Why is the exit code 255 instead of -1 in Perl?

Question

Why is it that when I shift the exit code, $?, in Perl by eight, I get 255 when I expect it to be -1?

Answer 1

Perl returns a subprocess exit code in the same manner as the C runtime library macro WEXITSTATUS, which has the following description in wait(2):

   WEXITSTATUS(status)
          evaluates to the least significant eight bits of the return code
          of the child which terminated, which may have been  set  as  the
          argument  to  a  call  to exit() or as the argument for a return
          statement in the main program.  This macro can only be evaluated
          if WIFEXITED returned non-zero.

The important part here is the least significant eight bits. This is why you are getting an exit code of 255. The perlvar man page describes $? as follows:

   $?      The status returned by the last pipe close, backtick (‘‘) com-
           mand, successful call to wait() or waitpid(), or from the sys-
           tem() operator.  This is just the 16-bit status word returned
           by the wait() system call (or else is made up to look like it).
           Thus, the exit value of the subprocess is really ("$? >> 8"),
           and "$? & 127" gives which signal, if any, the process died
           from, and "$? & 128" reports whether there was a core dump.

There is no special handling here for negative numbers in the exit code.

Answer 2

Which way are you shifting it? Please provide a code example.

also:

perldoc -f system

gives a very easy to understand example of what to do with $?

Also, http://www.gnu.org/s/libc/manual/html_node/Exit-Status.html

Exit values should be between 0 and 255. Your shifting combined with how negative values are actually stored by the computer should give some insight.

Answer 3

The exit status returned by 'wait()' is a 16-bit value. Of those 16 bits, the high-order 8 bits come from the low-order 8 bits of the value returned by 'exit()' — or the value returned from main(). If the program dies naturally, the low-order 8 bits of the 16 are all zero. If the program dies because of signal, the low-order 8 bits encode the signal number and a bit indicating whether a core dump happened. With a signal, the exit status is treated as zero — programs like the shell tend to interpret the low-order bits non-zero as a failure.

15      8 7      0   Bit Position
+-----------------+
|  exit  | signal |
+-----------------+

Most machines actually store the 16-bit value in a 32-bit integer, and that is handled with unsigned arithmetic. The higher-order 8 bits of the 16 may be all 1 if the process exited with 'exit(-1)', but that will appear as 255 when shifted right by 8 bits.

If you really want to convert the value to a signed quantity, you would have to do some bit-twiddling based on the 16th bit.

$status >>= 8;
($status & 0x80) ? -(0x100 - ($status & 0xFF)) : $status;

See also SO 774048 and SO 179565.