Strict Standards php error

Question

I have a small issue with my script.

I'm getting Strict Standards: Only variables should be passed by reference in

if( $checkDNS && ($domain = end(explode('@',$email, 2))) )

possible duplicate of [Strict Standards: Only variables should be passed by reference](http://stackoverflow.com/questions/2354609/strict-standards-only-variables-should-be-passed-by-reference) — Lorenz Meyer, Jun 21 '14 at 08:32

score 3 · Accepted Answer · answered Jan 21 '13 at 22:28

From the PHP manual:

This array is passed by reference because it is modified by the function. This means you must pass it a real variable and not a function returning an array because only actual variables may be passed by reference.

So you must use a variable in the end function:

$domain = explode('@',$email, 2);
if( $checkDNS && ($domain = end($domain)) )

score 2 · Answer 2 · answered Jan 21 '13 at 22:27

2

From the manual:

mixed end ( array &$array )

end takes the array by reference and move the internal pointer. Your array is the function output, so its unable to correctly modify the array by reference.

answered Jan 21 '13 at 22:27

datasage

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score 0 · Answer 3 · answered Jan 21 '13 at 22:33

Like the message says, end expects a variable because its parameter is a reference.

But since PHP 5.4 you can dereference arrays like that:

$domain = explode('@',$email, 2)[1];

Assuming that $email always contains @. You should assure that beforehand, otherwise end(...) would give you unexpected results too.

Strict Standards php error

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