How many processes will be created - c

Question

I have this piece of code and was wondering how many processes will be created. I am uncertain as I think because of the loop it will be 12 processes but also could be 8.
#include <unistd.h>
#include <sys/types.h>
int main() {
pid_t childpid;
int i;
childpid = fork();
for (i = 0; i < 3 && childpid == 0; i++) {
if (childpid == -1) {
perror("Failed to fork.");
return 1;
}
fprintf(stderr, "A\n");
childpid = fork();
if (childpid == 0) {
fprintf(stderr, "B\n");
childpid = fork();
fprintf(stderr, "C\n");
}
}
return 0;
}

Answer 1

Yes, you can use aptitude to evaluate how many processes are created, but I let the program decide it for me.
The child and parent are synchronized using semaphore, and a pcount variable is used to keep track of the number of created processes. pcount is incremented whenever childpid is evaluated to zero. Below are the additions to your program.
#include <semaphore.h>
#include <stdio.h>
#include <errno.h>
#include <stdlib.h>
#include <unistd.h>
#include <sys/mman.h>
int main(void)
{
/* Initialize and setup a semaphore */
sem_t* sema = mmap(NULL, sizeof(sem_t), PROT_READ | PROT_WRITE, MAP_SHARED | MAP_ANONYMOUS, -1, 0);
if (sema == MAP_FAILED)
exit(1);
if (sem_init(sema, 1, 1) != 0)
exit(1);
/* Initialize pcount */
int* pcount = mmap(NULL, sizeof(int), PROT_READ | PROT_WRITE, MAP_SHARED | MAP_ANONYMOUS, -1, 0);
if (pcount == MAP_FAILED)
exit(1);
*pcount = 1;
printf("pcount = %d\n", *pcount);
pid_t childpid;
int i;
childpid = fork();
if (childpid == 0) {
sem_wait(sema);
*pcount = *pcount + 1;
printf("pcount = %d\n", *pcount);
sem_post(sema);
}
for (i = 0; i < 3 && childpid == 0; i++) {
if (childpid == -1) {
perror("Failed to fork.");
return 1;
}
fprintf(stderr, "A\n");
childpid = fork();
if (childpid == 0) {
sem_wait(sema);
*pcount = *pcount + 1;
printf("pcount = %d\n", *pcount);
sem_post(sema);
fprintf(stderr, "B\n");
childpid = fork();
if (childpid == 0) {
sem_wait(sema);
*pcount = *pcount + 1;
printf("pcount = %d\n", *pcount);
sem_post(sema);
}
fprintf(stderr, "C\n");
}
}
return 0;
}
Terminal Session:
$ gcc SO.c -lpthread
$ ./a.out
pcount = 1
pcount = 2
A
pcount = 3
B
C
pcount = 4
C
A
pcount = 5
B
C
pcount = 6
C
A
pcount = 7
B
C
pcount = 8
C
So yes, 8 is the answer. Easy-Peasy :)

Answer 2

8 processes will be created.
Here is how it looks like after the first loop: