I'm trying to create a producer-consumer program, where the consumers must keep running until all the producers are finished, then consume what's left in the queue (if there's anything left) and then end. You can check my code bellow, I think I know where the problem (probably deadlock) is, but I don't know how to make it work properly.
#include<iostream>
#include<cstdlib>
#include <queue>
#include <thread>
#include <mutex>
#include <condition_variable>
using namespace std;
class Company{
public:
Company() : producers_done(false) {}
void start(int n_producers, int n_consumers); // start customer&producer threads
void stop(); // join all threads
void consumer();
void producer();
/* some other stuff */
private:
condition_variable cond;
mutex mut;
bool producers_done;
queue<int> products;
vector<thread> producers_threads;
vector<thread> consumers_threads;
/* some other stuff */
};
void Company::consumer(){
while(!products.empty()){
unique_lock<mutex> lock(mut);
while(products.empty() && !producers_done){
cond.wait(lock); // <- I think this is where the deadlock happens
}
if (products.empty()){
break;
}
products.pop();
cout << "Removed product " << products.size() << endl;
}
}
void Company::producer(){
while(true){
if((rand()%10) == 0){
break;
}
unique_lock<mutex> lock(mut);
products.push(1);
cout << "Added product " << products.size() << endl;
cond.notify_one();
}
}
void Company::stop(){
for(auto &producer_thread : producers_threads){
producer_thread.join();
}
unique_lock<mutex> lock(mut);
producers_done = true;
cout << "producers done" << endl;
cond.notify_all();
for(auto &consumer_thread : consumers_threads){
consumer_thread.join();
}
cout << "consumers done" << endl;
}
void Company::start(int n_producers, int n_consumers){
for(int i = 0; i<n_producers; ++i){
producers_threads.push_back(thread(&Company::producer, this));
}
for(int i = 0; i<n_consumers; ++i){
consumers_threads.push_back(thread(&Company::consumer, this));
}
}
int main(){
Company c;
c.start(2, 2);
c.stop();
return true;
}
I know, there are a lot of producer-consumer related questions here, and I've scrolled through at least 10 of them, but none provided answer to my issue.
When people use std::atomic along with std::mutex and std::condition_variable that results in deadlock in almost 100% of cases. This is because modifications to that atomic variable are not protected by the mutex and hence condition variable notifications get lost when that variable is updated after the mutex is locked but before condition variable wait in the consumer.
A fix would be to not use std::atomic and only modify and read producers_done while the mutex is held. E.g.:
void Company::consumer(){
for(;;){
unique_lock<mutex> lock(mut);
while(products.empty() && !producers_done)
cond.wait(lock);
if(products.empty())
break;
orders.pop();
}
}
Another error in the code is that in while(!products.empty()) it calls products.empty() without holding the mutex, resulting in a race condition.
The next error is keeping the mutex locked while waiting for the consumer threads to terminate. Fix:
{
unique_lock<mutex> lock(mut);
producers_done = true;
// mutex gets unlocked here.
}
cond.notify_all();
for(auto &consumer_thread : consumers_threads)
consumer_thread.join();
Related
Currently working on a project, im struggeling with threading and queue at the moment, the issue is that all threads take the same item in the queue.
Reproduceable example:
#include <iostream>
#include <queue>
#include <thread>
using namespace std;
void Test(queue<string> queue){
while (!queue.empty()) {
string proxy = queue.front();
cout << proxy << "\n";
queue.pop();
}
}
int main()
{
queue<string> queue;
queue.push("101.132.186.39:9090");
queue.push("95.85.24.83:8118");
queue.push("185.211.193.162:8080");
queue.push("87.106.37.89:8888");
queue.push("159.203.61.169:8080");
std::vector<std::thread> ThreadVector;
for (int i = 0; i <= 10; i++){
ThreadVector.emplace_back([&]() {Test(queue); });
}
for (auto& t : ThreadVector){
t.join();
}
ThreadVector.clear();
return 0;
}
You are giving each thread its own copy of the queue. I imagine that what you want is all the threads to work on the same queue and for that you will need to use some synchronization mechanism when multiple threads work on the shared queue as std queue is not thread safe.
edit: minor note: in your code you are spawning 11 threads not 10.
edit 2: OK, try this one to begin with:
std::mutex lock_work;
std::mutex lock_io;
void Test(queue<string>& queue){
while (!queue.empty()) {
string proxy;
{
std::lock_guard<std::mutex> lock(lock_work);
proxy = queue.front();
queue.pop();
}
{
std::lock_guard<std::mutex> lock(lock_io);
cout << proxy << "\n";
}
}
}
Look at this snippet:
void Test(std::queue<std::string> queue) { /* ... */ }
Here you pass a copy of the queue object to the thread.
