alarm() in child process

[adamdebek]

alarm() creates thread which wll send a signal, but fork() in multi-threaded process copy only thread from which was invoked, so alarm thread will not be present in that new process causing no arrival of alarm signal.

Reproduciton (ia32-generic-qemu):

#include <stdio.h>
#include <signal.h>
#include <unistd.h>
#include <stdlib.h>
#include <sys/wait.h>

volatile int n;

void handler(int signum) {
        printf("n: %d\n", ++n);
}

int main(void) {
    int i = 1;

    signal(SIGALRM, handler);

    alarm(5);
    while (i <= 8) {
        printf("tick %d\n", i++);
        sleep(1);
    }

    pid_t pid;
    pid = fork();
    if (pid < 0) {
        perror("fork");
        exit(EXIT_FAILURE);
    }
    else if (pid == 0) {
        alarm(5);
        i = 1;
        while (i <= 8) {
            printf("tick %d\n", i++);
            sleep(1);
        }

        exit(2);
    }

    int status;
    wait(&status);
    
    printf("status: %d\n", WEXITSTATUS(status));

    return 0;
}

In above reproduction handler is invoked once, but should be twice.

Furtermore as POSIX states:
If there is a previous alarm() request with time remaining, alarm() shall return a non-zero value that is the number of seconds until the previous request would have generated a SIGALRM signal. Otherwise, alarm() shall return 0.

Reproduction (ia32-generic-qemu):

#include <stdio.h>
#include <unistd.h>


int main(void) {
    unsigned int ret;
    alarm(1);
    ret = alarm(0);

    printf("ret: %u\n", ret);

    return 0;
}

The return value should be 1.