The following code is a simple thread game, that switches between threads causing the timer to decrease.
It works fine for 3 threads, causes and Abort(core dumped) for 4 threads, and causes a seg fault for 5 or more threads.
Anyone have any idea why this might be happening?
#include <stdio.h>
#include <stdlib.h>
#include <pthread.h>
#include <errno.h>
#include <assert.h>
int volatile num_of_threads;
int volatile time_per_round;
int volatile time_left;
int volatile turn_id;
int volatile thread_running;
int volatile can_check;
void * player (void * id_in){
int id= (int)id_in;
while(1){
if(can_check){
if (time_left<=0){
break;
}
can_check=0;
if(thread_running){
if(turn_id==id-1){
turn_id=random()%num_of_threads;
time_left--;
}
}
can_check=1;
}
}
pthread_exit(NULL);
}
int main(int argc, char *args[]){
int i;
int buffer;
pthread_t * threads =(pthread_t *)malloc(num_of_threads*sizeof(pthread_t));
thread_running=0;
num_of_threads=atoi(args[1]);
can_check=0;
time_per_round = atoi(args[2]);
time_left=time_per_round;
srandom(time(NULL));
//Create Threads
for (i=0;i<num_of_threads;i++){
do{
buffer=pthread_create(&threads[i],NULL,player,(void *)(i+1));
}while(buffer == EAGAIN);
}
can_check=1;
time_left=time_per_round;
turn_id=random()%num_of_threads;
thread_running=1;
for (i=0;i<num_of_threads;i++){
assert(!pthread_join(threads[i], NULL));
}
return 0;
}
See below on why you should not depend on volatile in pthreads. However, your specific problem is probably because you malloc your pthread array, based on the num_of_threads variable before you've actually set num_of_thread from argv[]:
pthread_t *threads = (pthread_t *)malloc (num_of_threads * sizeof (pthread_t));
thread_running = 0;
num_of_threads = atoi (args[1]);
So there's a very good chance you're writing beyond the end of the threads array. The num_of_threads variable will probably be zero on start-up which means you're not allocating what you think you are. Move the allocation to after the extraction of the arguments and that should fix it.
And now, for your viewing pleasure :-), my original rant on the unsafe use of volatile, which I still stand by.
Do not rely on volatile to protect your shared variables. The correct way to do this is with the pthread_mutex_blab_blah_blah calls.
Of particular note, examine this code segment:
if (can_check) {
if (time_left <= 0) {
break;
}
// URK!!
can_check=0;
URK!! is the point where your current thread may be switched out and another run, leading to the possibility that two threads can be running a critical section of code.
My advice is to forget the can_check altogether and just protect all the shared variables with a mutex, something like (from memory):
void *player (void * id_in) {
int id = (int)id_in;
while (1) {
pthread_mutex_lock (&mutex);
if (time_left <= 0) {
pthread_mutex_unlock (&mutex);
break;
}
if (thread_running) {
if (turn_id == id-1) {
turn_id = random() % num_of_threads;
time_left--;
}
}
pthread_mutex_unlock (&mutex);
}
pthread_exit(NULL);
}
Then put at file-level:
pthread_mutexattr_t mutexattr; // prob. not needed at file level.
pthread_mutex_t mutex;
and, in main, before starting any other threads:
pthread_mutexattr_init (&mutexattr);
// Change attributes if needed.
pthread_mutex_init (&mutex, &mutex_attr);
// Then run all you other stuff here, make sure you've joined with all threads.
pthread_mutex_destroy (&mutex);
Oh yeah, although I haven't done it, you should also check the return codes for all those mutex calls. I'm not going to add that since it'll clog up the answer with unnecessary detail, but it's good practice.
