Producer and Consumer Problem waiting threads

Question

I tried to code Producer and Consumer Problem but after producing up to a certain value -(val) produce thread ends. but the consumer thread's condition variable waiting for notification but there is no thread to notify so my program does not end. What changes should I made?

#include<bits/stdc++.h>
using namespace std;
mutex m;
vector<int>buffer;
condition_variable cv;
void producer(int val){
    while(val){
        std::unique_lock<mutex>lock(m);
        cv.wait(lock, []{ return buffer.size()<50; });
        buffer.push_back(val--);
        cout<<"Produced : "<<buffer.back()<<endl;
        lock.unlock();
        cv.notify_one();
      }
}
void consumer(){
    while(1){
        std::unique_lock<mutex>lock(m);
        cv.wait(lock, []{return buffer.size() > 0 ; });
        cout<<"Consumed : "<<buffer.back()<<endl;
        buffer.pop_back();
        lock.unlock();
        cv.notify_one();
    }
}

int main()
{
    int attempt=1;
     thread t1(producer,30);
    thread t2(consumer);
  
    t1.join();
     t2.join();
 

    return 0;

}

use another condition variable to notify the producer that the consumer is gone — 463035818_is_not_an_ai, Mar 25 '22 at 06:59
The producer will get over first when it reaches the Limit(Loop ends) but the consumer will be waiting for the notification. — VEMBU KARTHICK T, Mar 25 '22 at 07:08
Add a Boolean flag which is set when the producer ends and can be checked by the consumer to stop waiting — Alan Birtles, Mar 25 '22 at 07:13

score 1 · Answer 1 · answered Mar 25 '22 at 07:27

1

You need to have another variable which will indicate that your producer will produce no more so your consumer will know that

it doesn't need to wait for the producer anymore.
if it consumed the latest item it should stop the loop.

So the minimum modification of your code might look as follows:

using namespace std;
mutex m;
vector<int>buffer;
condition_variable cv;
bool producerExhausted = false;
void producer(int val){
    while(val){
        std::unique_lock<mutex>lock(m);
        cv.wait(lock, []{ return buffer.size()<50; });
        buffer.push_back(val--);
        cout<<"Produced : "<<buffer.back()<<endl;
        producerExhausted = val == 0;
        lock.unlock();
        cv.notify_one();
    }
}
void consumer(){
    while(1){
        std::unique_lock<mutex>lock(m);
        if(!producerExhausted)
            cv.wait(lock, []{return buffer.size() > 0 ; });
        cout<<"Consumed : "<<buffer.back()<<endl;
        buffer.pop_back();
        if(producerExhausted && buffer.empty())
            break;
        lock.unlock();
        cv.notify_one();
    }
}

answered Mar 25 '22 at 07:27

ixSci

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its simpler to `producerExhausted = true;` after the loop. In your code the condition `val == 0` is present twice, but no matter what the condition the producer can notify the consumer when its loop finished – 463035818_is_not_an_ai Mar 25 '22 at 08:08
@463035818_is_not_a_number no, you can't set it after the loop. It should be set inside the mutex, otherwise that variable is useless. – ixSci Mar 25 '22 at 08:10
oh right, one could use a different mutex, or `std::atomic`. Its additional cost, but that cost is only once and only when producer and consumer are finsihed anyhow – 463035818_is_not_an_ai Mar 25 '22 at 08:13
@463035818_is_not_a_number it would be an additional cost per iteration on `consumer` not once on `producer`, also it would require adding another notification after the loop ended. Yes, it is also a possible solution but I don't like it more than what I did. – ixSci Mar 25 '22 at 08:16
its a minor. I am only worried about `while(val)` and `producerExhausted = val == 0;` which could be done in one place. I didnt consider the issues you mention. Perhaps I'd use a `auto produce_more = [&]() { return val != 0; };` and call that in both places, just to have the stop condition only once. But yeah your code is fine as is – 463035818_is_not_an_ai Mar 25 '22 at 08:20

Producer and Consumer Problem waiting threads

1 Answers1