Efficiency of the multi producer-consumer execution - c++

I was trying to make a code with multi producers and consumers. I have created multi-threads for producer and consumer and used semaphores for synchronization. The code was working fine with single producer and consumer.
The problem which I am facing is that after some time of program execution, only the consumer1 and producer1 are participating in the process. I am not able to understand what happened to the other producers and consumers.
I would also like to know as how to make multi producer-consumer problem efficient? Efficient in the sense that all producer and consumer gets equal opportunity to produce and consume respectively?
C++ code(it includes a lot of C):
#include <iostream>
#include <pthread.h>
#include <semaphore.h>
#include <unistd.h>
#include <queue>
using namespace std;
sem_t empty;
sem_t full;
int cnt = 0;
pthread_mutex_t mutex = PTHREAD_MUTEX_INITIALIZER;
queue<int> q;
void *producer(void *a)
{
int *num = (int *)a;
while(1) {
sem_wait(&empty);
pthread_mutex_lock(&mutex);
cnt = cnt+1;
q.push(cnt);
cout<<cnt<<" item produced by producer "<<(*num+1)<<endl;
pthread_mutex_unlock(&mutex);
sem_post(&full);
sleep(1);
}
}
void *consumer(void *a)
{
int *num = (int *)a;
while(1) {
sem_wait(&full);
pthread_mutex_lock(&mutex);
cout<<q.front()<<" item consumed by consumer "<<(*num+1)<<endl;
q.pop();
pthread_mutex_unlock(&mutex);
sem_post(&empty);
sleep(1);
}
}
int main()
{
pthread_t p[5];
pthread_t c[5];
sem_init(&empty,0,5);
sem_init(&full,0,0);
int i;
for(i = 0; i < 5; i++) {
pthread_create(&p[i],NULL,producer,(void *)(&i));
}
for(i = 0; i < 5; i++) {
pthread_create(&c[i],NULL,consumer,(void *)(&i));
}
for(i = 0; i < 5; i++) {
pthread_join(p[i],NULL);
pthread_join(c[i],NULL);
}
}
Updated code:
#include <iostream>
#include <pthread.h>
#include <semaphore.h>
#include <unistd.h>
#include <queue>
#include <map>
using namespace std;
sem_t empty;
sem_t full;
int cnt = 0;
pthread_mutex_t mutex = PTHREAD_MUTEX_INITIALIZER;
map<pthread_t,int> mc,mp;
queue<int> q;
void *producer(void *a)
{
while(1) {
sem_wait(&empty);
pthread_mutex_lock(&mutex);
cnt = cnt+1;
q.push(cnt);
cout<<cnt<<" item produced by producer "<<mp[pthread_self()]<<endl;
pthread_mutex_unlock(&mutex);
sem_post(&full);
sleep(1);
}
}
void *consumer(void *a)
{
while(1) {
sem_wait(&full);
pthread_mutex_lock(&mutex);
cout<<q.front()<<" item consumed by consumer "<<mc[pthread_self()]<<endl;
q.pop();
pthread_mutex_unlock(&mutex);
sem_post(&empty);
sleep(1);
}
}
int main()
{
pthread_t p[5];
pthread_t c[5];
sem_init(&empty,0,5);
sem_init(&full,0,0);
int i;
pthread_mutex_lock(&mutex);
for(i = 0; i < 5; i++) {
pthread_create(&p[i],NULL,producer,NULL);
pthread_create(&c[i],NULL,consumer,NULL);
mc[c[i]] = i+1;
mp[p[i]] = i+1;
}
pthread_mutex_unlock(&mutex);
for(i = 0; i < 5; i++) {
pthread_join(p[i],NULL);
pthread_join(c[i],NULL);
}
}

Short answer
The threads do in fact execute with equal opportunity, but they just printout an identifier which is not theirs.
Detailed explanation
You keep in each thread a pointer num to the thread number. It's the pointer to that value which is saved and not the value itself. So all the threads point to the same counter, thinking to find there their own identifier.
