[EMAIL PROTECTED] wrote:
ok so $? states CHILD_ERROR or last status returned by the last ` ....`
command.
$! states yields the current value of errno
in shell if I say it will give me a true or false value.
cat /tmp/foo
if [ $? -eq 0 ]
then
echo yes command succeeded
else
echo no.
fi
In the shell any programs that you run are child processes so 'cat /tmp/foo'
is run as a child of the shell and sets $? when it ends.
In Perl if I wanted to represent this I would use $! for any notification
of any failures such as opening a file.
That is because open() is built in to perl.
But when I add the code in red
specifically $? >> 8, I get a exit value of 2 and when I change it to $?
127 I get a exit value of 0. I want to understand and know the code to
get a true (1) or false (0) value.
Programs that run on the command line return 0 on success or some non-zero
value on failure. Some programs return different non-zero values depending on
the type of failure but the C language defines two standard return values in
the stdlib.h file: EXIT_SUCCESS and EXIT_FAILURE.
perldoc -f system
[snip]
You can check all the failure possibilities by inspecting $? like
this:
if ($? == -1) {
print "failed to execute: $!\n";
}
elsif ($? & 127) {
printf "child died with signal %d, %s coredump\n",
($? & 127), ($? & 128) ? 'with' : 'without';
}
else {
printf "child exited with value %d\n", $? >> 8;
}
or more portably by using the W*() calls of the POSIX extension; see
perlport for more information.
John
--
use Perl;
program
fulfillment
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