Handling SIGCHLD, how to record the return values of children as they die

问题

void childSignalHandler(int signo) {
    int status;

    pid_t pid = wait(&status);

    struct PIDList* record = getRecordForPID(childlist, pid);
    if (record != NULL)
        record->returnValue = status;
}

Quick question:

I want this handler to, when a child dies (this app spawns lots of children), get their return value and record it (last three lines). Will this do it, or am I getting all this API business wrong?

Thank you for your time!

(also, linux API terminology is creepy as hell, check for dying children and whatnot)

回答1:

This should do the job, if you set your function as a handler for SIGCHLD.

However, SIGCHLD can be send to parent process not only after the child exited. Some other events are signaled this way as well (for instance when the child is stopped). See man wait(3) for detailed explanation.

回答2:

Note signals are not queued. If two children die in quick succession, you may only recieve one SIGCHLD. So, you should actually loop around calling waitpid() until there are no more exiting processes to handle:

int status;
pid_t pid;

while ((pid = waitpid(-1, &status, WNOHANG)) > 0) {
    if (WIFEXITED(status)) {
        struct PIDList *record = getRecordForPID(childlist, pid);

        if (record != NULL)
            record->returnValue = WEXITSTATUS(status);
    }
}

来源：https://stackoverflow.com/questions/5530904/handling-sigchld-how-to-record-the-return-values-of-children-as-they-die