Following script (argv.php3):

#!/usr/bin/php -f
<?
printf("argc: %d\nargv1-2:%s-%s\n",$argc,$argv[0],$argv[1]);
?>

when run as ./argv.php3 (without parameters), says:
argc: 1
argv1-2:-

when run as ./argv.php3 abc (1 parameter), says:

argc: 1
argv1-2:abc-

Shouldn't argc be different in two cases? When I do not put -f, script name is included in argv and argc is OK. 

Same effects if I run php -f argv.php3 (like the first case) and php argv.php3 (like the second case).