IO in C standard and C POSIX - c

Question

I have problem when testing difference IO function between C standard and C POSIX.
I wrote two program that read a file character by character passed as a parameter, and display each character as soon as it is read. These programs will consist of three processes that share the same descriptor and act identically (the identity of each process is shown with the character read).
The first program will be written with the POSIX standard functions (open, read)
int main(int argc, char **argv){
int fd; /* Descriptor */
char *fic = argv[1]; /* Le nom du fichier */
int end = 0;
/* Sémaphore */
struct sembuf operation;
int sem_id;
sem_id = semget(ftok(fic, 'S'), 1, 0666|IPC_CREAT);
semctl(sem_id, 0, SETVAL, 1);
if((fd = open(fic, O_RDONLY, 0666))==-1){
perror("open file\n");
}
int i = 0;
char c;
for(i=0; i<3; i++){
if(fork()==0){
/* Lire */
while(end==0){
operation.sem_num = 0;
operation.sem_op = -1;
semop(sem_id, &operation, 1);
if(read(fd, &c, sizeof(char))>0){
printf("[Pid=%d]%c\n", getpid(), c);
}
else{
end = 1;
break;
}
operation.sem_num = 0;
operation.sem_op = 1;
semop(sem_id, &operation, 1);
sleep(1);
}
return EXIT_SUCCESS;
}
else continue;
}
for(i=0; i<3; i++){
wait(NULL);
}
close(fd);
semctl(sem_id, 0, IPC_RMID, NULL);
return EXIT_SUCCESS;
}
The second program will be written with the C library functions (fopen, fgetc)
int main(int argc, char **argv){
FILE *fd; /* Descriptor */
char *fic = argv[1];
int end = 0;
/* Sémaphore */
struct sembuf operation;
int sem_id;
sem_id = semget(ftok(fic, 'S'), 1, 0666|IPC_CREAT);
semctl(sem_id, 0, SETVAL, 1);
if((fd = fopen(fic, "r"))==NULL){
perror("open file\n");
}
int i = 0;
char c;
for(i=0; i<3; i++){
if(fork()==0){
while(end==0){
operation.sem_num = 0;
operation.sem_op = -1;
semop(sem_id, &operation, 1);
if((c = fgetc(fd))!=EOF){
printf("[Pid=%d]%c\n", getpid(), c);
}
else{
end = 1;
break;
}
operation.sem_num = 0;
operation.sem_op = 1;
semop(sem_id, &operation, 1);
sleep(1);
}
return EXIT_SUCCESS;
}
else continue;
}
for(i=0; i<3; i++){
wait(NULL);
}
fclose(fd);
semctl(sem_id, 0, IPC_RMID, NULL);
return EXIT_SUCCESS;
}
In case of C POSIX, I found that 3 processes works in parallel using locking mechanism of semaphore.
But in case of C standard, there is only one process that read and display character in the screen.
Could anyone tell me the reason of this difference?
Thanks so much.

Answer 1

If I understand correctly, you run your programs and see output like this:
$ echo foo > /tmp/input
$ ./synchronized-read /tmp/input # PIDs differ; the readers "share" the input
[Pid=124] f
[Pid=123] o
[Pid=125] o
[Pid=123] # <-- (newline)
$ ./synchronized-fgetc /tmp/input # PIDs are all the same; one reader reads all
[Pid=567] f
[Pid=567] o
[Pid=567] o
[Pid=567] # <-- (newline)
The reason is that read operates directly on files, whereas fgetc operates on buffered standard IO streams.
So, in the first program, the readers each read a byte at a time off of the file directly. In the second program, the first reader to obtain the semaphore reads in the entire file — you didn't show us sample input but this fits your problem description — meaning the other readers immediately encounter EOF and exit.
If you modify the child logic in each to printf("[Pid=%d]EOF\n", getpid()) at end of input, you'll clearly see it.