s'\n", $arg1, $arg2);
   }
   elseif(is_int($arg1))
   {
      printf("One integer arg supplied: %d\n", $arg1);
   }
   else
   {
      printf("One non-integer arg supplied: '%s'\n", $arg1);
   }
}
header("Content-Type: text/plain");
example(2);
example(3.4);
example("test");
example(1,"test");
?>

bradgrafelman

Or, don't specify any required arguments at all:

function test() {
    $argc = func_num_args();
    $argv = func_get_args();
    echo 'You provided ' . $argc . ' arguments.<br/>';

echo 'They are, in order:<hr/>';
for($i = 0; $i < $argc; $i++) {
    echo $argv[$i] . '<br/>';
}
}

Piranha

Ok, then there are some ways to trick PHP at least. Thanks guys, I apreciate it 😃