My semaphore module is not working properly(Dining philosopher) - linux

I'm implementing a semaphore methods to understand synchronization and thread things.
By using my semaphore, I tried to solve the Dining Philosophers problem.
My plan was making deadlock situation first.
But I found that just only one philosopher eat repeatedly.
And I checked that my semaphore is working quite good by using other synchronization problems. I think there is some problem with grammar.
please let me know what is the problem.
Here is my code.
dinig.c (including main function)
#include "sem.h"
#include <stdio.h>
#include <stdlib.h>
#include <unistd.h>
static tsem_t *chopstick[5];
static tsem_t *updating;
static int update_status (int i, int eating)
{
static int status[5] = { 0, };
static int duplicated;
int idx;
int sum;
tsem_wait (updating);
status[i] = eating;
/* Check invalid state. */
duplicated = 0;
sum = 0;
for (idx = 0; idx < 5; idx++)
{
sum += status[idx];
if (status[idx] && status[(idx + 1) % 5])
duplicated++;
}
/* Avoid printing empty table. */
if (sum == 0)
{
tsem_signal (updating);
return 0;
}
for (idx = 0; idx < 5; idx++)
fprintf (stdout, "%3s ", status[idx] ? "EAT" : "...");
/* Stop on invalid state. */
if (sum > 2 || duplicated > 0)
{
fprintf (stdout, "invalid %d (duplicated:%d)!\n", sum, duplicated);
exit (1);
}
else
fprintf (stdout, "\n");
tsem_signal (updating);
return 0;
}
void *thread_func (void *arg)
{
int i = (int) (long) arg;
int k = (i + 1) % 5;
do
{
tsem_wait (chopstick[i]);
tsem_wait (chopstick[k]);
update_status (i, 1);
update_status (i, 0);
tsem_signal (chopstick[i]);
tsem_signal (chopstick[k]);
}
while (1);
return NULL;
}
int main (int argc,
char **argv)
{
int i;
for (i = 0; i < 5; i++)
chopstick[i] = tsem_new (1);
updating = tsem_new (1);
for (i = 0; i < 5; i++)
{
pthread_t tid;
pthread_create (&tid, NULL, thread_func, (void *) (long) i);
}
/* endless thinking and eating... */
while (1)
usleep (10000000);
return 0;
}
sem.c(including semaphore methods)
#include "sem.h"
.
sem.h(Header for sem.c)
#ifndef __SEM_H__
#define __SEM_H__
#include <pthread.h>
typedef struct test_semaphore tsem_t;
tsem_t *tsem_new (int value);
void tsem_free (tsem_t *sem);
void tsem_wait (tsem_t *sem);
int tsem_try_wait (tsem_t *sem);
void tsem_signal (tsem_t *sem);
#endif /* __SEM_H__ */
compile command
gcc sem.c dining.c -pthread -o dining

One problem is that in tsem_wait() you have the following code sequence outside of a lock:
while(sem->count <= 0)
continue;
There's no guarantee that the program will actually re-read sem->count - the compiler is free to produce machine code that does something like the following:
int temp = sem->count;
while(temp <= 0)
continue;
In fact, this will likely happen in an optimized build.
Try changing your busy wait loop to something like this so the count is checked while holding the lock:
void tsem_wait (tsem_t *sem)
{
pthread_mutex_lock(&(sem->mutexLock));
while (sem->count <= 0) {
pthread_mutex_unlock(&(sem->mutexLock));
usleep(1);
pthread_mutex_lock(&(sem->mutexLock));
}
// sem->mutexLock is still held here...
sem->count--;
pthread_mutex_unlock(&(sem->mutexLock));
}
Strictly speaking, you should do something similar for tsem_try_wait() (which you're not using yet).
Note that you might want to consider using a pthread_cond_t to make waiting on the counter changing more efficient.
Finally, your code to 'get' the chopsticks in thread_func() has the classic Dining Philosopher deadlock problem in the situation where each philosopher simultaneously acquires the 'left' chopstick (chopstick[i]) and ends up waiting forever to get the 'right' chopstick (chopstick[k]) since all the chopsticks are in some philosopher's left hand.

