On 2004-10-26, Peter Simons <[EMAIL PROTECTED]> wrote: > What will happen if I call getProcessExitCode for the same > process twice? Will that block? Cause an error? Or return > the same child's exit code again?
Assuming it is based on wait() or one of its derivatives, and I suspect it is, you cannot call it more than once for a single process. Once you have wait()ed for a process, the data about its termination is no longer tracked by the kernel, so you can no longer retrieve it. A Zomebie process is precisely one that is dead but has not yet been waited upon. They exist so that they can be waited upon. It should be pretty easy to install a SIGCHLD handler that reaps children whenever one dies. Such code is found all over the place in C. (And Python, and Perl, and... well everywhere except Java, probably.) -- John Goerzen Author, Foundations of Python Network Programming http://www.amazon.com/exec/obidos/tg/detail/-/1590593715 _______________________________________________ Glasgow-haskell-users mailing list [EMAIL PROTECTED] http://www.haskell.org/mailman/listinfo/glasgow-haskell-users
