Bug #28845 [Com]: warning on two TRUE expressions while calling include*() in a if-statement

Edit report at https://bugs.php.net/bug.php?id=28845&edit=1 ID: 28845 Comment by: melbetethiopia659 at gmail dot com Reported by: david dot rech at virusmedia dot de Summary: warning on two TRUE expressions while calling include*() in a if-statement Status: Closed Type: Bug Package: *Directory/Filesystem functions Operating System: Windows XP SP1 PHP Version: 4CVS-2004-06-19 (stable) Block user comment: N Private report: N New Comment: Melbet is a prominent online sports betting business in Ethiopia and despite the fact that Melbet is a relatively young company, sports betting enthusiasts have welcomed it because of the company's various events and high-quality services. The Melbet betting company has been granted full permission to operate in Ethiopia. Ethiopia is a thriving sports betting nation, with football being the most popular sport among bettors. https://playloudershop.com/ Previous Comments: ------------------------------------------------------------------------ [2004-10-06 09:02:44] david dot rech at virusmedia dot de Somehow fixed in 4.3.9, and I don't have a clue why. This was also reproduced by another developer. Did just nobody close the bug or anything? Dunno. Keep up the good work. ------------------------------------------------------------------------ [2004-06-19 20:04:43] david dot rech at virusmedia dot de Sorry, missed the closing brace in second if-statement... Must be: "Whereas that code, with the first expression in braces, prints "1" again. if( (include_once(__FILE__)) && is_integer(1) ) echo 1;" ------------------------------------------------------------------------ [2004-06-19 20:02:25] david dot rech at virusmedia dot de Description: ------------ I've notice some strange behaviour with include() and include_once() calls in an if-statement whereas two expressions are TRUE and logically combined with the '&&' operator. While if( true && true ) echo 1; would print out "1" as expected, this code however does not: if( include_once(__FILE__) && is_integer(1) ) echo 1; Whereas that code, with the first expression in braces, prints "1" again. if( (include_once(__FILE__) && is_integer(1) ) echo 1; In the example you should notice that the expression is still TRUE with other functions - even without braces around first expression. Also, somehow the result of the second expression gets copied to the first argument of the first expression as an integer. Look at this warning: Warning: main(1): failed to open stream: No such file or directory in [...] on line 12 That's what happens if second expression is true and include_once() was'nt put into braces. Try to negate the second expression - you will get a '0' referred as the argument of include_once() in the warning. So is this some strange magic, or did I've miss the point completely? Reproduce code: --------------- <?php // Expected behaviour if( true && true ) { // As expected - the expression is TRUE echo "1st: That may have worked\n"; } // Weird behaviour if( class_exists('stdClass') && is_integer(1) ) { // Again - the expression is TRUE echo "2nd: You don't see that, don't you?\n"; } if( @include_once(__FILE__) && is_integer(1) ) { // Expression seems to be FALSE and causes a weird warning without the @ echo "3rd: So you see this..?"; } if( (include_once(__FILE__)) && is_integer(1) ) { // TRUE echo "4rd: In fact, you see this..."; } ?> Expected result: ---------------- 1st: That may have worked 2nd: You don't see that, don't you? 3rd: So you see this..? 4rd: In fact, you see this... Actual result: -------------- 1st: That may have worked 2nd: You don't see that, don't you? 4rd: In fact, you see this... ------------------------------------------------------------------------ -- Edit this bug report at https://bugs.php.net/bug.php?id=28845&edit=1