Related

Pass a returned value from a thread to another thread

I'm learning posix thread programming and I'm struggling in the following program. So I've 2 threads with the ids: tid1 and tid2. I use the thread with tid1 to enter the values I want to sum each other into the second thread with tid2 but the result of the sum is not correct. So I've tried to display in tid2 the values I entered in tid1 but I realized that these are not correct too (didn't correspond to the entered values). Can you pleas tell me where I'm doing the error? The way I pass the pointer returned from a thread with tid1 to thread with tid2, is that correct?
Note: The values returned from tid1 are already correct because I checked it as you can also see in the main (line 56).
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <pthread.h>
void *entryValues(void *p)
{
int nbVal = *(int*)p;
int i = 0;
int val = 0;
int *tab;
while(i < nbVal){
printf("value %d: ", i);
scanf("%d", &val);
tab[i] = val;
i++;
}
pthread_exit(tab);
}
void *sumValues(void *p)
{
int *values;
int sum = 0, i = 0;
memcpy(values, (int*)p, sizeof((int*)p));
printf("%d\n", values[0]);
for (i = 0; i < sizeof(values); i++)
{
//printf("value %d: %d\n", i, values[i]);
sum += values[i];
}
printf("The sum of the entered values is: %d\n", sum);
return NULL;
}
int main(int argc, char **argv)
{
pthread_t tid1;
pthread_t tid2;
int nbValues = 0;
int *arr;
printf("Please enter the number of values you want to sum each other: ");
scanf("%d", &nbValues);
if(pthread_create(&tid1, NULL, entryValues, &nbValues)!= 0)
{
fprintf(stderr, "creation of thread failed\n");
return -1;
}
pthread_join(tid1, (void**)&arr);
//printf("array: %d\n", arr[0]);
if(pthread_create(&tid2, NULL, sumValues, arr) != 0)
{
fprintf(stderr, "creation of thread failed\n");
return -1;
}
pthread_join(tid2, NULL);
/* Waiting for the termination of tid */
printf("End of main thread\n");
return EXIT_SUCCESS;
}

