PHP 4.0 Bug #7591: continue doesn't work inside switch statement
| From: | joel at azursoft dot fr | Date: | Thu, 02 Nov 2000 14:57:15 +0000 |
| Subject: | PHP 4.0 Bug #7591: continue doesn't work inside switch statement | ||
| Groups: | php.dev | ||
| Request: | Send a blank email to php-dev+get-36788@lists.php.net to get a copy of this message | ||
From: joel@azursoft.fr
Operating system: linux glibc 2.1
PHP version: 4.0.3pl1
PHP Bug Type: Scripting Engine problem
Bug description: continue doesn't work inside switch statement
When placed inside a switch statement located inside a for
loop, the continue doesn't work as usual in C:
the following code display the following output:
<?
for ( $i=0; $i<10; $i++ ) {
switch ( $i ) {
case 5: continue;
default:
echo "$i<br>";
}
echo "Should be printed only if i != 5 and now i == $i<br>";
}
?>
//output:
0
Should be printed only if i != 5 and now i == 0
1
Should be printed only if i != 5 and now i == 1
2
Should be printed only if i != 5 and now i == 2
3
Should be printed only if i != 5 and now i == 3
4
Should be printed only if i != 5 and now i == 4
Should be printed only if i != 5 and now i == 5
6
Should be printed only if i != 5 and now i == 6
7
Should be printed only if i != 5 and now i == 7
8
Should be printed only if i != 5 and now i == 8
9
Should be printed only if i != 5 and now i == 9
// End of output
The line :
Should be printed only if i != 5 and now i == 5
should not appear
--
Edit Bug report at: http://bugs.php.net/?id=7591&edit=1