Passing a variable as reference without making it mandatory

[mp96brbj]

I'm writing a function where I want to pass a variable as reference, but I want it to be an optional argument.

This is what I want to do:

function foo($str1, &$str2='')

$str2 should be optional, but when supplied, I want to modify it. However, the

=''

part makes PHP spit out a parse error. If I only write

&$str2

, PHP will complain about "missing argument 2..."

I want $str2 to behave like the third argument in ereg() does. How can I do that?

[jerdo]

Not sure but I believe I did this in a script once. Try putting a line at the beginning your function definition like this:

if (!$str2)
{
  $str2 = "";
}

Then just call it without the =''

[Weedpacket]

None too sure, either, but would

$str = func_num_args()==2 ? &func_get_arg(1) : null

(and a function signature with only one argument) work?

[mp96brbj]

Hi and thanks for replying.
Unfortunately, none of those solutions worked.

jerdo's solution still produced a warning when the second argument was not supplied.

Weedpacket's solution didn't work because PHP won't let me make aliases for a function execution.

Still, it must be possible in someway.

[barand]

Does this work?

function foo($str1, $str2="") {

}

call with

$x = foo($str);

or

$x = foo($str, &$anotherstr);

[rinjani]

Passing by reference requires that you pass a $variable to the function. So I don't think you can hard code it that way.
Can you declare an empty value for the $str2 before you call the function?