Why are my semaphores not working as expected? - c

Question

I need to print the following sentence to the console "My name is Bond, James Bond" alternating the words using semaphores.
Whenever I print the words using line breaks "\n", everything prints in the expected order, however if I don't use line breaks everything prints out of order.
This is my code:
#include <stdio.h>
#include <stdlib.h>
#include <sys/mman.h>
#include <sys/stat.h>
#include <sys/types.h>
#include <fcntl.h>
#include <wait.h>
#include <unistd.h>
#include <semaphore.h>
#include <string.h>
int main(){
pid_t pid;
int i = 0;
sem_t *sem[4];
char * names[4] = {"/sem_ex07_1", "/sem_ex07_2", "/sem_ex07_3", "/sem_ex07_4"};
for(i = 0; i < 4; i++){
sem_unlink(names[i]);
}
for(i = 0; i < 4; i++){
if((sem[i] = sem_open(names[i], O_CREAT | O_EXCL, 0644, 0)) == SEM_FAILED){
printf("sem_open() error\n");
sem_unlink(names[i]);
exit(1);
}
}
sem_post(sem[0]);
for(i = 0; i < 3; i++){
pid = fork();
if(pid == 0){
break;
}
}
if(pid == 0){
if(i == 0){
sem_wait(sem[0]);
printf("My");
sem_post(sem[1]);
sem_wait(sem[0]);
printf("Bond, ");
sem_post(sem[1]);
} else if(i == 1){
sem_wait(sem[1]);
printf("name ");
sem_post(sem[2]);
sem_wait(sem[1]);
printf("James ");
sem_post(sem[2]);
} else if(i == 2){
sem_wait(sem[2]);
printf("is ");
sem_post(sem[0]);
sem_wait(sem[2]);
printf("Bond.\n");
sem_post(sem[3]);
}
} else if(pid > 0){
sem_wait(sem[3]);
for(i = 0; i < 4; i++){
sem_unlink(names[i]);
}
}
return 0;
}
Why is my output only correct if I add a line break on each printf?

Answer 1

Why is my output only correct if I add a line break on each printf?
printf() is mostly by default line buffered. stdio functions which use streams (instead of file descriptors like write() does) use a buffer.
The buffer can be set via setvbuf(3).
See also Disable buffering for stdin and stdout using setvbuf()