> The exception occurs when you have a $SIG{CHLD} handler. That makes the system think you've handled it already.

Not true. The program below forks, the child exits immediately. You will see that the signal handler is fired, but the child process stays around (as a zombie) until it's reaped by the waitpid call. 

$SIG{CHLD} = sub {print("SIGCHLD\n")};

exit unless $pid = fork;
sleep 1;
system "ps -p $pid";
waitpid $pid, 0;
system "ps -p $pid";
__END__
SIGCHLD
  PID TTY          TIME CMD
29260 pts/32   00:00:00 perl <defunct>
SIGCHLD
  PID TTY          TIME CMD
SIGCHLD

[download]
The signal handler fires three times in total: one for the child process, and once each for the calls to system.

It's only when you explicitely set $SIG{CHLD} to IGNORE that child processes stay behind.

It's noteworthy to remember that POSIX does not specify what should happen if $SIG{CHLD} is ignored. (Which makes it a joy to program platform independently on Unix)

In reply to Re^2: Possible waitpid problem - pid reuse? by Anonymous Monk
in thread Possible waitpid problem - pid reuse? by tsteiger

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