This copy is local to each thread, so it gets destroyed after every thread exits so in the end your program does not have any effect on the actual queue object that resides in the main() function.
To fix this, you need to either make the parameter take a reference or a pointer:
void Test(std::queue<std::string>& queue) { /* ... */ }
This makes the parameter directly refer to the queue object present inside main() instead of creating a copy.
Now, the above code is still not correct since queue is prone to data-race and neither std::queue nor std::cout is thread-safe and can get interrupted by another thread while currently being accessed by one. To prevent this, use a std::mutex:
// ...
#include <mutex>
// ...
// The mutex protects the 'queue' object from being subjected to data-race amongst different threads
// Additionally 'io_mut' is used to protect the streaming operations done with 'std::cout'
std::mutex mut, io_mut;
void Test(std::queue<std::string>& queue) {
std::queue<std::string> tmp;
{
// Swap the actual object with a local temporary object while being protected by the mutex
std::lock_guard<std::mutex> lock(mut);
std::swap(tmp, queue);
}
while (!tmp.empty()) {
std::string proxy = tmp.front();
{
// Call to 'std::cout' needs to be synchronized
std::lock_guard<std::mutex> lock(io_mut);
std::cout << proxy << "\n";
}
tmp.pop();
}
}
This synchronizes each thread call and prevents access from any other threads while queue is still being accessed by a thread.
Edit:
Alternatively, it'd be much faster in my opinion to make each thread wait until one of them receives a notification of your push to std::queue. You can do this through the use of std::condition_variable:
// ...
#include <mutex>
#include <condition_variable>
// ...
std::mutex mut1, mut2;
std::condition_variable cond;
void Test(std::queue<std::string>& queue, std::chrono::milliseconds timeout = std::chrono::milliseconds{10}) {
std::unique_lock<std::mutex> lock(mut1);
// Wait until 'queue' is not empty...
cond.wait(lock, [queue] { return queue.empty(); });
while (!queue.empty()) {
std::string proxy = std::move(queue.front());
std::cout << proxy << "\n";
queue.pop();
}
}
// ...
int main() {
std::queue<string> queue;
std::vector<std::thread> ThreadVector;
for (int i = 0; i <= 10; i++)
ThreadVector.emplace_back([&]() { Test(queue); });
// Notify the vectors of each 'push()' call to 'queue'
{
std::unique_lock<std::mutex> lock(mut2);
queue.push("101.132.186.39:9090");
cond.notify_one();
}
{
std::unique_lock<std::mutex> lock(mut2);
queue.push("95.85.24.83:8118");
cond.notify_one();
}
{
std::unique_lock<std::mutex> lock(mut2);
queue.push("185.211.193.162:8080");
cond.notify_one();
}
{
std::unique_lock<std::mutex> lock(mut2);
queue.push("87.106.37.89:8888");
cond.notify_one();
}
{
std::unique_lock<std::mutex> lock(mut2);
queue.push("159.203.61.169:8080");
cond.notify_one();
}
for (auto& t : ThreadVector)
t.join();
ThreadVector.clear();
}
For below program, thread Pool always picks the same thread ID 0x7000095f9000! Why so?
Should every push condi.notify_one() wake up all threads same time? What could be the reason same thread ID get picked?
Computer supports 3 threads.
Any other info on using function objects would be helpful!!
