Doc #53674 [Bgs]: Example 1 does not work, contradicts Description

Edit report at http://bugs.php.net/bug.php?id=53674&edit=1 ID: 53674 User updated by: thomas dot edwards at gmail dot com Reported by: thomas dot edwards at gmail dot com Summary: Example 1 does not work, contradicts Description Status: Bogus Type: Documentation Problem Package: Calendar related Operating System: Mac OS X PHP Version: 5.3.5 Block user comment: N Private report: N New Comment: In the description it says that the first variable ($calendar) is an integer, but in Example 1 it’s clearly a string. And when you run the code, neither a string or integer seems to work. Previous Comments: ------------------------------------------------------------------------ [2011-01-06 17:32:10] danbrown@php.net Since when is 31 not an integer? ------------------------------------------------------------------------ [2011-01-06 17:27:22] thomas dot edwards at gmail dot com Description: ------------ --- From manual page: http://www.php.net/function.cal-days-in-month#Examples --- Description: int cal_days_in_month ( int $calendar , int $month , int $year ) Example 1: <?php $num = cal_days_in_month(CAL_GREGORIAN, 8, 2003); // 31 echo "There was $num days in August 2003"; ?> Clearly not a integer! Does not work! Test script: --------------- $num = cal_days_in_month(CAL_GREGORIAN, 8, 2003); // 31 echo "There was $num days in August 2003"; ------------------------------------------------------------------------ -- Edit this bug report at http://bugs.php.net/bug.php?id=53674&edit=1