Related
Considering the following code:
#define _XOPEN_SOURCE 600
#define _DEFAULT_SOURCE
#include <pthread.h>
#include <stdatomic.h>
#include <stdint.h>
#include <unistd.h>
#include <stdio.h>
#include <stdlib.h>
#define ENTRY_NUM 100
struct value {
pthread_mutex_t mutex;
int i;
};
struct entry {
atomic_uintptr_t val;
};
struct entry entries[ENTRY_NUM];
void* thread1(void *arg)
{
for (int i = 0; i != ENTRY_NUM; ++i) {
struct value *val = (struct value*) atomic_load(&entries[i].val);
if (val == NULL)
continue;
pthread_mutex_lock(&val->mutex);
printf("%d\n", val->i);
pthread_mutex_unlock(&val->mutex);
}
return NULL;
}
void* thread2(void *arg)
{
/*
* Do some costy operations before continuing.
*/
usleep(1);
for (int i = 0; i != ENTRY_NUM; ++i) {
struct value *val = (struct value*) atomic_load(&entries[i].val);
pthread_mutex_lock(&val->mutex);
atomic_store(&entries[i].val, (uintptr_t) NULL);
pthread_mutex_unlock(&val->mutex);
pthread_mutex_destroy(&val->mutex);
free(val);
}
return NULL;
}
int main() {
for (int i = 0; i != ENTRY_NUM; ++i) {
struct value *val = malloc(sizeof(struct value));
pthread_mutex_init(&val->mutex, NULL);
val->i = i;
atomic_store(&entries[i].val, (uintptr_t) val);
}
pthread_t ids[2];
pthread_create(&ids[0], NULL, thread1, NULL);
pthread_create(&ids[1], NULL, thread2, NULL);
pthread_join(ids[0], NULL);
pthread_join(ids[1], NULL);
return 0;
}
Suppose in function thread1, entries[i].val is loaded, then the scheduler schedule the process to sleep.
Then thread2 awakes from usleep, since ((struct val*) entries[0].val)->mutex aren't locked, thread2 locks it, stores NULL to entries[0].val and free the original of entries[0].val.
Now, is that a race condition? If so, how to avoid this without locking entries or entries[0]?
You are correct my friend, there is indeed a race condition in such code.
Let me open by overall and saying that thread is prawn to race conditions by definition and this is also correct for any other library implemented thread which is unacknowledged by your compiler ahead of compilation time.
Regarding your specific example, yes as you've explained yourself since we can not assume when does your scheduler go into action, thread1 could atomic load your entries, context switch to thread2, which will then free these entries before thread1 gets processor time again. How do you prevent or avoid such race conditions? avoid accessing them without locking them, even though atomic load is an "atomic read" you are logically allowing other threads to access these entries. the entire code scope of both thread1 and thread2 should be protected with a mutex. despite using atomic_load, you are just guaranteeing that at that atomic time, no other accesses to that entry will be made, but during the time between the atomic_load and your first calling to pthread_mutex_lock context switches can indeed occur! as you've mentioned yourself, this is both bad practice and logically wrong. - hence as I've already stated, you should protect the entire scope with pthread_mutex_lock
In general, as I've stated in the beginning of this, considering that your compiler is unaware to the concept of threads during compilation times, it is very sensitive to race conditions that you may not even be aware of, - e.g.: when accessing different areas of some shared memory, the compiler does not take into consideration that other threads may exist and accesses some memory as it desires, and may affect different areas of memories during, even though logically the code itself doesn't, and the "correctness" of the code is valid.
there was some paper published on this called
Threads Cannot be Implemented as a Library by Hans-J Boehm I highly suggest you read it, I promise it will increase your understanding of race conditions and threads using pthread at general!
I am new in threads in c. My code has a thread that increasing a counter and occasionally(randomly) another thread reads that counter. I used mutex in this code but my code always gives me value equal to 1. Although I used pthread_mutex_unlock but it seems the value becomes lock for ever. what should I do to solve the problem?