Named Pipe, Communication between 2 children

I have a problem with my code. I want to make communication between 2 children process. One of them is a server, which opens a file and sends each letter to the second process. The second process is counting letters and it should make a new file and save results. I have problems with the last step because the first process gonna finish faster than the second, what causes the end of the program. I have no idea how fix it. Looking for some tips :).
Here you got result.
My code:
#include <stdio.h>
#include <stdlib.h>
#include <unistd.h>
#include <sys/types.h>
#include <sys/stat.h>
#include <errno.h>
#include <fcntl.h>
#include <signal.h>
#include <string.h>
//stale
#define FIFO "my_fifo"
#define SIZE 26
//zmienne globalne
int desk; //deskryptor pliku
int tab[SIZE];
//prototypy funkcji
void parentKillAll();
void server(FILE * file);
void client();
void cleanUp(FILE * file);
int checkEntryData(int argc, char *argv);
void replaceTabWithZero(int * tab);
void countLetters(int * tab, char ch);
void saveResults(int * tab, char *title);
void showTab(int * tab);
int main(int argc, char *argv[]) {
if (!checkEntryData(argc, argv[1]))
return 1;
replaceTabWithZero(tab);
FILE *file = fopen(argv[1], "r");
umask(0);
mkfifo(FIFO, 0666);
if (file) {
if (fork() == 0) {
server(file);
exit(0);
} else if (fork() == 0) {
client();
saveResults(tab, strcat(argv[1], "Result"));
showTab(tab);
exit(0);
} else {
cleanUp(file);
parentKillAll();
}
} else {
perror("Error");
}
return 0;
}
void parentKillAll() {
sleep(1);
kill(0, SIGKILL);
exit(0);
}
void server(FILE * file) {
char ch;
while ((ch = fgetc(file)) != EOF) {
desk = open(FIFO, O_WRONLY);
write(desk, &ch, 1);
}
}
void client() {
char ch;
while (1) {
desk = open(FIFO, O_RDONLY);
read(desk, &ch, 1);
countLetters(tab, ch);
printf("%c", ch);
}
}
void cleanUp(FILE *file) {
wait(0);
fclose(file);
close(desk);
}
int checkEntryData(int argc, char *argv) {
if (argc < 2) {
fprintf(stderr, "Nie poprawna ilosc argumentow\n");
return 0;
}
if (access(argv, F_OK)) {
fprintf(stderr, "Podany plik \'%s\' nie istnieje\n", argv);
return 0;
}
if (access(argv, R_OK)) {
fprintf(stderr, "Brak uprawnien do odczytu pliku \'%s\'\n", argv);
return 0;
}
return 1;
}
void replaceTabWithZero(int * tab) {
for (int i = 0; i < SIZE; i++)
tab[i] = 0;
}
void countLetters(int *tab, char ch) {
int chVal = ch;
if (chVal > 92)
chVal -= 32;
if (chVal > 64 && chVal < 91)
tab[chVal-65] += 1;
}
void saveResults(int *tab, char * title) {
FILE *plik = fopen(title, "w");
if (plik) {
for (int i = 0; i < SIZE; i++)
fprintf(plik, "%c - %d\n", (i+97), tab[i]);
} else {
perror("Error");
}
fclose(plik);
}
void showTab(int * tab) {
for (int i = 0; i < SIZE; i++)
printf("\n%d", tab[i]);
}
The real problem is that the client process can never finish, because it runs an infinite while(1) loop without any exit conditions.
You should rewrite it so that it exits after reading all available data:
void client() {
char ch;
// Open the fifo only once, instead of once per character
desk = open(FIFO, O_RDONLY);
// Loop until there is no more data to read
while(read(desk, &ch, 1) > 0) {
countLetters(tab, ch);
printf("%c", ch);
}
}
This is technically sufficient to make it work, but you should also look into a series of other issues:
You should have two wait(0) calls so that you wait for both processes, and you shouldn't try to kill anything.
The server process should only be opening the fifo once, not once per character.
You should be comparing fgetc output to EOF before forcing the value into a char. Since you do it after, running your program on a ISO-8859-1 terminal will cause it to confuse EOF and the letter ΓΏ
You are using strcat on argv[1], even though you don't know how much space that array has. You should use your own buffer of a known length.
You should check the return value of all your system calls to ensure they succeed. Checking with access and then assuming it'll be fine is not as good since calls can fail for other reasons.
Canonical Unix behavior is to exit with 0 for success, and >= 1 for error.
It's good practice to use a larger buffer (e.g. 65536 bytes instead of 1) when using read/write directly. stdio functions like fgetc already uses a larger buffer behind the scenes.
Using a named pipe obviously works, but since you spawn both processes it would be more natural to use an unnamed one.

mutex and its effect on execution time (and cpu usage)

