printing odd and even number printing alternately using threads in C++

Question

Odd even number printing using thread I came across this question and wanted to discuss solution in C++ . What I can think of using 2 binary semaphores odd and even semaphore. even semaphore initialized to 1 and odd initialized to 0.

**T1 thread function** 
funOdd()
{  
  wait(even)  
  print odd;  
  signal(odd)  
}


**T2 thread function**
funEven()  
{  
  wait(odd)  
  print even  
  signal(even)  
}  

In addition to this if my functions are generating only number and there is a third thread T3 which is going to print those numbers then what should be ideal design ? I used an array where odd number will be placed at odd place and even number will be place at even position. T3 will read from this array this will avoid any thread saftey over this array and if T3 does not find any index then it will wait till that index gets populated. Another solution can be to use a queue which will have a mutex which can be used by T1 and T2 while insertion.

Please comment on this solution and how can i make it more efficient.

Edit to make problem much clear: Overall problem is that I have two producers (T1,T2) and a single consumer (T3), and my producers are interdependent.

Answer 1

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 < 100;)
    {
        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 < 100;)
    {
        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;
}

Answer 2

Solution using condition variable.

#include<iostream>
#include<thread>
#include<mutex>
using namespace std;
mutex oddevenMu;
condition_variable condVar;
int number = 1;

void printEvenOdd(bool isEven, int maxnubmer)
{
    unique_lock<mutex> ul(oddevenMu);
    while (number < maxnubmer)
    {
        condVar.wait(ul, [&]() {return number % 2 == isEven;});
        cout << number++ << " ";
        condVar.notify_all();
    }

}

int main(string args[])
{
    thread oddThread(printEvenOdd, false, 100);
    thread evenThread(printEvenOdd, true, 100);
    oddThread.join();
    evenThread.join();
    return 0;
}

Answer 3

 #include  <stdio.h>
 #include  <stdlib.h>
 #include  <iostream>
 #include  <pthread.h>
 #include  <semaphore.h>

  sem_t sem;
  sem_t sem2;
  using namespace std ;

int count = 1;

void increment(int x)
{
    cout << "called by thread : " << x << "count is : " << count ++ << "\n";
}

void *printAltmessage1(void *thread_value)
{
    for(int m=0; m < (*(int *)thread_value); m++)
    {
        if (sem_wait(&sem) == 0)
        {
            cout << " Thread printAltmessage1 is executed" <<"\n";  
            increment(1);
            sem_post(&sem2);
        }
    }
}

void *printAltmessage2(void *thread_value)
{
    for(int m=0; m < (*(int *)thread_value); m++)
    {
        if (sem_wait(&sem2) == 0)
        {
            cout << " Thread printAltmessage2 is executed" <<"\n";
            increment(2);  
            sem_post(&sem);
        }
    }
}

int main()
{
     sem_init(&sem,0, 1);
     sem_init(&sem2,0, 0);
     pthread_t threads[2];
     int x =8;
     for(int i=0;i<2;i++)
     {
          if(i==0)
          int rc =pthread_create(&threads[i],NULL,printAltmessage1,(void*)&x);
          else
          int rc =pthread_create(&threads[i],NULL,printAltmessage2,(void*)&x);
      }
      pthread_exit(NULL);
      return 0;
}

Answer 4

This is the easiest solution you can refer:

#include<iostream>
#include<mutex>
#include<pthread.h>
#include<cstdlib>
int count=0;
using namespace std;
mutex m;
void* printEven(void *a)
{
   while(1)
   {
       m.lock();
       if(count%2==0)
       {
          cout<<" I am Even"<<count<<endl;
          count++;
       }
       if(count==100)
           break;
       m.unlock();
   }
}
void* printOdd(void *b)
{
    while(1)
    {
       m.lock();
       if(count%2!=0)
       {
           cout<<"I am odd"<<count<<endl;
           count++;
       }
       if(count>100)
          break;
       m.unlock();
    }
 }
 int main()
 {
     int *ptr = new int();
     pthread_t thread1, thread2;
     pthread_attr_t attr;
     pthread_attr_init(&attr);
     pthread_create(&thread1,&attr,&printEven,NULL);
     pthread_create(&thread2,&attr,&printOdd, NULL);
     pthread_join(thread1,&ptr);
     pthread_join(thread2,&ptr);
     delete ptr;
 }

Answer 5

This is simple solution using single function.

