Question

Passing by reference:

<?php
$str = "test    \n";
trim(&$str);
echo "-" . "$str" . "-";
?>

output is:

-test
-

but when I do

<?php
$str = "test    \n";
$str = trim($str);
echo "-" . "$str" . "-";
?>

the output is:

-test-

Why can’t I pass this by reference?

Answer 1

Because trim() does not expect a reference and thus does not modify the string that was passed to it. Passing a reference only makes sense if the function expects one – and then you do not have the choice of not passing a reference since what matters if the function definition contains a reference argument or not.

What you are trying to do, call-time pass-by-reference is deprecated in PHP since a long time. Besides that, even if it wasn’t deprecated it would only work for functions that actually modify the argument.

Note: There is no reference sign on a function call – only on function definitions. Function definitions alone are enough to correctly pass the argument by reference. As of PHP 5.3.0, you will get a warning saying that “call-time pass-by-reference” is deprecated when you use & in foo(&$a);.