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fork() -> defunct

Posted on 1998-02-17
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With a fork command I can create a child process.
But when the child process "finish", it's not realy terminated, why it go in state defunct.
How can I realy terminate a child process without this state?

The problems still in the number of childs defunct that can I have. I maked a par test and the number is a number terminated. I my situation I need create some child that make some jobs in "background", but with this situation I reach at one point that I can't create more child (about 8000 time).
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Question by:pedimina
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dhm earned 400 total points
ID: 1296018
When a process fork()s off a child and the child exits, it goes into defunct (also known as "zombie") state until the the parent retrieves the child's exit code with the wait(2) system call.  (There are several flavors of wait() which vary in the amount of information they return and in the way you specify which child process you're interested in.)  If the programmer has forgotten to wait() for child processes he starts, you'll see these defunct process start to accumulate.  They get cleaned up when the parent finishes because the defunct children (all the parent's children, actually) are adopted by process 1, the init process.  One of init's jobs is to wait() for any processes that become its children.

So, in direct answer to your question, if you execute this section of code in your parent process, you'll collect your children's exit codes and clean up the defunct processes:

pid_t       child;
int       exit_status;

while ((child = waitpid( (pid_t)-1, &exit_status, WNOHANG )) != (pid_t)-1) {
      printf( "child %d exited with status %d\n",
            child, WEXITSTATUS(exit_status) );
}

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