Everytime you access *num, you get access not to the value that i had when you launched the thread, but its current value.
Unfortunately, in every loop of main(), you reuse the variable i. So the last loop, you'll set i back to 0, and wait for the first threads to join. But all these threads loop forever, so the loop will hardly get a chance to go beyond this initial 0 value. So that every thread thinks it's the number *num+1 that is 1 at this moment.
Note by the way that you create a race condition as someone pointed out in the comments: all the consumer and producer threads dereference the pointer, accessing to the same variable in a mutex-protected region. This is ok. But while they are reading the variable, the main thread still happily can change the shared variable outside of any lock. This is definitively a risk of race.
Workaround
std::thread would allow you to pass i by walue, so that each thread has its own unaltered copy of is id.
With pthreads you have to pass a pointer to a value. Unfortunately, even if you'd do a local copy of the value pointed at, right at the start of the thread, you'd still be in a race condition.
A quick workaround to observe which thread is really doing the work would be to printout as well the result of pthread_self() (see here how to do it). Or to store the ids in an array of int, and pass to each thread the address to a unique element in that array.

Related

C++ pthread deadlock solution

I am trying to use p_thread to print out numbers in order like:
0
1
2
3
4
Without using a global variable, just local variable, but I met deadlock and I don't know how to fix it yet.
This is my code:
#include <pthread.h>
#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <fcntl.h>
static pthread_mutex_t bsem; // Mutex semaphore
static pthread_cond_t waitTurn = PTHREAD_COND_INITIALIZER; // Condition variable to control the turn
//static int turn; // Index to control access to the turn array
static int nthreads; // Number of threads from input
struct SFE{
int turn;
int thread;
};
void *thread_function(void *void_ptr_argv)
{
SFE *threadNum = (SFE *) void_ptr_argv;
pthread_mutex_lock(&bsem);
// if its not our turn then wait
while(threadNum->turn != threadNum->thread){
pthread_cond_wait(&waitTurn, &bsem);
}
pthread_mutex_unlock(&bsem);
std::cout << "I am Thread " << threadNum->turn << std::endl;
pthread_mutex_lock(&bsem);
threadNum->turn++;
pthread_cond_broadcast(&waitTurn);
pthread_mutex_unlock(&bsem);
return nullptr;
}
int main()
{
std::cin >> nthreads;
pthread_mutex_init(&bsem, NULL); // Initialize bsem to 1
pthread_t *tid= new pthread_t[nthreads];
SFE threadNumber;
threadNumber.turn = 0;
for(int i=0;i<nthreads;i++)
{
// initialize the thread number here (remember to follow the rules from the specifications of the assignment)
threadNumber.thread = i;
pthread_create(&tid[i], nullptr, thread_function, (void*)&threadNumber);
}
for(int i = 0; i < nthreads; i++)
{
pthread_join(tid[i], nullptr);
}
return 0;
}
I am expecting a simple way to solve my problem
Think about threadNum->thread. threadNum is an unique object, thus each thread gets an unspecified number threadNum->thread between 0 and nthreads. Only the last thread gets the correct number nthreads - 1 .
You should update
struct SFE{
int *turn;
int thread;
};
Allocate the array SFE threadNum[nthread] and pass &threadNum[i] to i-th thread.
int turn = 0;
for(int i=0;i<nthreads;i++)
{
// initialize the thread number here (remember to follow the rules from the specifications of the assignment)
threadNumber[i].turn = &turn;
threadNumber[i].thread = i;
pthread_create(&tid[i], nullptr, thread_function, &threadNumber[i]);
}
Your code is C-ism. This code w/o iostream and new is a clean C. If you use C++, all will be much easier.

unable to implement list operations using thread

Thread newbie here. In the following code, I want to add elements to the global list using one thread and search for random elements using another thread.
