Print Odd Even in Separate threads in C++| TechTrendings

This is a popular question asked for experience programmer while testing multithreading knowledge. We can do it in two ways.

Approach 1: Using lock mechanism

#include<iostream>
#include<thread>
#include<mutex>

using namespace std;

std::mutex m;
int count = 0;

void printEven()
{
    cout<< "Entered Even\n"<<endl;
    while(count <= 10)
    {
       m.lock();
       if(count%2 == 0)
           cout<< count++ << "";
       m.unlock();
     }
}

void printOdd()
{
    cout<< "Entered odd\n"<<endl;
    while(count < 10)
    {
       m.lock();
       if(count%2 == 1)
           cout<< count++ << "";
       m.unlock();
     }
}

int main()
{
    std::thread t1(printOdd);
    std::thread t2(printEven);
    t1.join();
    t2.join();
    return 0;
}

Approach 2: Using Condition variable

#include <iostream>
#include <thread>
#include <mutex>
#include <condition_variable>

std::mutex mu;
std::condition_variable cond;
int count = 1;

void PrintOdd()
{
    for(; count < 10;)
    {
        std::unique_lock<std::mutex> locker(mu);
        cond.wait(locker,[](){ return (count%2 == 1); });
        std::cout << "From Odd:    " << count << std::endl;
        count++;
        locker.unlock();
        cond.notify_all();
    }

}

void PrintEven()
{
    for(; count < 10;)
    {
        std::unique_lock<std::mutex> locker(mu);
        cond.wait(locker,[](){ return (count%2 == 0); });
        std::cout << "From Even: " << count << std::endl;
        count++;
        locker.unlock();
        cond.notify_all();
    }
}

int main()
{
    std::thread t1(PrintOdd);
    std::thread t2(PrintEven);
    t1.join();
    t2.join();
    return 0;
}