O/P
Checking if not empty
Not Empty
0x700009576000 0
Checking if not empty
Checking if not empty
Checking if not empty
Not Empty
0x7000095f9000 1
Checking if not empty
Not Empty
0x7000095f9000 2
Checking if not empty
Not Empty
0x7000095f9000 3
Checking if not empty
Not Empty
0x7000095f9000 4
Checking if not empty
Not Empty
0x7000095f9000 5
Checking if not empty
Code
#include <iostream>
#include <vector>
#include <queue>
#include <thread>
#include <condition_variable>
#include <chrono>
using namespace std;
class TestClass{
public:
void producer(int i) {
unique_lock<mutex> lockGuard(mtx);
Q.push(i);
cond.notify_all();
}
void consumer() {
{
unique_lock<mutex> lockGuard(mtx);
cout << "Checking if not empty" << endl;
cond.wait(lockGuard, [this]() {
return !Q.empty();
});
cout << "Not Empty" << endl;
cout << this_thread::get_id()<<" "<<Q.front()<<endl;
Q.pop();
}
};
void consumerMain() {
while(1) {
consumer();
std::this_thread::sleep_for(chrono::seconds(1));
}
}
private:
mutex mtx;
condition_variable cond;
queue<int> Q;
};
int main()
{
std::vector<std::thread> vecOfThreads;
std::function<void(TestClass&)> func = [&](TestClass &obj) {
while(1) {
obj.consumer();
}
};
unsigned MAX_THREADS = std::thread::hardware_concurrency()-1;
TestClass obj;
for(int i=0; i<MAX_THREADS; i++) {
std::thread th1(func, std::ref(obj));
vecOfThreads.emplace_back(std::move(th1));
}
for(int i=0; i<4*MAX_THREADS/2; i++) {
obj.producer(i);
}
for (std::thread & th : vecOfThreads)
{
if (th.joinable())
th.join();
}
return 0;
}
Any other info on using function objects would be helpful!! Thanks in advance!!
Any other pointers?
The very short unlocking of the mutex that happens in the consumer threads will in your case most probably let the running thread acquire the lock again, and again and again.
If you instead simulate some work being done after the workload has been picked from the queue by calling consumerMain (which sleeps a little) instead of consumer, you would likely see different threads picking up the workload.
while(1) {
obj.consumerMain();
}
I am learning condition variables in C++11 and wrote this program based on a sample code.
The goal is to accumulate in a vector the first ten natural integers that are generated by a producer and pushed into the vector by a consumer. However it does not work since, for example on some runs, the vector only contains 1, 7 and 10.
#include <mutex>
#include <condition_variable>
#include<vector>
#include <iostream>
#include <cstdio>
std::mutex mut;
#define MAX 10
int counter;
bool isIncremented = false;
std::vector<int> vec;
std::condition_variable condvar;
void producer() {
while (counter < MAX) {
std::lock_guard<std::mutex> lg(mut);
++counter;
isIncremented = true;
condvar.notify_one();
}
}
void consumer() {
while (true) {
std::unique_lock<std::mutex> ul(mut);
condvar.wait(ul, [] { return isIncremented; });
vec.push_back(counter);
isIncremented = false;
if (counter >= MAX) {
break;
}
}
}
int main(int argc, char *argv[]) {
std::thread t1(consumer);
std::thread t2(producer);
t2.join();
t1.join();
for (auto i : vec) {
std::cout << i << ", ";
}
std::cout << std::endl;
// Expected output: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10,
// Example of actual output: 1, 7, 10,
std::cout << "Press enter to quit";
getchar();
return 0;
}
The problem is that you only remember the last number your producer produced. And your producer never waits until the consumer has consumed what it produced. If your producer thread gets to do more than one iteration of its loop before the consumer thread gets to run (which is not unlikely since the loop doesn't do much), the consumer will only see the last number the producer produced and only push that one into the vector…
To solve this problem, either use a second condition variable to make the producer wait for someone to pick up the last result it produced, or use something that can store more than one result between producer and consumer, or a combination thereof…
Note: Notifying a condition variable is not a blocking call. If it were, it would have to ask you to hand over the mutex so it can internally release it or you'd end up in a deadlock. notify_one() will just wake up one of the threads that are waiting on the condition variable and return. The wait call that the woken thread was blocking on will reacquire the mutex before it returns. In your case, it's not unlikely that the consumer thread be woken and then fail to reacquire the mutex and block again right away because your producer thread is still holding on to the mutex when it's calling notify_one(). Thus, as a general rule of thumb, you want to release the mutex associated with a condition variable should you be holding it before you call notify…
A side note, apparently you used the lock_guard<> in producer, but unique_lock in consumer. In the consumer, the unique_lock also doesn't seem to guard the share resource exclusively.
Below is a modified code that uses unique_lock in both producer and consumer, that guard against shared resource counter.
The code adds a sleep in the producer so that the consumer can be notified of the counter change.
Output seems to be as expected.