#include <list>
#include <algorithm>
#include <mutex>
#include <thread>
using namespace std;
list<int> some_list;
mutex some_mutex;
void add_to_list(int new_value)
{
lock_guard<mutex> guard(some_mutex);
some_list.push_back(new_value);
}
bool list_contains(int value_to_find)
{
lock_guard<mutex> guard(some_mutex);
return find(begin(some_list), end(some_list), value_to_find) != end(some_list);
}
int main()
{
for(int i = 0; i < 100; ++i)
{
// Add i to some_list through one thread only
thread t(add_to_list, i);
t.detach();
// Search elements in different thread
thread t2(list_contains, i);
t2.detach();
}
return 0;
}
However, when I pass i along with add_to_list then that many no of threads are created. I want to add the elements to the list using single thread only. How to do this? Do, I need to pass vector of elements to the thread instead of variable?
First of all, don't use detach(). In the code above the program starts 200 threads and ends (when main() returns) without waiting for any of them to finish. Use join() instead, this will allow you to wait for the thread completion properly.
Then, if you want add_to_list to be done sequentially, then do that in a separate loop.
For example:
int main() {
vector<thread> threads;
// Add i to some_list through one thread only
threads.emplace_back([] {
for (int i = 0; i < 100; ++i) {
add_to_list(i);
}
});
for (int i = 0; i < 100; ++i) {
// Search elements in different threads
threads.emplace_back([i] {
list_contains(i);
});
}
// Join all threads (waits for their completion)
for (auto& t : threads) {
t.join();
}
}

How to iterate through boost thread specific pointers

I have a multi-thread application. Each thread initializes a struct data type in its own local storage. Some elements are being added to the vectors inside the struct type variables. At the end of the program, I would like to iterate through these thread local storages and add all the results together. How can I iterate through the thread specific pointer so that I can add all the results from the multi threads together ?
Thanks in advance.
boost::thread_specific_ptr<testStruct> tss;
size_t x = 10;
void callable(string str, int x) {
if(!tss.get()){
tss.reset(new testStruct);
(*tss).xInt.resize(x, 0);
}
// Assign some values to the vector elements after doing some calculations
}
Example:
#include <iostream>
#include <vector>
#include <boost/thread/mutex.hpp>
#include <boost/thread/tss.hpp>
#include <boost/thread.hpp>
#include <boost/asio.hpp>
#include <boost/bind.hpp>
#define NR_THREAD 4
#define SAMPLE_SIZE 500
using namespace std;
static bool busy = false;
struct testStruct{
vector<int> intVector;
};
boost::asio::io_service ioService;
boost::thread_specific_ptr<testStruct> tsp;
boost::condition_variable cond;
boost::mutex mut;
void callable(int x) {
if(!tsp.get()){
tsp.reset(new testStruct);
}
(*tsp).intVector.push_back(x);
if (x + 1 == SAMPLE_SIZE){
busy = true;
cond.notify_all();
}
}
int main() {
boost::thread_group threads;
size_t (boost::asio::io_service::*run)() = &boost::asio::io_service::run;
boost::asio::io_service::work work(ioService);
for (short int i = 0; i < NR_THREAD; ++i) {
threads.create_thread(boost::bind(run, &ioService));
}
size_t iterations = 10;
for (int i = 0; i < iterations; i++) {
busy = false;
for (short int j = 0; j < SAMPLE_SIZE; ++j) {
ioService.post(boost::bind(callable, j));
}
// all threads need to finish the job for the next iteration
boost::unique_lock<boost::mutex> lock(mut);
while (!busy) {
cond.wait(lock);
}
cout << "Iteration: " << i << endl;
}
vector<int> sum(SAMPLE_SIZE, 0); // sum up all the values from thread local storages
work.~work();
threads.join_all();
return 0;
}
So, after I haven given some thought to this issue, I have come up with such a solution:
void accumulateTLS(size_t idxThread){
if (idxThread == nr_threads) // Suspend all the threads till all of them are called and waiting here
{
busy = true;
}
boost::unique_lock<boost::mutex> lock(mut);
while (!busy)
{
cond.wait(lock);
}
// Accumulate the variables using thread specific pointer
cond.notify_one();
}
With boost io_service, the callable function can be changed after the threads are initialized. So, after I have done all the calculations, I am sending jobs(as many as the number of threads) to the io service again with callable function accumulateTLS(idxThread). The N jobs are sent to N threads and the accumulation process is done inside accumulateTLS method.
