C Program to lock file using semaphores.

Write a C Program to lock file using semaphores.
Using semaphores, We can control access to files, shared memory and other things. The basic functionality of a semaphore is that you can either set it, check it, or wait until it clears then set it (“test-n-set”). In C semaphores functions defined in the sys/sem library. Read more about C Programming Language . and read the C Programming Language (2nd Edition) by K and R.

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#include <stdio.h>
#include <sys/file.h>
#include <error.h>
#include <sys/sem.h>
#define MAXBUF 100
#define KEY 1216
#define SEQFILE "seq_file"
int semid,fd;
void my_lock(int);
void my_unlock(int);
union semnum
{
int val;
struct semid_ds *buf;
short *array;
}arg;
int main()
{
int child, i,n, pid, seqno;
char buff[MAXBUF+1];
pid=getpid();
if((semid=semget(KEY, 1, IPC_CREAT | 0666))= = -1)
{
perror("semget");
exit(1);
}
arg.val=1;
if(semctl(semid,0,SETVAL,arg)<0)
perror("semctl");
if((fd=open(SEQFILE,2))<0)
{
perror("open");
exit(1);
}
pid=getpid();
for(i=0;i<2;i++)
{
my_lock(fd);
lseek(fd,01,0);
if((n=read(fd,buff,MAXBUF))<0)
{
perror("read");
exit(1);
}
printf("pid:%d, Seq no:%dn", pid, seqno);
seqno++;
sprintf(buff,"%dn", seqno);
n=strlen(buff);
lseek(fd,01,0);
if(write(fd,buff,n)!=n)
{
perror("write");
exit(1);
}
sleep(1);
my_unlock(fd);
}
}
void my_lock(int fd)
{
struct sembuff sbuf=(0, -1, 0);
if(semop(semid, &sbuf, 1)= =0)
printf("Locking: Resource…n");
else
printf("Error in Lockn");
}
void my_unlock(int fd)
{
struct sembuff sbuf=(0, 1, 0);
if(semop(semid, &sbuf, 1)= =0)
printf("UnLocking: Resource…n");
else
printf("Error in Unlockn");
}
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C Program to solve the producer consumer problem

Write a C Program to solve the producer consumer problem with two processes using semaphores.
Producer-consumer problem is the standard example of multiple process synchronization problem. The problem occurs when concurrently producer and consumer tries to fill the data and pick the data when it is full or empty. producer consumer problem is also known as bounded-buffer problem. In this program We use the semaphores, to solve the problem.Read more about C Programming Language . and read the C Programming Language (2nd Edition) by K and R.

/***********************************************************
* You can use all the programs on www.c-program-example.com
* for personal and learning purposes. For permissions to use the
* programs for commercial purposes,
* contact info@c-program-example.com
* To find more C programs, do visit www.c-program-example.com
* and browse!
*
* Happy Coding
***********************************************************/

#include <stdio.h>
#include <stdlib.h>
#include <unistd.h>
#include <time.h>
#include <sys/types.h>
#include <sys/ipc.h>
#include <sys/sem.h>


#define NUM_LOOPS 20
int main(int argc, char* argv[])
{
int sem_set_id;
union semun sem_val;
int child_pid;
int i;
struct sembuf sem_op;
int rc;
struct timespec delay;


sem_set_id = semget(IPC_PRIVATE, 1, 0600);
if (sem_set_id == -1) {
perror("main: semget");
exit(1);
}
printf("Semaphore set created,
semaphore set id '%d'.n", sem_set_id);


sem_val.val = 0;
rc = semctl(sem_set_id, 0, SETVAL, sem_val);
child_pid = fork();
switch (child_pid) {
case -1:
perror("fork");
exit(1);
case 0:
for (i=0; i<NUM_LOOPS; i++) {
sem_op.sem_num = 0;
sem_op.sem_op = -1;
sem_op.sem_flg = 0;
semop(sem_set_id, &sem_op, 1);
printf("consumer: '%d'n", i);
fflush(stdout);
sleep(3);
}
break;
default:
for (i=0; i<NUM_LOOPS; i++)
{
printf("producer: '%d'n", i);
fflush(stdout);
sem_op.sem_num = 0;
sem_op.sem_op = 1;
sem_op.sem_flg = 0;
semop(sem_set_id, &sem_op, 1);
sleep(2);
if (rand() > 3*(RAND_MAX/4))
{
delay.tv_sec = 0;
delay.tv_nsec = 10;
nanosleep(&delay, NULL);
}
}
break;
}


return 0;
}
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