#include <unistd.h>
#include <stdio.h>
#include <stdlib.h>
#include <sys/types.h>
#include <sys/stat.h>
#include <sys/fcntl.h>
#include <pthread.h>
///////////////////////////
long int count=0;
int RunFlag=0;
pthread_mutex_t mtx;
void *readfun(void *arg)
{
while (RunFlag==0)
{
pthread_mutex_lock(&mtx);
count=count+1;
pthread_mutex_unlock(&mtx);
}
}
void *printfun(void *arg)
{
int RadnTime=0;
for (int j=0;j<4;j++)
{
RadnTime=rand()%3+1;
sleep(RadnTime);
printf("The current counter after %d seconds is: ",RadnTime);
pthread_mutex_lock(&mtx);
printf("%d\n",count);
pthread_mutex_unlock(&mtx);
}
}
void main ()
{
pthread_t thread1;
pthread_t thread2;
pthread_mutex_init(&mtx, NULL);
pthread_create(&thread1,NULL,readfun,NULL);
pthread_create(&thread2,NULL,printfun,NULL);
//stop the counter
RunFlag=1;
pthread_exit(NULL);
}
You are setting RunFlag immediately after the two threads are created, so readfun barely has any time to execute. RunFlag=1; should be at the end of printfun.
As far as I know, reading from and writing to RunFlag isn't guaranteed to be atomic, so access to it should be protected as well. I don't see a problem happening here (given the values in question), but you are entering undefined behaviour territory.
Your functions don't return anything even though they are declared as returning void*. Add return NULL; to them.
Finally, the second %d should be %ld since count is a long int.
Note that
pthread_mutex_t mtx;
pthread_mutex_init(&mtx, NULL);
can be replaced with
pthread_mutex_t mtx = PTHREAD_MUTEX_INITIALIZER;
1) You also have data race as two are accessing RunFlag. So, you need to protect it with synchronization primitive (such as mutex or use a semaphore).
But you don't need the RunFlag for this purpose. Since you can use an infinite loop in readfun() thread and then simply call _exit() from printfun() thread to exit the whole process.
2) Thread functions must return a pointer to satisfy by pthread_create()'s requirement. But you thread functions don't return anything. You can add return NULL; at the end of both thread functions.
Be aware of signed integer overflow as count could potentially overflow.
I'm fairly new to threads in C. For this program I need to declare a thread which I pass in a for loop thats meant to print out the printfs from the thread.
I can't seem to get it to print in correct order. Here's my code:
#include <stdio.h>
#include <stdlib.h>
#include <pthread.h>
#define NUM_THREADS 16
void *thread(void *thread_id) {
int id = *((int *) thread_id);
printf("Hello from thread %d\n", id);
return NULL;
}
int main() {
pthread_t threads[NUM_THREADS];
for (int i = 0; i < NUM_THREADS; i++) {
int code = pthread_create(&threads[i], NULL, thread, &i);
if (code != 0) {
fprintf(stderr, "pthread_create failed!\n");
return EXIT_FAILURE;
}
}
return EXIT_SUCCESS;
}
//gcc -o main main.c -lpthread
That's the classic example of understanding multi-threading.
The threads are running concurrently, scheduled by OS scheduler.
There is no such thing as "correct order" when we are talking about running in parallel.
Also, there is such thing as buffers flushing for stdout output. Means, when you "printf" something, it is not promised it will happen immediately, but after reaching some buffer limit/timeout.
Also, if you want to do the work in the "correct order", means wait until the first thread finishes it's work before staring next one, consider using "join":
http://man7.org/linux/man-pages/man3/pthread_join.3.html
UPD:
passing pointer to thread_id is also incorrect in this case, as a thread may print id that doesn't belong to him (thanks Kevin)
I have the function display.c :
/* DO NOT EDIT THIS FILE!!! */
#include <stdio.h>
#include <unistd.h>
#include "display.h"
void display(char *str)
{
char *p;
for (p=str; *p; p++)
{
write(1, p, 1);
usleep(100);
}
}
and display.h is:
/* DO NOT EDIT THIS FILE!!! */
#ifndef __CEID_OS_DISPLAY_H__
#define __CEID_OS_DISPLAY_H__
void display(char *);
#endif
My task is to use pthreads in order to have the following output:
abcd
abcd
abcd
..
