I am having issues getting my project to work and was hoping someone could assist. The guidelines are as follows:
You will use the
pthread package to create 4 producer threads and 4 consumer threads. Each producer thread inserts
character ‘X’ into a buffer of size 10,000,000 characters. Each consumer thread removes the most recently-inserted
character from the buffer. Each thread then repeats the process
So far my code looks like this:
#include <stdlib.h>
#include <stdio.h>
#include <unistd.h>
#include <sys/ipc.h>
#include <sys/sem.h>
#include <semaphore.h>
#include <pthread.h>
#define N 10000000
sem_t mutex;
sem_t full;
sem_t empty;
typedef struct
{
char const* buf[N];
char in;
char out;
} bufferItems;
bufferItems sharedBuffer;
void *producer(void *arg) {
while(1) {
sem_wait(&empty);
sem_wait(&mutex);
sharedBuffer.buf[sharedBuffer.in] = "X";
sharedBuffer.in = (sharedBuffer.in+1)%N;
printf("Producer\n");
sem_post(&mutex);
sem_post(&full);
}
}
void *consumer(void *arg){
while(1){
sem_wait(&full);
sem_wait(&mutex);
sharedBuffer.buf[sharedBuffer.out] = NULL;
sharedBuffer.out = (sharedBuffer.out+1)%N;
printf("Consumer\n");
sem_post(&mutex);
sem_post(&empty);
}
}
int main(void) {
sem_init(&mutex, 0, 0);
sem_init(&full, 0, 0);
sem_init(&empty, 0, N);
pthread_t p;
pthread_t c;
// create four producer threads
for(int t=0; t<4; t++){
printf("In main: creating producer thread %d\n", t);
int err = pthread_create(&p,NULL,producer,NULL);
if (err){
printf("ERROR from pthread_create() on producer thread %d\n", err);
exit(-1);
}
}
// create four consumer threads
for(int t=0; t<4; t++){
printf("In main: creating consumer thread %d\n", t);
int err = pthread_create(&c,NULL,consumer,NULL);
if (err){
printf("ERROR; from pthread_create() on consumer thread %d\n", err);
exit(-1);
}
}
}
But the output I get when I run it is this:
In main: creating producer thread 0
In main: creating producer thread 1
In main: creating producer thread 2
In main: creating producer thread 3
In main: creating consumer thread 0
In main: creating consumer thread 1
In main: creating consumer thread 2
In main: creating consumer thread 3
It seems as though the child threads are either not executing, or the semaphores are not working properly so there is deadlock. Any recommendation on how to make my code better or work properly would be greatly appreciated.
You need to call pthread_join in your main thread after creating all the child threads. pthread_join waits for the specified thread to terminate. Otherwise the main thread will exit and take out all the child threads prematurely.
Related
I learned the concept of semaphore. And I'm trying to implement it.
I've been trying to implement it for over 19 hours, but I can't do it, so I'm writing to ask for your help.
It checks the progress of the current two threads, just as it does with CV, and if both threads output entered, it can resume the subsequent operation.
Below is the full text of the code.
`
#include <stdio.h>
#include <unistd.h>
#include <assert.h>
#include <semaphore.h>
#include <pthread.h>
#include <stdlib.h>
void *child1(void *arg) {
printf("child thread 1 entered!\n");
// call semaphoreshere here
printf("child thread 1 exits!\n");
return NULL;
}
void *child2(void *arg) {
printf("child thread 2: entered!\n");
// call semaphores here
printf("child thread 2: exits\n");
return NULL;
}
int main(int argc, char *argv[]) {
pthread_t p1, p2;
printf("parent thread: begin\n");
// init semaphores here
// // sem_init(&empty, 0, 0);
// // sem_init(&full, 0, 0); //Code tried but not working properly
pthread_create(&p1, NULL, child1, NULL);
pthread_create(&p2, NULL, child2, NULL);
pthread_join(p1, NULL);
pthread_join(p2, NULL);
printf("parent thread: end\n");
return 0;
}
`
Using ONLY TWO semaphores within the code, this attempts to control the thread's internal execution order so that both threads must output the ~~entered message before it can exit.
The execution result I want is as follows.
