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Home/ Questions/Q 8859551
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Editorial Team
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Editorial Team
Asked: June 14, 20262026-06-14T15:05:35+00:00 2026-06-14T15:05:35+00:00

I have the following source code, which I am sending signals from parent to

  • 0

I have the following source code, which I am sending signals from parent to child:

sigset_t sig_m1, sig_m2, sig_null;
int signal_flag=0;

void start_signalset();

void sig_func(int signr) {
  printf("%d\n", signr, n);

  start_signalset();
}

void start_signalset() {
  if(signal(SIGUSR2, sig_func) == SIG_ERR) {
    exit(0);
  }

 if(signal(SIGUSR1, sig_func) == SIG_ERR) {
    exit(0);
  }
}

void wait_for_parents() {
    while(signal_flag == 0) {
        sigsuspend(&sig_null);
    }
}


int main(){
    int result,pt_pid;
    start_signalset();
    pt_pid=getpid(); 

    result = fork();
    if(result==-1){ 

        printf("Can't fork child\n");
        exit(-1);
    } else if (result == 0) {

            wait_for_parents();

    } else {
       kill(result,SIGUSR2);
       kill(result,SIGUSR2);
       kill(result,SIGUSR1);
       kill(result,SIGUSR2);

        signal_flag = 1;
    }    
    return 0; 
}

I see: 31, 31, 31, 30, but I was expecting to see 31, 31, 30, 31. Where I have an error? I think there is some problem with synchronization. However, I do not understand how to fix it, and I’m not sure that the problem exists.

Regards, Denis.

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-14T15:05:36+00:00Added an answer on June 14, 2026 at 3:05 pm

    You are sending signals so quickly that the signal handler processing SIGUSR1 gets interrupted by the next signal and processes that first before it returns.

    • signal 2 arrives, prints
    • signal 2 arrives, prints
    • signal 1 arrives, gets interrupted before it can print because another signal 2 arrives
    • signal 2 arrives, prints
    • after last signal handler returned signal handler for SIGUSR1 continues execution and prints.

    Either send your parent some sort or acknowledgement, make your parent sleep a while or make your signal handler return faster (store signal arrival in a buffer. Invoking printf is quite a long call.

    When signals for a process arrive a random thread of that process gets interrupted and begins executing the signal handler. If the signal handler returns the thread will continue execution as usual. Signal handlers can also be interrupted!

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