Only variables should be passed by reference

Question

// Other variables
$MAX_FILENAME_LENGTH = 260;
$file_name = $_FILES[$upload_name][\'name\'];
//echo \"testing-\".$file_name.\"
\";
//$file_name = strtolower         


        

Answer 1

PHP complains because end() expects a reference to something that it wants to change (which can be a variable only). You however pass the result of explode() directly to end() without saving it to a variable first. At the moment when explode() returns your value, it exists only in memory and no variable points to it. You cannot create a reference to something (or to something unknown in the memory), that does not exists.

Or in other words: PHP does not know, if the value you give him is the direct value or just a pointer to the value (a pointer is also a variable (integer), which stores the offset of the memory, where the actual value resides). So PHP expects here a pointer (reference) always.

But since this is still just a notice (not even deprecated) in PHP 7, you can savely ignore notices and use the ignore-operator instead of completely deactivating error reporting for notices:

$file_extension = @end(explode('.', $file_name));