I have never set $! in a program to set the exit status

Of course you don't set $! to set the exit status!! You set $! to zero before a call to system(), in order to clear it. If system returns no error, but $! is set afterwards, you know that there must have been an error indeed. If you would not clear $! before the call to system(), you would not know afterwards whether there was an error in system(), or in some earlier function. For example, in the two following cases (on my platform), system returns the same value in both cases, and $! is also set in both cases, although only the first system() call actually failed:

# On Windows 2000, ActiveState Perl 5.8.8 or 5.10

# case one:
opendir(FOO,'.');
system('this_program_does_not_exist');
# $? == 256, $! != 0

# case two:
opendir(BAR,'this_directory_does_not_exist');
system('perl -e "exit 1"');
# $? == 256, $! != 0
[download]

But if you explicitly reset $!, you can distinguish between those cases:

# On Windows 2000, ActiveState Perl 5.8.8 or 5.10

# case one:
opendir(FOO,'.');
$!=0;
system('this_program_does_not_exist');
# $? == 256, $! != 0

# case two:
opendir(BAR,'this_directory_does_not_exist');
$!=0;
system('perl -e "exit 1"');
# $? == 256, $! == 0
[download]