P.S. instead of work.~work(), work.reset() should be used.

Making threads redo a print function in order

This is a home assignment.
Have to print a string(given as input) in small chunks(Size given as input) by multiple threads one at a time in order 1,2,3,1,2,3,1,2(number of threads is given as input).
A thread does this printing function on creation and I want it to redo it after all the other threads. I face two problems:
1. Threads don't print in fixed order(mine gave 1,3,2,4 see output)
2. Threads need to re print till the entire string is exhausted.
This is what I tried...
#include<iostream>
#include<mutex>
#include<thread>
#include<string>
#include<vector>
#include<condition_variable>
#include<chrono>
using namespace std;
class circularPrint{
public:
int pos;
string message;
int nCharsPerPrint;
mutex mu;
condition_variable cv;
circularPrint(){
pos=0;
}
void shared_print(int threadID){
unique_lock<mutex> locker(mu);
if(pos+nCharsPerPrint<message.size())
cout<<"Thread"<<threadID<<" : "<<message.substr(pos,nCharsPerPrint)<<endl;
else if(pos<message.size())
cout<<"Thread"<<threadID<<" : "<<message.substr(pos)<<endl;
pos+=nCharsPerPrint;
}
};
void f(circularPrint &obj,int threadID){
obj.shared_print(threadID);
}
int main(){
circularPrint obj;
cout<<"\nMessage : ";
cin>>obj.message;
cout<<"\nChars : ";
cin>>obj.nCharsPerPrint;
int nthreads;
cout<<"\nThreads : ";
cin>>nthreads;
vector<thread> threads;
for(int count=1;count<=nthreads;++count)
{
threads.push_back(thread(f,ref(obj),count));
}
for(int count=0;count<nthreads;++count)
{
if(threads[count].joinable())
threads[count].join();
}
return 0;
}
Why would you want to multithread a method that can only be executed once at a time?
Anyway, something like this below? Be aware that the take and print use different locks and that there is a chance the output does not show in the expected order (hence, the why question above).
#include <iostream>
#include <mutex>
#include <thread>
#include <string>
#include <vector>
#include <algorithm>
using namespace std;
class circularPrint
{
public:
int pos;
string message;
int nCharsPerPrint;
mutex takeLock;
mutex printLock;
circularPrint() {
pos = 0;
}
string take(int count) {
lock_guard<mutex> locker(takeLock);
count = std::min(count, (int)message.size() - pos);
string substring = message.substr(pos, count);
pos += count;
return substring;
}
void print(int threadID, string& message) {
lock_guard<mutex> locker(printLock);
cout << "Thread" << threadID << " : " << message << endl;
}
void loop(int threadID) {
string message;
while((message = take(nCharsPerPrint)).size() > 0) {
print(threadID, message);
}
}
};
void f(circularPrint &obj, int threadID)
{
obj.loop(threadID);
}
int main()
{
circularPrint obj;
//cout << "\nMessage : ";
//cin >> obj.message;
//cout << "\nChars : ";
//cin >> obj.nCharsPerPrint;
int nthreads;
//cout << "\nThreads : ";
//cin >> nthreads;
nthreads = 4;
obj.message = "123456789012345";
obj.nCharsPerPrint = 2;
vector<thread> threads;
for (int count = 1; count <= nthreads; ++count)
threads.push_back(thread(f, ref(obj), count));
for (int count = 0; count < nthreads; ++count) {
if (threads[count].joinable())
threads[count].join();
}
return 0;
}
Currently each thread exits after printing one message - but you need more messages than threads, so each thread will need to do more than one message.
How about putting an infinite loop around your current locked section, and breaking out when there are no characters left to print?
