Communication between two pthreads to display - multithreading

I have looked at other questions and still cant seem to understand this concept. I have created four threads and each thread needs to communicate with the same thread. (Min and Display), (Max and Display), and so on..
I know I need to use:
pthread_mutex_t myLock1 = PTHREAD_MUTEX_INITIALIZER;
pthread_mutex_lock(&myLock1);
(some global variable) = ....;
pthread_mutex_unlock(&myLock1);
Here is one of the threads I have created:
pthread_create(&minThread, NULL, (void *) mini, (void *) numbers);
pthread_join(minThread, (void *) &min);
void *mini(void *numbs1)
{
int *numbers = (int *) numbs1;
min = malloc(sizeof(int));
*min = (numbers[0]);
for (i = 0; i < length; i++) //Computing min value
{
if ( numbers[i] < *min )
{
*min = numbers[i];
}
}
printf("Min has been computed.\n");
pthread_exit(min);
}
I am now supposed to incorporate the use of another thread (disThread) to display the variable I have computed in minThread (min) and other similar threads that have done computations using those mutex system calls I have specified. How can I set this up correctly? Any tips are appreciated to further my understanding.

Turns out that using the mutex system calls are pretty simple. For anyone else who runs into a similar problem, the thread functions void *mini and void *disi follow this format:
void *mini(void *numbs1)
{
pthread_mutex_lock(&theLock);
int *numbers = (int *) numbs1;
min = malloc(sizeof(int));
*min = (numbers[0]);
for (i = 0; i < LENGTH; i++) //Computing min value
{
if ( numbers[i] < *min )
{
*min = numbers[i];
}
}
pthread_mutex_unlock(&theLock);
printf("Min has been computed.\n");
pthread_exit(min);
}
void *disi()
{
pthread_mutex_lock(&theLock);
x = *min; //Displaying min value (x initialized as global variable)
pthread_mutex_unlock(&theLock);
printf("The Minimum value is %d\n", x);
}
A mutex lock and mutex unlock should cover the range of a shared variable between two threads. This is so there is no interference between the two threads working simultaneously.

Related

Try to compare 2 methods to implement bounded blocking queue

bounded blocking queue is famous, of course. There are mostly 2 methods to implement it. I try to understand which way is better:
Method 1: use counting semaphore
void *producer(void *arg) {
int i;
for (i = 0; i < loops; i++) {
sem_wait(&empty);
sem_wait(&mutex);
put(i);
sem_post(&mutex);
sem_post(&full);
}
}
void *consumer(void *arg) {
int i;
for (i = 0; i < loops; i++) {
sem_wait(&full);
sem_wait(&mutex);
int tmp = get();
sem_post(&mutex);
sem_post(&empty);
printf("%d\n", tmp);
}
}
Method 2: classic monitor pattern
class BoundedBuffer {
private:
int buffer[MAX];
int fill, use;
int fullEntries;
pthread_mutex_t monitor; // monitor lock
pthread_cond_t empty;
pthread_cond_t full;
public:
BoundedBuffer() {
use = fill = fullEntries = 0;
}
void produce(int element) {
pthread_mutex_lock(&monitor);
while (fullEntries == MAX)
pthread_cond_wait(&empty, &monitor);
//do something
pthread_cond_signal(&full);
pthread_mutex_unlock(&monitor);
}
int consume() {
pthread_mutex_lock(&monitor);
while (fullEntries == 0)
pthread_cond_wait(&full, &monitor);
//do something
pthread_cond_signal(&empty);
pthread_mutex_unlock(&monitor);
return tmp;
}
}
I understand the 2nd method can solve a lot of other problems. But how to compare these 2 methods? Looks like they can both fulfill the task.
Is there any link on detailed comparision?
Appreciate your help.
Thanks.
The big difference between those two methods is that the first one does not use pthread_ specific functions (semaphores are not part of pthread) and as such is not guaranteed to work in multithreaded enviornment.
In particular, semaphores do not protect memory ordering, so things written in one thread might not be readable on another. Mutexes are suitable for multi-thread message queue.

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.

