PHP Fatal Error. Does empty() try to alter the results passed into it?

Ran into a strange problem in PHP today and I'm wondering if someone can explain it. While comparing two arrays I initially tried something like this:

echo empty(array_diff( ar开发者_JS百科ray('foo','bar') , array('bar','foo') ))

This results in the following error:

Fatal Error: Can't use function return value in write context

Rewriting this as...

$dif = array_diff( array('foo','bar') , array('bar','foo') );
echo empty($dif);

...works perfectly. Empty should just be evaluating the value passed in to it, not writing to it, so what's going wrong here? Tested in both PHP 5.2.10 and PHP 5.3.2.

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I've resolved the issue by using !count() instead of empty(), but I'm curious why it doesn't work in the first place. Is empty() trying to alter the result from array_diff?

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Check the manual on empty():

Note: empty() only checks variables as anything else will result in a parse error. In other words, the following will not work: empty(trim($name)).

empty(), like e.g. echo() and die(), is a language construct, and therefore has different rules than a normal function (in which your example would work fine).