(You may then find that the first thread does all the work; you can hack that by putting a zero-length sleep outside the locked section, or by making all the threads wait for some single signal to start, or just live with it.)
EDIT: Hadn't properly realised that you wanted to assign work to specific threads (which is normally a really bad idea). But if each thread knows its ID, and how many there are, it can figure out which characters it is supposed to print. Then all it has to do is wait till all the preceding characters have been printed (which it can tell using pos), do its work, then repeat until it has no work left to do and exit.
The only tricky bit is waiting for the preceding work to finish. You can do that with a busy wait (bad), a busy wait with a sleep in it (also bad), or a condition variable (better).
You need inter thread synchronization, each thread doing a loop "print, send a message to next one, wait for a message (from the last thread)".
You can use semaphores, events, messages or something similar.
Something as:
#include <string>
#include <iostream>
#include <condition_variable>
#include <thread>
#include <unistd.h>
using namespace std;
// Parameters passed to a thread.
struct ThreadParameters {
string message; // to print.
volatile bool *exit; // set when the thread should exit.
condition_variable* input; // condition to wait before printing.
condition_variable* output; // condition to set after printing.
};
class CircularPrint {
public:
CircularPrint(int nb_threads) {
nb_threads_ = nb_threads;
condition_variables_ = new condition_variable[nb_threads];
thread_parameters_ = new ThreadParameters[nb_threads];
threads_ = new thread*[nb_threads];
exit_ = false;
for (int i = 0; i < nb_threads; ++i) {
thread_parameters_[i].message = to_string(i + 1);
thread_parameters_[i].exit = &exit_;
// Wait 'your' condition
thread_parameters_[i].input = &condition_variables_[i];
// Then set next one (of first one if you are the last).
thread_parameters_[i].output =
&condition_variables_[(i + 1) % nb_threads];
threads_[i] = new thread(Thread, &thread_parameters_[i]);
}
// Start the dance, free the first thread.
condition_variables_[0].notify_all();
}
~CircularPrint() {
// Ask threads to exit.
exit_ = true;
// Wait for all threads to end.
for (int i = 0; i < nb_threads_; ++i) {
threads_[i]->join();
delete threads_[i];
}
delete[] condition_variables_;
delete[] thread_parameters_;
delete[] threads_;
}
static void Thread(ThreadParameters* params) {
for (;;) {
if (*params->exit) {
return;
}
{
// Wait the mutex. We don't really care, by condition variables
// need a mutex.
// Though the mutex will be useful for the real assignement.
unique_lock<mutex> lock(mutex_);
// Wait for the input condition variable (frees the mutex before waiting).
params->input->wait(lock);
}
cout << params->message << endl;
// Free next thread.
params->output->notify_all();
}
}
private:
int nb_threads_;
condition_variable* condition_variables_;
ThreadParameters* thread_parameters_;
thread** threads_;
bool exit_;
static mutex mutex_;
};
mutex CircularPrint::mutex_;
int main() {
CircularPrint printer(10);
sleep(3);
return 0;
}
using vector<shared_ptr<...>> would be more elegant than just arrays, though this works:
g++ -std=c++11 -o test test.cc -pthread -Wl,--no-as-needed
./test

Extend the life of threads with synchronization (C++11)

I have a program with a function which takes a pointer as arg, and a main. The main is creating n threads, each of them running the function on different memory areas depending on the passed arg. Threads are then joined, the main performs some data mixing between the area and creates n new threads which do the the same operation as the old ones.
To improve the program I would like to keep the threads alive, removing the long time necessary to create them. Threads should sleep when the main is working and notified when they have to come up again. At the same way the main should wait when threads are working as it did with join.
I cannot end up with a strong implementation of this, always falling in a deadlock.