POSIX semaphore with related processes running threads

I have an assignment to implement Producer consumer problem in a convoluted way(may be to test my understanding). The parent process should set up a shared memory. The unnamed semaphores(for empty count and filled count) should be initialized and a mutex should be initialized. Then two child processes are created, a producer child and a consumer child. Each child process should create a new thread which should do the job.
PS: I have read that the semaphore's should be kept in a shared memory as they would be shared by different processes.
Please provide some hints, or suggest changes.
So far, I have done this:
struct shmarea
{
unsigned short int read;
unsigned short int max_size;
char scratch[3][50];
unsigned short int write;
sem_t sem1;// Empty slot semaphore
sem_t sem2;// Filled slot Semaphore
};
void *thread_read(void* args);
void *thread_write(void *args);
pthread_mutex_t work_mutex;
struct shmarea *shma;
int main()
{
int fork_value,i=0,shmid;
printf("Parent process id is %d\n\n",getpid());
int res1,res2;
key_t key;
char *path = "/tmp";
int id = 'S';
key = ftok(path, id);
shmid = shmget(key,getpagesize(),IPC_CREAT|0666);
printf("Parent:Shared Memory id = %d\n",id);
shma = shmat(shmid,0,0);
shma->read = 0;
shma->max_size = 3;
shma->write = 0;
pthread_t a_thread;
pthread_t b_thread;
void *thread_result1,*thread_result2;
res1 = sem_init(&(shma->sem1),1,3);//Initializing empty slot sempahore
res2 = sem_init(&(shma->sem2),1,0);//Initializing filled slot sempahore
res1 = pthread_mutex_init(&work_mutex,NULL);
while(i<2)
{
fork_value = fork();
if(fork_value > 0)
{
i++;
}
if(fork_value == 0)
{
if(i==0)
{
printf("***0***\n");
//sem_t sem1temp = shma->sem1;
char ch;int res;
res= pthread_create(&a_thread,NULL,thread_write,NULL);
}
if(i==1)
{
printf("***1***\n");
//sem_t sem2temp = shma->sem2;
int res;
char ch;
res= pthread_create(&b_thread,NULL,thread_read,NULL);
}
}
}
int wait_V,status;
res1 = pthread_join(a_thread,&thread_result1);
res2 = pthread_join(b_thread,&thread_result2);
}
void *thread_read(void *args)
{
while(1)
{
sem_wait(&(shma->sem2));
pthread_mutex_lock(&work_mutex);
printf("The buf read from consumer:%s\n",shma->scratch[shma->read]);
shma->read = (shma->read+1)%shma->max_size;
pthread_mutex_unlock(&work_mutex);
sem_post(&(shma->sem1));
}
}
void *thread_write(void *args)
{
char buf[50];
while(1)
{
sem_wait(&(shma->sem1));
pthread_mutex_lock(&work_mutex);
read(STDIN_FILENO,buf,sizeof(buf));
strcpy(shma->scratch[shma->write],buf);
shma->write = (shma->write+1)%shma->max_size;
pthread_mutex_unlock(&work_mutex);
sem_post(&(shma->sem2));
}
}
(1) Your biggest problem by far is that you have managed to write a fork bomb. Because you don't exit either child in the fork loop each child is going to fall through and loop around and create their own children until you crash or bring the system down. You want something more like this:
while(i < 2)
{
fork_value = fork();
if(fork_value > 0)
i++;
if(fork_value == 0)
{
if(i==0)
{
printf("0 child is pid %d\n", getpid());
int res;
res = pthread_create(&a_thread,NULL,thread_write,NULL);
res = pthread_join(a_thread,&thread_result1);
exit(0);
}
if(i==1)
{
printf("1 child is pid %d\n", getpid());
int res;
res = pthread_create(&b_thread,NULL,thread_read,NULL);
res = pthread_join(b_thread,&thread_result2);
exit(0);
}
}
}
for (i = 0; i < 2; ++i)
wait(NULL);
Notice the wait on the children which you neglected.
(2) Always check your return codes. They are like safety belts, a bit of a drag but so helpful when you crash. (Yes, I didn't take my advice here but you should.)
(3) These names are awful.
unsigned short int read;
unsigned short int write;
Stay away from naming variables after system calls. It's confusing and just asking for trouble.
(4) Terminology wise, processes with a common ancestor, like these, are related. The parent can open shared memory and other resources and pass it on to the children. Unrelated processes would, for example, multiple instances of program launched from different terminals. They can share resources but not in the "inherited" way forked processes do.
It's late and didn't get around to looking at what you are doing with the threads and such but this should get you started.

implementing a barrier issue

static int barrier_counter = 0;
pthread_cond_t condition;
pthread_mutex_t local_lock;
int init_barrier(int n) {
if (n < 0) {
return -1;
}
barrier_counter = n;
pthread_mutex_init(&local_lock, NULL);
pthread_cond_init(&condition, NULL);
return 0;
}
int barrier() {
pthread_mutex_trylock(&local_lock);
barrier_counter--;
printf("inside barrier befor the while n is : %d \n",barrier_counter );
while (0 < barrier_counter) {
printf("inside the barrier n is : %d\n", barrier_counter);
pthread_cond_wait(&condition,&local_lock);
}
printf("befor bordcast : %d \n",barrier_counter );
pthread_cond_broadcast(&condition);
printf("done the brodcast when n is : %d \n",barrier_counter );
pthread_mutex_unlock(&local_lock);
return 0;
}
we tried to implement barrier but somehow we dont get to do broadcast and we finish anyway.
we dont even get deadlock, there are few threads that are actually waiting but not the amount we specified in init_barrier.
pthread_cond_wait() requires a locked mutex. Your call to pthread_mutex_trylock() might fail, and so continue without acquiring the mutex.
I suggest you use pthread_mutex_lock() instead.
Also, unless you're looking to re-implement barriers, You should use pthread_barrier_init() and pthread_barrier_wait().

pthread_join skipping first thread

So I have the following code, some of it left out so it's easier to understand.
for (unsigned int t = 0; t < NUM_THREADS; t++)
{
if (pthread_create(&threads[t], NULL, thread_run, (void*) &threadData) != 0)
{
perror("pthread_create");
}//end if
}
for (unsigned int z = 0; z < NUM_THREADS; z++)
{
if (pthread_join(threads[z], NULL) != 0)
{
perror("pthread_join");
}
}
My Problem is the join function, it is skipping the first thread I create, and continuing on. The current solution I have is adding an extra thread and not making the first one do any work.
Any ideas why this might be happening?
IMO there is not pthreads problem; you are just creating NUM_THREADS + 1 threads and join only first NUM_THREADS of them.

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