..
Note that I must not edit the file display.c or the file display.c. I have to use mutexes in order to succeed the output that is shown above.
The following block of code is my closest attempt to finally reach the result I want:
#include <stdio.h>
#include <stdlib.h>
#include <sys/wait.h>
#include <sys/types.h>
#include <sys/mman.h>
#include <unistd.h>
#include <pthread.h>
#include "display.h"
pthread_t mythread1;
pthread_t mythread2;
pthread_mutex_t m1, m2;
void *ab(void *arg)
{
pthread_mutex_lock(&m1);
display("ab");
pthread_mutex_unlock(&m1);
}
void *cd(void *arg)
{
pthread_mutex_lock(&m1);
display("cd\n");
pthread_mutex_unlock(&m1);
}
int main(int argc, char *argv[])
{
pthread_mutex_init(&m1, NULL);
pthread_mutex_init(&m2, NULL);
int i;
for(i=0;i<10;i++)
{
pthread_create(&mythread1, NULL, ab, NULL);
pthread_create(&mythread2, NULL, cd, NULL);
}
pthread_join(mythread1, NULL);
pthread_join(mythread2, NULL);
pthread_mutex_destroy(&m1);
pthread_mutex_destroy(&m2);
return EXIT_SUCCESS;
}
The output of the code above is something like this:
abcd
abcd
abcd
abcd
ababcd
cd
abcd
abcd
abcd
abcd
As you can see "ab" and "cd\n" are never mixed but every time I run the code the output differs. I want to make sure that every time I run this code the output will be:
abcd
abcd
abcd
for ten times.
I am really stuck with this since I can't find any solution from the things I already know.
A mutex cannot (by itself) solve your problem. It can prevent your two threads from running concurrently, but it cannot force them to take turns.
You can do this with a condition variable in addition to the mutex, or with a pair of semaphores. Either way, the key is to maintain at all times a sense of which thread's turn it is.
Myself, I think the semaphore approach is easier to understand and code. Each semaphore is primarily associated with a different thread. That thread must lock the semaphore to proceed. When it finishes one iteration it unlocks the other semaphore to allow the other thread to proceed, and loops back to try to lock its own semaphore again (which it cannot yet do). The other thread works the same way, but with the semaphore roles reversed. Roughly, that would be:
sem_t sem1;
sem_t sem2;
// ...
void *thread1_do(void *arg) {
int result;
do {
result = sem_wait(&sem1);
// do something
result = sem_post(&sem2);
} while (!done);
}
void *thread2_do(void *arg) {
int result;
do {
result = sem_wait(&sem2);
// do something else
result = sem_post(&sem1);
} while (!done);
}
Semaphore initialization, error checking, etc. omitted for brevity.
Updated to add:
Since you now add that you must use mutexes (presumably in a non-trivial way) the next best way to go is to introduce a condition variable (to be used together with the mutex) and an ordinary shared variable to track which thread's turn it is. Each thread then waits on the condition variable to obtain the mutex, under protection of the mutex checks the shared variable to see whether it is its turn, and if so, proceeds. Roughly, that would be:
pthread_mutex_t mutex = PTHREAD_MUTEX_INITIALIZER;
pthread_cond_t cond = PTHREAD_COND_INITIALIZER;
int whose_turn = 1;
// ...
void *thread1_do(void *arg) {
int result;
result = pthread_mutex_lock(&mutex);
while (1) {
if (whose_turn == 1) {
// do something
whose_turn = 2; // it is thread 2's turn next
}
// break from the loop if finished
result = pthread_cond_broadcast(&cond);
result = pthread_cond_wait(&cond, &mutex);
}
result = pthread_mutex_unlock(&mutex);
}
void *thread1_do(void *arg) {
int result;
result = pthread_mutex_lock(&mutex);
while (1) {
if (whose_turn == 2) {
// do something else
whose_turn = 1; // it is thread 1's turn next
}
// break from the loop if finished
result = pthread_cond_broadcast(&cond);
result = pthread_cond_wait(&cond, &mutex);
}
result = pthread_mutex_unlock(&mutex);
}
Error checking is again omitted for brevity.