Simple baseline code, any hints about how to modify this would be much appreciated
#include <thread>
#include <climits>
...
void myfunc(void * p) {
do_something(p);
}
int main(){
void * myp[n_threads] {a_location, another_location,...};
std::thread mythread[n_threads];
for (unsigned long int j=0; j < ULONG_MAX; j++) {
for (unsigned int i=0; i < n_threads; i++) {
mythread[i] = std::thread(myfunc, myp[i]);
}
for (unsigned int i=0; i < n_threads; i++) {
mythread[i].join();
}
mix_data(myp);
}
return 0;
}
Here is a possible approach using only classes from the C++11 Standard Library. Basically, each thread you create has an associated command queue (encapsulated in std::packaged_task<> objects) which it continuously check. If the queue is empty, the thread will just wait on a condition variable (std::condition_variable).
While data races are avoided through the use of std::mutex and std::unique_lock<> RAII wrappers, the main thread can wait for a particular job to be terminated by storing the std::future<> object associated to each submitted std::packaged_tast<> and call wait() on it.
Below is a simple program that follows this design. Comments should be sufficient to explain what it does:
#include <thread>
#include <iostream>
#include <sstream>
#include <future>
#include <queue>
#include <condition_variable>
#include <mutex>
// Convenience type definition
using job = std::packaged_task<void()>;
// Some data associated to each thread.
struct thread_data
{
int id; // Could use thread::id, but this is filled before the thread is started
std::thread t; // The thread object
std::queue<job> jobs; // The job queue
std::condition_variable cv; // The condition variable to wait for threads
std::mutex m; // Mutex used for avoiding data races
bool stop = false; // When set, this flag tells the thread that it should exit
};
// The thread function executed by each thread
void thread_func(thread_data* pData)
{
std::unique_lock<std::mutex> l(pData->m, std::defer_lock);
while (true)
{
l.lock();
// Wait until the queue won't be empty or stop is signaled
pData->cv.wait(l, [pData] () {
return (pData->stop || !pData->jobs.empty());
});
// Stop was signaled, let's exit the thread
if (pData->stop) { return; }
// Pop one task from the queue...
job j = std::move(pData->jobs.front());
pData->jobs.pop();
l.unlock();
// Execute the task!
j();
}
}
// Function that creates a simple task
job create_task(int id, int jobNumber)
{
job j([id, jobNumber] ()
{
std::stringstream s;
s << "Hello " << id << "." << jobNumber << std::endl;
std::cout << s.str();
});
return j;
}
int main()
{
const int numThreads = 4;
const int numJobsPerThread = 10;
std::vector<std::future<void>> futures;
// Create all the threads (will be waiting for jobs)
thread_data threads[numThreads];
int tdi = 0;
for (auto& td : threads)
{
td.id = tdi++;
td.t = std::thread(thread_func, &td);
}
//=================================================
// Start assigning jobs to each thread...
for (auto& td : threads)
{
for (int i = 0; i < numJobsPerThread; i++)
{
job j = create_task(td.id, i);
futures.push_back(j.get_future());
std::unique_lock<std::mutex> l(td.m);
td.jobs.push(std::move(j));
}
// Notify the thread that there is work do to...
td.cv.notify_one();
}
// Wait for all the tasks to be completed...
for (auto& f : futures) { f.wait(); }
futures.clear();
//=================================================
// Here the main thread does something...
std::cin.get();
// ...done!
//=================================================
//=================================================
// Posts some new tasks...
for (auto& td : threads)
{
for (int i = 0; i < numJobsPerThread; i++)
{
job j = create_task(td.id, i);
futures.push_back(j.get_future());
std::unique_lock<std::mutex> l(td.m);
td.jobs.push(std::move(j));
}
// Notify the thread that there is work do to...
td.cv.notify_one();
}
// Wait for all the tasks to be completed...
for (auto& f : futures) { f.wait(); }
futures.clear();
// Send stop signal to all threads and join them...
for (auto& td : threads)
{
std::unique_lock<std::mutex> l(td.m);
td.stop = true;
td.cv.notify_one();
}
// Join all the threads
for (auto& td : threads) { td.t.join(); }
}
The concept you want is the threadpool. This SO question deals with existing implementations.