I wrote a very simple test program to examine efficiency of pthread mutex. But I'm not able to analyse the results I get. (I can see 4 CPUs in Linux System Monitor and that's why I have at least 4 active threads, because I want to keep all of them busy.) The existence of mutex is not necessary in the code.
#include <stdio.h>
#include <stdlib.h>
#include <pthread.h>
pthread_mutex_t lock1, lock2, lock3, lock4;
void do_sth() { /* just open a files, read it and copy to another file */
int i;
for (i = 0; i < 1; i++) {
FILE* fp = fopen("(2) Catching Fire.txt", "r");
if (fp == NULL) {
fprintf(stderr, "could not open file\n");
exit(1);
}
char filename[20];
sprintf(filename, "a%d", (int)pthread_self());
FILE* wfp = fopen(filename, "w");
if (wfp == NULL) {
fprintf(stderr, "could not open file for write\n");
exit(1);
}
int c;
while (c = fgetc(fp) != EOF) {
c++;
fputc(c, wfp);
}
close(fp);
close(wfp);
}
}
void* routine1(void* param) {
pthread_mutex_lock(&lock1);
do_sth();
pthread_mutex_unlock(&lock1);
}
void* routine2(void* param) {
pthread_mutex_lock(&lock2);
do_sth();
pthread_mutex_unlock(&lock2);
}
void* routine3(void* param) {
pthread_mutex_lock(&lock3);
do_sth();
pthread_mutex_unlock(&lock3);
}
void* routine4(void* param) {
pthread_mutex_lock(&lock4);
do_sth();
pthread_mutex_unlock(&lock4);
}
int main(int argc, char** argv) {
int i ;
pthread_mutex_init(&lock1, 0);
pthread_mutex_init(&lock2, 0);
pthread_mutex_init(&lock3, 0);
pthread_mutex_init(&lock4, 0);
pthread_t thread1[4];
pthread_t thread2[4];
pthread_t thread3[4];
pthread_t thread4[4];
for (i = 0; i < 4; i++)
pthread_create(&thread1[i], NULL, routine1, NULL);
for (i = 0; i < 4; i++)
pthread_create(&thread2[i], NULL, routine2, NULL);
for (i = 0; i < 4; i++)
pthread_create(&thread3[i], NULL, routine3, NULL);
for (i = 0; i < 4; i++)
pthread_create(&thread4[i], NULL, routine4, NULL);
for (i = 0; i < 4; i++)
pthread_join(thread1[i], NULL);
for (i = 0; i < 4; i++)
pthread_join(thread2[i], NULL);
for (i = 0; i < 4; i++)
pthread_join(thread3[i], NULL);
for (i = 0; i < 4; i++)
pthread_join(thread4[i], NULL);
printf("Hello, World!\n");
}
I execute this program in two ways, with and without all the mutex. and I measure time of execution (using time ./a.out) and average cpu load (using htop). here is the results:
first: when I use htop, I can see that loadavg of the system considerably increases when I do not use any mutex in the code. I have no idea why this happens. (is 4 active threads not enough to get the most out of 4 CPUs?)
second: It takes (a little) less time for the program to execute with all those mutex than without it. why does it happen? I mean, it should take some time to sleep and wake up a thread.
edit: I guess, when I use locks I put other threads to sleep and it eliminates a lot of context-switch (saving some time), could this be the reason?
You are using one lock per thread, so that's why when you use all the mutexes you don't see an increase in the execution time of the application: dosth() is not actually being protected from concurrent execution.
Since all the threads are working on the same file, all they should be accessing it using the same lock (otherwise you will have incorrect results: all the threads trying to modify the file at the same time).
Try running again the experiments using just one global lock.

Linux: POSIX Timers ignore Thread Priority

I have a Thread which has to run every millisecond. When no other thread of the program is active, everything is fine. But if more than 3 other threads are running, the Timer-Thread is only called less than 100 times per second (on my test machine).
It seems that the priority settings of the Timer are ignored.
I have tested this with Kernel Versions 3.12 and 3.18.
Test code, which prints some values after 10000 calls of the timer thread (so normally after 10 seconds):
#define NTHREADS 3
#include <sched.h>
#include <pthread.h>
#include <signal.h>
timer_t timer;
unsigned long long val = 0;
pthread_attr_t attrHigh, attrLow;
void TimerTestThread()
{
val++;
if(val >= 10000)
printf("%i ", val);
}
void BusyThread()
{
int a;
while(1)
{
a++;
}
}
int main()
{
pthread_attr_init(&attrHigh);
pthread_attr_setinheritsched(&attrHigh, PTHREAD_EXPLICIT_SCHED);
pthread_attr_setschedpolicy(&attrHigh, SCHED_FIFO);
struct sched_param paramHigh;
paramHigh.sched_priority = 90;
pthread_attr_setschedparam(&attrHigh, &paramHigh);
pthread_attr_init(&attrLow);
pthread_attr_setinheritsched(&attrLow, PTHREAD_EXPLICIT_SCHED);
pthread_attr_setschedpolicy(&attrLow, SCHED_FIFO);
struct sched_param paramLow;
paramLow.sched_priority = 1;
pthread_attr_setschedparam(&attrLow, &paramLow);
struct sigevent evp;
evp.sigev_notify = SIGEV_THREAD;
evp.sigev_notify_function = TimerTestThread;
evp.sigev_notify_attributes = &attrHigh;
struct itimerspec value;
value.it_interval.tv_sec = 0; // Interval
value.it_interval.tv_nsec = 1000000;
value.it_value.tv_sec = 0; // Initial Expiration
value.it_value.tv_nsec = 1000000;
int i;
pthread_t threads[NTHREADS];
for(i=0; i<NTHREADS;i++)
{
pthread_create(&(threads[i]), &attrLow, BusyThread, NULL);
}
if(timer_create(CLOCK_MONOTONIC, &evp, &timer) != 0)
{
i = 5;
}
if(timer_settime(timer, 0, &value, NULL) != 0)
{
i = 6;
}
while(1);
}
I do not understand why the behavior is like this. Maybe you see something i missed.
EDIT: Corrected a silly source copy error