#include <mutex>
#include <condition_variable>
#include<vector>
#include <iostream>
#include <cstdio>
#include <thread>
#include <chrono>
std::mutex mut;
#define MAX 10
int counter = 0;
bool isIncremented = false;
std::vector<int> vec;
std::condition_variable condvar;
void producer() {
while (counter < MAX) {
std::unique_lock<std::mutex> lg(mut);
++counter;
isIncremented = true;
lg.unlock();
condvar.notify_one();
std::this_thread::sleep_for(std::chrono::milliseconds(10));
}
}
void consumer() {
while (true) {
std::unique_lock<std::mutex> ul(mut);
condvar.wait(ul, [] { return isIncremented; });
vec.push_back(counter);
isIncremented = false;
if (counter >= MAX) {
break;
}
ul.unlock();
}
}
int main(int argc, char *argv[]) {
std::thread t1(consumer);
std::thread t2(producer);
t2.join();
t1.join();
for (auto i : vec) {
std::cout << i << ", ";
}
std::cout << std::endl;
return 0;
}
Using #MichaelKenzel suggestions from the answer, here is a working example. std::queue is used in order to store more than one result between producer and consumer.
#include<mutex>
#include<condition_variable>
#include<vector>
#include<iostream>
#include<cstdio>
#include<thread>
#include<queue>
std::mutex mut;
#define MAX 10
int counter;
std::queue<int> data_queue;
std::vector<int> vec;
std::condition_variable condvar;
void producer()
{
while (counter < MAX)
{
++counter;
std::lock_guard<std::mutex> lg(mut);
data_queue.push(counter);
condvar.notify_one();
}
}
void consumer()
{
while (true)
{
std::unique_lock<std::mutex> ul(mut);
condvar.wait(ul, [] { return !data_queue.empty(); });
int data = data_queue.front();
data_queue.pop();
ul.unlock();
vec.push_back(data);
if (data >= MAX)
{
break;
}
}
}
int main(int argc, char *argv[])
{
std::thread t1(consumer);
std::thread t2(producer);
t2.join();
t1.join();
for (auto i : vec)
{
std::cout << i << ", ";
}
std::cout << std::endl;
return 0;
}
The thread1 function does not seem to get executed
#include <iostream>
#include <fstream>
#include <thread>
#include <condition_variable>
#include <queue>
std::condition_variable cv;
std::mutex mu;
std::queue<int> queue;
bool ready;
static void thread1() {
while(!ready) {std::this_thread::sleep_for(std::chrono::milliseconds(10));}
while(ready && queue.size() <= 4) {
std::unique_lock<std::mutex> lk(mu);
cv.wait(lk, [&]{return !queue.empty();});
queue.push(2);
}
}
int main() {
ready = false;
std::thread t(thread1);
while(queue.size() <= 4) {
{
std::lock_guard<std::mutex> lk(mu);
queue.push(1);
}
ready = true;
cv.notify_one();
}
t.join();
for(int i = 0; i <= queue.size(); i++) {
int a = queue.front();
std::cout << a << std::endl;
queue.pop();
}
return 0;
}
On my Mac the output is 1 2 1 2 but in my ubuntu its 1 1 1. I'm compiling with g++ -std=c++11 -pthread -o thread.out thread.cpp && ./thread.out. Am I missing something?
This:
for(int i = 0; i <= queue.size(); i++) {
int a = queue.front();
std::cout << a << std::endl;
queue.pop();
}
Is undefined behavior. A for loop that goes from 0 to size runs size+1 times. I would suggest that you write this in the more idiomatic style for a queue:
while(!queue.empty()) {
int a = queue.front();
std::cout << a << std::endl;
queue.pop();
}
When I run this on coliru, which I assume runs some kind of *nix machine, I get 4 1's: http://coliru.stacked-crooked.com/a/8de5b01e87e8549e.
Again, you haven't specified anything that would force each thread to run a certain amount of times. You only (try to*) cause an invariant where the queue will reach size 4, either way. It just happens to be that on the machines that we ran it on, thread 2 never manages to acquire the mutex.
This example will be more interesting if you add more work or even (just for pedagogical purposes) delays at various points. Simulating that the two threads are actually doing work. If you add sleeps at various points you can ensure that the two threads alternate, though depending where you add them you may see your invariant of 4 elements in the thread break!
*Note that even your 4 element invariant on the queue, is not really an invariant. It is possible (though very unlikely) that both threads pass the while condition at the exact same moment, when there are 3 elements in the queue. One acquires the lock first and pushes, and then the other. So you can end up with 5 elements in the queue! (as you can see, asynchronous programming is tricky). In particular you really need to check the queue size when you have the lock in order for this to work.