The idea is to have a container for a number of thread instances. Each instance is associated with a function which polls a task queue, and when a task is available, pulls it and run it. Once the task is over (if it terminates, but that's another problem), the thread simply loop over to the task queue.
So you need a synchronized queue, a thread class which implements the loop on the queue, an interface for the task objects, and maybe a class to drive the whole thing (the pool class).
Alternatively, you could make a very specialized thread class for the task it has to perform (with only the memory area as a parameter for instance). This requires a notification mechanism for the threads to indicate that they are done with the current iteration.
The thread main function would be a loop on that specific task, and at the end of one iteration, the thread signals its end, and wait on condition variables to start the next loop. In essence, you would be inlining the task code within the thread, dropping the need of a queue altogether.
using namespace std;
// semaphore class based on C++11 features
class semaphore {
private:
mutex mMutex;
condition_variable v;
int mV;
public:
semaphore(int v): mV(v){}
void signal(int count=1){
unique_lock lock(mMutex);
mV+=count;
if (mV > 0) mCond.notify_all();
}
void wait(int count = 1){
unique_lock lock(mMutex);
mV-= count;
while (mV < 0)
mCond.wait(lock);
}
};
template <typename Task>
class TaskThread {
thread mThread;
Task *mTask;
semaphore *mSemStarting, *mSemFinished;
volatile bool mRunning;
public:
TaskThread(Task *task, semaphore *start, semaphore *finish):
mTask(task), mRunning(true),
mSemStart(start), mSemFinished(finish),
mThread(&TaskThread<Task>::psrun){}
~TaskThread(){ mThread.join(); }
void run(){
do {
(*mTask)();
mSemFinished->signal();
mSemStart->wait();
} while (mRunning);
}
void finish() { // end the thread after the current loop
mRunning = false;
}
private:
static void psrun(TaskThread<Task> *self){ self->run();}
};
classcMyTask {
public:
MyTask(){}
void operator()(){
// some code here
}
};
int main(){
MyTask task1;
MyTask task2;
semaphore start(2), finished(0);
TaskThread<MyTask> t1(&task1, &start, &finished);
TaskThread<MyTask> t2(&task2, &start, &finished);
for (int i = 0; i < 10; i++){
finished.wait(2);
start.signal(2);
}
t1.finish();
t2.finish();
}
The proposed (crude) implementation above relies on the Task type which must provide the operator() (ie. a functor like class). I said you could incorporate the task code directly in the thread function body earlier, but since I don't know it, I kept it as abstract as I could. There's one condition variable for the start of threads, and one for their end, both encapsulated in semaphore instances.
Seeing the other answer proposing the use of boost::barrier, I can only support this idea: make sure to replace my semaphore class with that class if possible, the reason being that it is better to rely on well tested and maintained external code rather than a self implemented solution for the same feature set.
All in all, both approaches are valid, but the former gives up a tiny bit of performance in favor of flexibility. If the task to be performed takes a sufficiently long time, the management and queue synchronization cost becomes negligible.
Update: code fixed and tested. Replaced a simple condition variable by a semaphore.
It can easily be achieved using a barrier (just a convenience wrapper over a conditional variable and a counter). It basically blocks until all N threads have reached the "barrier". It then "recycles" again. Boost provides an implementation.
void myfunc(void * p, boost::barrier& start_barrier, boost::barrier& end_barrier) {
while (!stop_condition) // You'll need to tell them to stop somehow
{
start_barrier.wait ();
do_something(p);
end_barrier.wait ();
}
}
int main(){
void * myp[n_threads] {a_location, another_location,...};
boost::barrier start_barrier (n_threads + 1); // child threads + main thread
boost::barrier end_barrier (n_threads + 1); // child threads + main thread
std::thread mythread[n_threads];
for (unsigned int i=0; i < n_threads; i++) {
mythread[i] = std::thread(myfunc, myp[i], start_barrier, end_barrier);
}
start_barrier.wait (); // first unblock the threads
for (unsigned long int j=0; j < ULONG_MAX; j++) {
end_barrier.wait (); // mix_data must not execute before the threads are done
mix_data(myp);
start_barrier.wait (); // threads must not start new iteration before mix_data is done
}
return 0;
}
The following is a simple compiling and working code performing some random stuffs. It implements aleguna's concept of barrier. The task length of each thread is different so it is really necessary to have a strong synchronization mechanism. I will try to do a pool on the same tasks and benchmark the result, and then maybe with futures as pointed out by Andy Prowl.