Conditional variable and rwlock deadlock

I have a simple threaded program which use a conditional variable and a rwlock. I've been staring at it for hours trying different approaches. The problem is that a thread or more stops at the rwlock after a while although it is not locked for writing. Maybe I miss something about how those locks work or how they are implemented.
#include <stdio.h>
#include <pthread.h>
#include <stdlib.h>
#include <time.h>
#include <math.h>
#include <unistd.h>
//global variables
pthread_mutex_t mutex;
pthread_cond_t cond;
pthread_rwlock_t rwlock;
int counter;
int listLength = 1;
void* worker(void* arg){
do {
usleep(200);
printf("Before rwlock\n");
pthread_rwlock_rdlock(&rwlock);
printf("Before mutex\n");
pthread_mutex_lock(&mutex);
printf("Afer mutex\n");
counter++;
//signal the main
if (counter == 5 ||
(listLength < 5 && counter == listLength)){
printf("Signal main\n");
pthread_cond_signal(&cond);
counter = 0;
}
pthread_mutex_unlock(&mutex);
pthread_rwlock_unlock(&rwlock);
} while(listLength != 0);
return NULL;
}
int main(int argc, char* argv[]){
if (argc != 2){
perror("Invalid number of args");
exit(1);
}
//get arguments
int workers = atoi(argv[1]);
//initialize sync vars
pthread_rwlockattr_t attr;
pthread_rwlockattr_setkind_np(&attr,
PTHREAD_RWLOCK_PREFER_WRITER_NONRECURSIVE_NP);
pthread_mutex_init(&mutex, NULL);
pthread_cond_init(&cond, NULL);
pthread_rwlock_init(&rwlock, &attr);
counter = 0;
//create threads
pthread_t threadArray[workers];
int threadOrder[workers];
for (int i = 0; i < workers; i++){
threadOrder[i] = i;
if (pthread_create(&threadArray[i], NULL,
worker, &threadOrder[i]) != 0){
perror("Cannot create thread");
exit(1);
}
}
while(listLength != 0) {
//wait for signal and lock the list
pthread_mutex_lock(&mutex);
while (pthread_cond_wait(&cond, &mutex) != 0);
pthread_rwlock_wrlock(&rwlock);
printf("In write lock\n");
pthread_mutex_unlock(&mutex);
pthread_rwlock_unlock(&rwlock);
printf("release wrlock\n");
}
//join the threads
for (int i = 0; i < workers; i++){
if (pthread_join(threadArray[i], NULL) !=0){
perror("Cannot join thread");
exit(1);
}
}
//release resources
pthread_mutex_destroy(&mutex);
pthread_cond_destroy(&cond);
pthread_rwlock_destroy(&rwlock);
return 0;
}
Looks like this code has several inconsistencies in it.
You're using mutex together with rwlock which means that all the threads of this kind are always locked. If you remove the rwlock code - it won't change the behaviour.
I cannot see the pthread_rwlock_init() call, and suppose you've called it in another place. Anyway pay attention you do call it and you don't call it twice or more times with the same rowlock object.
The same applies to pthread_rwlockattr_destroy()
I cannot see the reason why pthread_rwlock_rdlock() would block without write lock. Be sure you don't do it. Or else you could do a mutual lock of your mutex

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