I was able to solve this by making the second thread wait on a separate predicate on a separate conditional variable. I'm not sure if queue.size() is thread safe.
#include <iostream>
#include <fstream>
#include <thread>
#include <condition_variable>
#include <queue>
std::condition_variable cv;
std::condition_variable cv2;
std::mutex mu;
std::queue<int> queue;
bool tick;
bool tock;
static void thread1() {
while(queue.size() < 6) {
std::unique_lock<std::mutex> lk(mu);
cv2.wait(lk, []{return tock;});
queue.push(1);
tock = false;
tick = true;
cv.notify_one();
}
}
int main() {
tick = false;
tock = true;
std::thread t(thread1);
while(queue.size() < 6) {
std::unique_lock<std::mutex> lk(mu);
cv.wait(lk, []{return tick;});
queue.push(2);
tick = false;
tock = true;
cv2.notify_one();
}
t.join();
while(!queue.empty()) {
int r = queue.front();
queue.pop();
std::cout << r << std::endl;
}
return 0;
}
So I have two threads where they share the same variable, 'counter'. I want to synchronize my threads by only continuing execution once both threads have reached that point. Unfortunately I enter a deadlock state as my thread isn't changing it's checking variable. The way I have it is:
volatile int counter = 0;
Thread() {
- some calculations -
counter++;
while(counter != 2) {
std::this_thread::yield();
}
counter = 0;
- rest of the calculations -
}
The idea is that since I have 2 threads, once they reach that point - at different times - they will increment the counter. If the counter isn't equal to 2, then the thread that reached there first will have to wait until the other has incremented the counter so that they are synced up. Does anyone know where the issue lies here?
To add more information about the problem, I have two threads which perform half of the operations on an array. Once they are done, I want to make sure that they both have completed finish their calculations. Once they are, I can signal the printer thread to wake up and perform it's operation of printing and clearing the array. If I do this before both threads have completed, there will be issues.
Pseudo code:
Thread() {
getLock()
1/2 of the calculations on array
releaseLock()
wait for both to finish - this is the issue
wake up printer thread
}
In situations like this, you must use an atomic counter.
std::atomic_uint counter = 0;
In the given example, there is also no sign that counter got initialized.
You are probably looking for std::conditional_variable: A conditional variable allows one thread to signal to another thread. Because it doesn't look like you are using the counter, and you're only using it for synchronisation, here is some code from another answer (disclaimer: it's one of my answers) that shows std::conditional_variable processing logic on different threads, and performing synchronisation around a value:
unsigned int accountAmount;
std::mutex mx;
std::condition_variable cv;
void depositMoney()
{
// go to the bank etc...
// wait in line...
{
std::unique_lock<std::mutex> lock(mx);
std::cout << "Depositing money" << std::endl;
accountAmount += 5000;
}
// Notify others we're finished
cv.notify_all();
}
void withdrawMoney()
{
std::unique_lock<std::mutex> lock(mx);
// Wait until we know the money is there
cv.wait(lock);
std::cout << "Withdrawing money" << std::endl;
accountAmount -= 2000;
}
int main()
{
accountAmount = 0;
// Run both threads simultaneously:
std::thread deposit(&depositMoney);
std::thread withdraw(&withdrawMoney);
// Wait for both threads to finish
deposit.join();
withdraw.join();
std::cout << "All transactions processed. Final amount: " << accountAmount << std::endl;
return 0;
}
I would look into using a countdown latch. The idea is to have one or more threads block until the desired operation is completed. In this case you want to wait until both threads are finished modifying the array.
Here is a simple example:
#include <condition_variable>
#include <mutex>
#include <thread>
class countdown_latch
{
public:
countdown_latch(int count)
: count_(count)
{
}
void wait()
{
std::unique_lock<std::mutex> lock(mutex_);
while (count_ > 0)
condition_variable_.wait(lock);
}
void countdown()
{
std::lock_guard<std::mutex> lock(mutex_);
--count_;
if (count_ == 0)
condition_variable_.notify_all();
}
private:
int count_;
std::mutex mutex_;
std::condition_variable condition_variable_;
};
and usage would look like this
std::atomic<int> result = 0;
countdown_latch latch(2);
void perform_work()
{
++result;
latch.countdown();
}
int main()
{
std::thread t1(perform_work);
std::thread t2(perform_work);
latch.wait();
std::cout << "result = " << result;
t1.join();
t2.join();
}