>>>
parent thread: begin
child thread 1 entered!
child thread 2: entered!
child thread 2: exits
child thread 1 exits!
parent thread: end
>>>
parent thread: begin
child thread 2 entered!
child thread 1: entered!
child thread 1: exits
child thread 2 exits!
parent thread: end
Like this, I want to implement only the role of monitoring each other to see if they've entered.
I'd appreciate your help.
thanks.
What you want to do (in semaphoore-thinking) is:
thread 1 waits till thread 2 is done before it exits.
thread 2 waits till thread 1 is done before it exits.
Which leads to the following code, which I modified a bit because I cannot stand global variables.
#include <stdio.h>
#include <unistd.h>
#include <assert.h>
#include <semaphore.h>
#include <pthread.h>
#include <stdlib.h>
typedef struct SharedContext_tag {
sem_t t1done;
sem_t t2done;
} SharedContext_t;
void *child1(void* arg) {
SharedContext_t* ctx = (SharedContext_t*) arg;
printf("child thread 1: entered!\n");
sem_post(&ctx->t1done);
sem_wait(&ctx->t2done);
printf("child thread 1 exits!\n");
return NULL;
}
void *child2(void* arg) {
SharedContext_t* ctx = (SharedContext_t*) arg;
printf("child thread 2: entered!\n");
sem_post(&ctx->t2done);
sem_wait(&ctx->t1done);
printf("child thread 2: exits!\n");
return NULL;
}
int main(int argc, const char* argv[]) {
pthread_t p1;
pthread_t p2;
SharedContext_t context;
sem_init(&context.t1done, 0, 0);
sem_init(&context.t2done, 0, 0);
printf("parent thread: begin\n");
pthread_create(&p1, NULL, child1, &context);
pthread_create(&p2, NULL, child2, &context);
pthread_join(p1, NULL);
pthread_join(p2, NULL);
printf("parent thread: end\n");
sem_close(&context.t1done);
sem_close(&context.t2done);
return 0;
}
On my machine at this time (being careful here!), the output is as required:
> ./sema
parent thread: begin
child thread 1: entered!
child thread 2: entered!
child thread 2: exits!
child thread 1 exits!
parent thread: end
In order for it to work, you need to link against the real-time library, librt. You do so by adding -pthread to your compile command.
> clang-13 -g -O0 -pthread -o sema sema.c
i have little trouble with pthread priority.I have set scheduler FIFO. I have 2 thread, each have diffretnt priority. On my project when i push A or B, one thread start working. Thats fine, but when one thread working and push again B(priority15) and then A(priority 20) my assumption is that thread A jumps before B in stack queue and after first thread will by next thread A with priority 20 and then thread B (15). Is that problem FIFO or something else?
PS: I dont want use semaphore, i want to solve it just with priority and scheduler. Thanks
#include <stdio.h>
#include<pthread.h>
#include<semaphore.h>
#include<unistd.h>
#include <time.h>
#include <sched.h>
sem_t mutex;
void* levelOnefunction(void *a)
{
sem_wait(&mutex);
int* b = (int *) a;
printf("Thread! next main!\n");
printf("sched prior:%d\n",*b);
sleep(3);
sem_post(&mutex);
return 0;
}
int main()
{
char s;
pthread_t t;
pthread_attr_t tattr;
struct sched_param param;
sem_init(&mutex,0,1);
pthread_attr_init(&tattr);
if(pthread_attr_setschedpolicy(&tattr,SCHED_FIFO)!=0)
printf("ERROR FIFO!\n");
//pthread_setschedparam(t,SCHED_FIFO,¶m);
if(pthread_attr_getschedparam(&tattr,¶m)!=0)
printf("ERROR attr get sheduler!\n");
/* int k =pthread_attr_setinheritsched(&tattr,PTHREAD_EXPLICIT_SCHED);
if (k!=0)
printf("ERROR\n"); */
printf("Initial priority is %d \n",param.sched_priority);
int min = sched_get_priority_min(SCHED_FIFO);
int max = sched_get_priority_max(SCHED_FIFO);
printf("MIN - %d ---> MAX - %d\n",min,max);
scanf("%c",&s);
while (s != '\0')
{
if(s=='a')
{
printf("a\n");
param.sched_priority=20;
if(pthread_attr_setschedparam(&tattr,¶m)!=0)
printf("ERROR attr_setschedul!\n");
printf("update priority is:%d\n",param.sched_priority);
if(pthread_create(&t,&tattr,levelOnefunction,(void *)¶m.sched_priority)!=0)
printf("ERROR thread1\n");
pthread_join(t,NULL);
printf("main finish!\n");
printf("%c end\n\n",s);
}
else if(s=='b')
{
printf("b\n");
param.sched_priority=15;
if(pthread_attr_setschedparam(&tattr,¶m)!=0)
printf("ERROR attr_setschedul!\n");
if(pthread_create(&t,&tattr,levelOnefunction,(void *)¶m.sched_priority)!=0)
printf("ERROR Thread 2 ");
pthread_join(t,NULL);
printf("main finish!\n");
printf("%c end\n\n",s);
}
else if(s=='\n')
{
}
else if (s!='a' && s!='b')
{
printf("bad key\n");
}
scanf("%c",&s);
}
sem_destroy(&mutex);
return 0;
}
You are calling pthread_join() in the main thread immediately after creating each child thread.