Note in particular that when a thread waits on a condition variable, it releases the associated mutex. It reaquires the mutex before returning from the wait. Note also that each checks at each iteration whether it is its turn to proceed. This is necessary because spurious wakeups from waiting on a condition variable are possible.
You can use a conditional variable to take turns between threads:
pthread_cond_t cond = PTHREAD_COND_INITIALIZER;
int turn = 0;
void *ab(void *arg)
{
pthread_mutex_lock(&m1);
while (turn != 0)
pthread_cond_wait(&cond, &m1);
display("ab");
turn = 1;
pthread_mutex_unlock(&m1);
pthread_cond_signal(&cond);
}
void *cd(void *arg)
{
pthread_mutex_lock(&m1);
while (turn != 1)
pthread_cond_wait(&cond, &m1);
display("cd\n");
turn = 0;
pthread_mutex_unlock(&m1);
pthread_cond_signal(&cond);
}
Another problem is, you are joining with the last two pair of threads created in main() thread, which are not necessarily the ones get executed as last. If threads created early are not completed, then you are destroying the mutex m1 while it might be in use and exiting whole process.
Consider this approach to the issue:
for(i=0;i<10;i++)
{
ab();
cd();
}
This achieves your goals completely, given the shown code. The problem with your example is that you effectively prevent any synchronization and that seems to be your goal even!
Assuming that before the output, you actually want to do something useful which takes CPU time, the answer is that you will have to change the code of display(), which is simply not suitable for parallelization. Proper concurrent code is designed to work independent of other code, in particular it shouldn't compete for resources (locks) with other calls and it shouldn't rely on the order it finishes.
In summary, you can't learn much from this, it's a bad example. In order to improve the code (the one you don't want to change, but that's your problem), consider that the resource the different threads compete for is stdout. If they each wrote to their own buffer, you could create the threads, wait for them to finish and only then reorder their results for output.
I am trying to use eventfd for synchronization b/w two threads. Please look below code. In this if main thread acquires a lock first it wont release unless I uncomment sleep after write function. It is true if thread gets the lock first. Please let me know how to handle without using sleep after write.
#include <pthread.h>
#include <stdint.h>
#include <stdio.h>
int event_fd;
uint64_t counter;
void * Thread1(void *p)
{
printf("\n%s eventfd = %d\n",(char*)p, event_fd);
while(1)
{
read(event_fd, &counter,sizeof(counter));
printf("\n %s function counter = %llu\n",(char*)p,counter);
sleep(1);
write(event_fd,&counter,sizeof(counter));
//sleep(1);
}
return NULL;
}
void main()
{
char str[]="In Thread1";
int ret;
pthread_t p_th;
printf("Events demonstration pid = %d sizeof counter %lu\n ",getpid(),sizeof(counter));
event_fd=eventfd(1,0);
printf("event_fd %d\n",event_fd);
pthread_create(&p_th,NULL,Thread1, str);
while(1)
{
read(event_fd, &counter,sizeof(counter));
printf("\n In main function counter = %llu\n",counter);
sleep(1);
write(event_fd,&counter,sizeof(counter));
//sleep(1);
}
pthread_exit (NULL);
}
Firstly, they aren't locks. They are eventfd semaphores.
Secondly, they contain no provision for fairness amongst threads. So if you omit the sleep() clause, the semaphore will be grabbed again the moment you release it by the next read. eventfd is more often used in a producer consumer environment where one end is doing the write and the other the read.