Bug #17323 Updated: posix_isatty wants a long instead of resource

ID: 17323 Updated by: mfischer@php.net Reported By: polone@townnews.com -Status: Open +Status: Critical Bug Type: POSIX related Operating System: Linux PHP Version: 4.2.0 Assigned To: mfischer New Comment: Critical before 4.3 release. Best thing is probably to use the old-style code, if someone has a better idea, speak up. Previous Comments: ------------------------------------------------------------------------ [2002-05-22 03:54:11] mfischer@php.net Reopening, problem not completely fixed, assigning to me. ------------------------------------------------------------------------ [2002-05-20 19:31:52] mfischer@php.net This bug has been fixed in CVS. You can grab a snapshot of the CVS version at http://snaps.php.net/. In case this was a documentation problem, the fix will show up soon at http://www.php.net/manual/. In case this was a PHP.net website problem, the change will show up on the PHP.net site and on the mirror sites. Thank you for the report, and for helping us make PHP better. Will also be in 4.2.2 (if there will be a release). ------------------------------------------------------------------------ [2002-05-20 19:08:47] polone@townnews.com The bug seems rather simple. Using command-line PHP, the following fragment should work: #!/usr/bin/php -q <?php $fd = fopen('php://stdout','w'); if (posix_isatty($fd)) print "yes, it is a terminal\n"; else print "no, it is not a terminal\n"; ?> However, it appears that the function definition requires a long passed, instead of a resource. The error message reported is: dns:root-/usr/bin> ./test.php PHP Warning: posix_isatty() expects parameter 1 to be long, resource given in /usr/bin/test.php on line 4 <br /> <b>Warning</b>: posix_isatty() expects parameter 1 to be long, resource given in <b>/usr/bin/test.php</b> on line <b>4</b ><br /> no, it is not a terminal A workaround appears to be the following: #!/usr/bin/php -q <?php $fd = fopen('php://stdout','w'); settype($fd,'int'); if (posix_isatty($fd)) print "yes, it is a terminal\n"; else print "no, it is not a terminal\n"; ?> The explicit type-cast fixes the problem. Maybe I have settings in the php.ini file wrong, but I don't think this function should behave like this. Perhaps it requires just a change to the source. This is also broke with posix_ttyname(). Am I suppose to acquire a file descriptor another way? Regards, Patrick O'Lone ------------------------------------------------------------------------ -- Edit this bug report at http://bugs.php.net/?id=17323&edit=1