This means that you never have more than one child thread created at a time - after creating one child thread, the main thread will then block in pthread_join() until that child thread is complete. Only then does it call scanf() again and potentially create another child thread.
Thread priorities don't come into it.
So I'm very new to the whole concept of pthread in C but please hear me out. I have the following code:
#include <stdio.h>
#include <stdlib.h>
#include <stdbool.h>
#include <unistd.h>
#include <errno.h>
#include <pthread.h>
#include <unistd.h>
static pthread_mutex_t mutex = PTHREAD_MUTEX_INITIALIZER;
static pthread_cond_t endCond = PTHREAD_COND_INITIALIZER;
static pthread_cond_t startCond = PTHREAD_COND_INITIALIZER;
void * threadThingy(void * n){
pthread_cond_wait(&startCond, &mutex);
printf("%d: RAND: %d\n", *((int*)n), rand());
//Lock mutex before broadcasting to main thread
pthread_mutex_lock(&mutex);
pthread_cond_broadcast(&endCond);
pthread_mutex_unlock(&mutex);
free(n);
fflush(stdout);
return 0;
}
int main(void){
printf("Starting\n");
pthread_t threads[100];
int i = 0;
while(i < 10){
int *arg = malloc(sizeof(int));
*arg = i;
pthread_create(&threads[i], NULL, threadThingy, arg);
i++;
}
pthread_mutex_lock(&mutex);
pthread_cond_broadcast(&startCond);
int finished = 0;
while(finished <= 100){
pthread_cond_wait(&endCond, &mutex);
//Lock mutex so no other requests can come in
pthread_mutex_lock(&mutex);
finished++;
int *arg = malloc(sizeof(int));
*arg = 11;
pthread_create(threads[i], NULL, threadThingy, arg);
i++;
pthread_cond_broadcast(&startCond);
pthread_mutex_unlock(&mutex);
}
printf("Stopping\n");
sleep(1000);
}
The whole goal is to run (only) 10 threads simultaneously of the 100. My idea was to start 10 threads, than wait until one is finished and start another one. So I let the program wait until a thread returns, then I start a new one so the thread that just returned gets replaced. What have I missed? Because now I only get this as an output:
Starting
0: RAND: 1804289383
As mentioned by Lavigne958, in function threadThingy() there is deadlock caused by pthread_cond_wait() as it will acquire the lock. Again, you are trying to lock it in next line. This is causing deadlock.
There are a few things need to check:
You need to lock the mutex before calling pthread_cond_wait().
If you solve the above issue, using multiple condition variable with the same mutex may cause further deadlock.
If you are not joining the threads, it will be better to create detached threads using PTHREAD_CREATE_DETACHED attribute.
The problem of N number of threads running simultaneously can be solved with one semaphore OR one condition variable(and one mutex). Example with semaphore is given below.
