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Re: System call + signals = bad return code?

by papidave (Pilgrim)
on Sep 29, 2007 at 00:19 UTC ( #641646=note: print w/replies, xml ) Need Help??

in reply to System call + signals = bad return code?


The following note applies to unix-like system calls only. My mojo does not apply to windows.

As tye apparently explained to you, most SysV-style (and by exension, linux) system calls are interruptible. The wait(2) and waitpid(2) calls are examples.

Since your database connection to MySQL needs to use sockets to communicate with the database, you may receive signals (SIGPOLL is common with asynchronous I/O, for example). Likewise, if you have a second process in the background, it can send a SIGCLD when it exits. And timers send SIGALRM.

The fun part of this is that Unix only stacks one of each signal for calls to the handler -- so your signal handler might only get called once even when two processes die. This doesn't just apply to $SIG{FOO} handlers, it also applies to the implicit handler in wait() calls. I haven't (yet) walked the Perl source to satisfy my curiosity, but I expect that since it uses waitpid(), it's trapping that case reasonably well.

Some test cases I ran using system() and alarm() shows that the $! value sometimes (but not always) gets set to "interrupted system call" when this occurs. It depends on where you are in the system call when the signal arrives, I think. In any event, my general rule is to trust the value of ( $? >> 8 ), which gives the return status of the child process. YMMV, especially if you were to create more than one child process -- i don't know which one would end up in $?.

I have definitely seen the case where the signal arrives and the child process continues running to completion long after the return code is "returned." In C, I avoid the whole thing by calling popen() and reading until end-of-file. I don't know if you can apply that technique to wget in perl using open my $fh, '-|', $cmd or not.

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