#include <iostream>
#include <thread>
#include <mutex>
#include <condition_variable>
#include <chrono>
#include <complex>
#include <random>
const unsigned int n_threads=4; //varying this will not (almost) change the total amount of work
const unsigned int task_length=30000/n_threads;
const float task_length_variation=task_length/n_threads;
unsigned int rep=1000; //repetitions of tasks
class t_chronometer{
private:
std::chrono::steady_clock::time_point _t;
public:
t_chronometer(): _t(std::chrono::steady_clock::now()) {;}
void reset() {_t = std::chrono::steady_clock::now();}
double get_now() {return std::chrono::duration_cast<std::chrono::duration<double>>(std::chrono::steady_clock::now() - _t).count();}
double get_now_ms() {return
std::chrono::duration_cast<std::chrono::duration<double,std::milli>>(std::chrono::steady_clock::now() - _t).count();}
};
class t_barrier {
private:
std::mutex m_mutex;
std::condition_variable m_cond;
unsigned int m_threshold;
unsigned int m_count;
unsigned int m_generation;
public:
t_barrier(unsigned int count):
m_threshold(count),
m_count(count),
m_generation(0) {
}
bool wait() {
std::unique_lock<std::mutex> lock(m_mutex);
unsigned int gen = m_generation;
if (--m_count == 0)
{
m_generation++;
m_count = m_threshold;
m_cond.notify_all();
return true;
}
while (gen == m_generation)
m_cond.wait(lock);
return false;
}
};
using namespace std;
void do_something(complex<double> * c, unsigned int max) {
complex<double> a(1.,0.);
complex<double> b(1.,0.);
for (unsigned int i = 0; i<max; i++) {
a *= polar(1.,2.*M_PI*i/max);
b *= polar(1.,4.*M_PI*i/max);
*(c)+=a+b;
}
}
bool done=false;
void task(complex<double> * c, unsigned int max, t_barrier* start_barrier, t_barrier* end_barrier) {
while (!done) {
start_barrier->wait ();
do_something(c,max);
end_barrier->wait ();
}
cout << "task finished" << endl;
}
int main() {
t_chronometer t;
std::default_random_engine gen;
std::normal_distribution<double> dis(.0,1000.0);
complex<double> cpx[n_threads];
for (unsigned int i=0; i < n_threads; i++) {
cpx[i] = complex<double>(dis(gen), dis(gen));
}
t_barrier start_barrier (n_threads + 1); // child threads + main thread
t_barrier end_barrier (n_threads + 1); // child threads + main thread
std::thread mythread[n_threads];
unsigned long int sum=0;
for (unsigned int i=0; i < n_threads; i++) {
unsigned int max = task_length + i * task_length_variation;
cout << i+1 << "th task length: " << max << endl;
mythread[i] = std::thread(task, &cpx[i], max, &start_barrier, &end_barrier);
sum+=max;
}
cout << "total task length " << sum << endl;
complex<double> c(0,0);
for (unsigned long int j=1; j < rep+1; j++) {
start_barrier.wait (); //give to the threads the missing call to start
if (j==rep) done=true;
end_barrier.wait (); //wait for the call from each tread
if (j%100==0) cout << "cycle: " << j << endl;
for (unsigned int i=0; i<n_threads; i++) {
c+=cpx[i];
}
}
for (unsigned int i=0; i < n_threads; i++) {
mythread[i].join();
}
cout << "result: " << c << " it took: " << t.get_now() << " s." << endl;
return 0;
}