#include <stdio.h>
#include <pthread.h>
#include <semaphore.h>
#include <unistd.h>
sem_t mysem;
#define NUM_CONCURRENT_THREADS 4
#define MAX_THREADS 40
void *thread(void *arg)
{
printf("Thread id %ld: started\n", pthread_self());
sleep(5); // Do some work
printf("Thread id %ld: Exiting\n", pthread_self());
sem_post(&mysem);
return NULL;
}
int main()
{
pthread_t t[MAX_THREADS];
pthread_attr_t attr;
int rc, i = 0;
sem_init(&mysem, 0, NUM_CONCURRENT_THREADS);
rc = pthread_attr_init(&attr);
rc = pthread_attr_setdetachstate(&attr, PTHREAD_CREATE_DETACHED);
printf("\nParent begin\n");
while(i < MAX_THREADS)
{
sem_wait(&mysem);
pthread_create(&t[i], &attr, thread, NULL);
i++;
}
printf("\nParent end.\n");
sem_destroy(&mysem);
return 0;
}
Please check blog Tech Easy for more information on threads.
in the function that your threads run, you start by waiting on a condition but you forgot to take the mutex before. So you first must take the mutex before waiting on the condition.
you have what we call a deadlock.
What happens is:
the first thread wakes up (the pthread_con_wait function acquires the lock for you already)
then you try to acquire the lock again => deadlock because you already own the lock so you kinda deadlock yourself.
I'm playing with a fairly simple C example. The program creates two threads and starts them in parallel. Each thread is designed to modify a global variable using a Mutex, and print out the value.
#include <stdio.h>
#include <stdlib.h>
#include <unistd.h>
#include <pthread.h>
int A=10;
pthread_mutex_t M;
void *codice_Th1(void *arg) {
int i;
for (i=0; i<10;i++){
pthread_mutex_lock(&M);
printf("Thread %s: ", (char *)arg);
A++;
printf("A = %d \n", A);
pthread_mutex_unlock(&M);
sleep(1);
}
pthread_exit(0);
}
void *codice_Th2(void *arg) {
int i;
for (i=0; i<10;i++){
pthread_mutex_lock(&M);
printf("Thread %s: ", (char *)arg);
A--;
printf("A = %d \n", A);
pthread_mutex_unlock(&M);
sleep(1);
}
pthread_exit(0);
}
The main() simply creates the thread, and join the main thread with thread 1 and 2.
int main(){
pthread_t th1, th2;
...
}
What bothers me, is that I get the following output
Thread th1: Thread th2: A = 11
A = 10
Thread th1: A = 11
Thread th2: A = 10
Thread th1: Thread th2: A = 11
A = 10
Thread th1: Thread th2: A = 11
A = 10
Thread th2: Thread th1: A = 9
A = 10
Thread th1: A = 11
Thread th2: A = 10
whereas I would expect every line to execute the printf statements in sequence, given they are inside a mutex.
In other words, I can't understand the ouput
Thread th2: Thread th1: A = 9
I would expect always something similar to
Thread NAME: A = VALUE
Am I missing something?
Never mind, I believe I found the issue. I did not initialize the Mutex with pthread_mutex_init(&M, NULL); before using it.
Setting
int main(){
pthread_t th1, th2;
int ret;
pthread_mutex_init(&M, NULL);
fixed the issue. I assume using pthread_mutex_init is a requirement. Unfortunately, skipping the mutex initialization din't produce any warning or error. The script silently compiled.
I have been dealing with a problem in a thread pool for some days now. I tried all types of different things but I can't seem to solve the issue. I have made a simple version that reproduces the problem.
Code:
#include <unistd.h>
#include <signal.h>
#include <stdio.h>
#include <stdlib.h>
#include <pthread.h>
#include <time.h>
struct bsem_t bsem;
pthread_t threads[2];
/* Binary semaphore */
typedef struct bsem_t {
pthread_mutex_t mutex;
pthread_cond_t cond;
int v;
} bsem_t;
void bsem_post(bsem_t *bsem) {
pthread_mutex_lock(&bsem->mutex);
bsem->v = 1;
pthread_cond_broadcast(&bsem->cond);
pthread_mutex_unlock(&bsem->mutex);
}
void bsem_wait(bsem_t *bsem) {
pthread_mutex_lock(&bsem->mutex);
while (bsem->v != 1) {
pthread_cond_wait(&bsem->cond, &bsem->mutex);
}
bsem->v = 0;
pthread_mutex_unlock(&bsem->mutex);
}
/* Being called by each thread on SIGUSR1 */
void thread_exit(){
printf("%u: pthread_exit()\n", (int)pthread_self());
pthread_exit(NULL);
}
/* Startpoint for each thread */
void thread_do(){
struct sigaction act;
act.sa_handler = thread_exit;
sigaction(SIGUSR1, &act, NULL);
while(1){
bsem_wait(&bsem); // Each thread is blocked here
puts("Passed semaphore");
}
}
/* Main */
int main(){
bsem.v = 0;
pthread_create(&threads[0], NULL, (void *)thread_do, NULL);
pthread_create(&threads[1], NULL, (void *)thread_do, NULL);
pthread_detach(threads[0]);
pthread_detach(threads[1]);
puts("Created threads");
sleep(2);
pthread_kill(threads[0], SIGUSR1);
pthread_kill(threads[1], SIGUSR1);
puts("Killed threads");
sleep(10);
return 0;
}
What the code does, is create two threads. Both threads wait on a binary semaphore (bsem_wait). Then while they are waiting I send a SIGUSR1 signal to both resulting on pthread_exit() being executed on each thread. On my terminal it shows that everything goes as planned..
Output:
Created threads
Killed threads
2695145216: pthread_exit()
2686752512: pthread_exit()
The problem
Although the output seems correct, using pstree shows that only one of two threads die. The other thread stays alive until the whole program exits. Why is this?
Update
Replacing my custom binary semaphore with a normal semaphore seems to solve this for no apparent reason..
#include <unistd.h>
#include <signal.h>
#include <stdio.h>
#include <stdlib.h>
#include <pthread.h>
#include <time.h>
#include <semaphore.h>
sem_t sem;
pthread_t threads[2];
/* Caller thread will exit */
void thread_exit(){
printf("%u: pthread_exit()\n", (int)pthread_self());
pthread_exit(NULL);
}
/* Startpoint for each thread */
void* thread_do(){
struct sigaction act;
act.sa_handler = thread_exit;
sigaction(SIGUSR1, &act, NULL);
while(1){
sem_wait(&sem); // Each thread is blocked here
puts("Passed semaphore");
}
}
/* Main */
int main(){
sem_init(&sem, 0, 0); // Normal semaphore
pthread_create(&threads[0], NULL, thread_do, NULL);
pthread_create(&threads[1], NULL, thread_do, NULL);
pthread_detach(threads[0]);
pthread_detach(threads[1]);
puts("Created threads in pool");
sleep(2);
//PROBLEM
pthread_kill(threads[0], SIGUSR1);
pthread_kill(threads[1], SIGUSR1);
puts("Destroyed pool");
sleep(10);
return 0;
}
You can't get there from here
pthread_exit() is not listed in the "signal safe funtions" of the signal(7) man page.
rewrite your code to have the pthread_exit call outside of the signal handler.
So the issue seemed to be a deadlock!
The problem is that each thread is waiting inside the bsem_wait of the binary semaphore in different locations:
void bsem_wait(bsem_t *bsem) {
pthread_mutex_lock(&bsem->mutex); // THREAD 2 BLOCKED HERE
while (bsem->v != 1) {
pthread_cond_wait(&bsem->cond, &bsem->mutex); // THREAD 1 WAITING HERE
}
bsem->v = 0;
pthread_mutex_unlock(&bsem->mutex);
}
Thread 1 in this case is the fastest thread. Thread 2 is the slower one. When I run the signal to kill a thread, the waiting thread unblocks and exits as expected. The problem is that it never unlocks the mutex. So the blocked thread (2) remains blocked forever. For some reason, the thread won't be terminated because it is waiting on the mutex.
Just adding an unblock before exiting, solves the issue:
void thread_exit(){
printf("%u: pthread_exit()\n", (int)pthread_self());
pthread_mutex_unlock(&mutex); // NEW CODE
pthread_exit(NULL);
}
This ofcourse is a hack to demonstrate what is happening and shouldn't be used. I am going to follow Jasen's advice and get rid of the signal handler all together and solve it some other way. Namely I have to assure that the thread goes through the whole bsem_wait!