#include <iostream>
#include <thread>
#include <condition_variable>
using namespace std;

mutex mu;
condition_variable cond;
int count = 1;

void PrintOddAndEven(bool even, int n){
    while(count < n){
        unique_lock<mutex> lk(mu);
        cond.wait(lk, [&](){return count%2 == even;});
        cout << count++ << " ";
        lk.unlock();
        cond.notify_all();
    }
}

int main() {
    int n = 10;
    thread t1(PrintOddAndEven, true, n);
    thread t2(PrintOddAndEven, false, n);

    t1.join();
    t2.join();
    return 0;
}

Answer 6

    #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" << 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;
    }

Answer 7

#include "threadFunc.hpp"
#include <iostream>
#include <thread>
#include <mutex>
#include <condition_variable>

using namespace std;

mutex t1;
condition_variable cond;


int number = 11;
int count = 0;

void printEven()
{
    while(1)
    {
        unique_lock<mutex> ul(t1);
        if(count< number)
        {
            if(count % 2 != 0)
            {
                cond.wait(ul);
            }

            cout<<count<<" : printed by thread"<<this_thread::get_id()<<endl;
            count++;
        }
        if(count > number)
            break;

        ul.unlock();
        cond.notify_all();
    }

}

void printOdd()
{
    while(1)
    {
        unique_lock<mutex> ul(t1);
        if(count< number)
        {
            if(count % 2 == 0)
            {
                cond.wait(ul);
            }

            cout<<count<<" : printed by thread"<<this_thread::get_id()<<endl;
            count++;
        }
        if(count > number)
            break;

        ul.unlock();
        cond.notify_all();
    }

}

Answer 8

I fail to understand why you want to use three separate threads for a serial behavior. But I will answer anyway:)

One solution would be to use a modified producer/consumer pattern with a prioritized queue between producers and consumers. The sort operation on the queue would depend on the integer value of the posted message. The consumer would peek an element in the queue and check if it is the next expected element. If not, it would sleep/wait.

A bit of code:

class Elt implements Comparable<Elt> {
  int value;
  Elt(value) { this.value=value; }
  int compare(Elt elt);
}

class EltQueue extends PriorityBlockingQueue<Elt> { // you shouldn't inherit colelctions, has-a is better, but to make it short
  static EltQueue getInstance(); // singleton pattern
}

class Consumer{
  Elt prevElt = new Elt(-1);
  void work()
  {
    Elt elt = EltQueue.getInstance().peek();
    if (elt.getValue() == prevElt.getValue()+1)) {
      EltQueue.getInstance().poll();
      //do work on Elt
    }
  }
}

class Producer {
  int n=0; // or 1!
  void work() {
    EltQueue.getInstance().put(new Elt(n+=2));
  }
}

Answer 9

As a first thing, the two functions should a least contain a loop, (unless you just want a single number)

A more standard solution (which remaps your idea) is to have a global structure containing a a mutex, and two condition variables (odd and even) plus a return value, and another condition for the printing. than use a uique_lock to handle the synchronization.

IN PSEUDOCODE:

struct global_t
{
    mutex mtx;
    int value = {0};
    condition_variable be_odd, be_even, print_it;
    bool bye = {false};

    global_t() { be_odd.notify(); }
} global;

void odd_generator()
{
    int my_odd = 1;
    for(;;)
    {
        unique_lock lock(global.mtx);
        if(global.bye) return;
        global.be_odd.wait(lock);
        global_value = my_odd; my_odd+=2;
        global.print_it.notify();
        if(my_odd > 100) bye=true;
    } //let RAII to manage wait states and unlocking
};

void even_generator()
{ /* same as odd, with inverted roles */ }

void printer()
{
    for(;;)
    {
        unique_lock lock(global.mtx);
        if(bye) return;
        global.ptint_it.wait(lock);
        std::cout << global.value << std::endl;
        ((global.value & 1)? global.be_even: global.be_odd).notify();
    }
}


int main()
{
    thread oddt(odd_generator), event(even_generator), printt(printer);
    oddt.join(), event.join(), printer.join();
}

Note that, apart didactic purpose, this solution adds no value respect to a simple loop printing the value of a counter, since there will never be real concurrency.

Note also (to avoid globals) that you can wrap everything into a class (making the actual main a class method) and instantate that class on the stack inside the new main.

Answer 10

Solution is based on C++11 critical code section aka mutex.

Here's the working code, followed by an explanation.

Tested and working on VS2013:

using namespace std;
#include <iostream>
#include <string>
#include <thread>
#include <mutex>

std::mutex mtx;

void oddAndEven(int n, int end);

int main()
{
std::thread odd(oddAndEven, 1, 10);
std::thread Even(oddAndEven, 2, 10);

odd.join();
Even.join();

return 0;
}



void oddAndEven(int n, int end){
int x = n;
for (; x < end;){
    mtx.lock();
    std::cout << n << " - " << x << endl;
    x += 2;
    mtx.unlock();
    std::this_thread::yield();
    continue;
 }
}

i.e:

Thread odd goes to method oddAndEven with starting number 1 thus he is the odd. He is the first to acquire the lock which is the mtx.lock().

Meanwhile, thread Even tries to acquire the lock too but thread odd acquired it first so thread Even waits.

Back to thread odd (which has the lock), he prints the number 1 and releases the lock with mtx.unlock(). At this moment, we want thread Even to acquire lock and to print 2 so we notify thread Even by writing std::this_thread::yield(). Then thread Even does the same.

etc etc etc.

Answer 11

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

std::mutex mu;
unsigned int change = 0;

void printConsecutiveNumbers(int start, int end,unsigned int consecutive)
{
    int x = start;
    while (x < end)
    {
        //each thread has check there time is coming or not
        if (change % consecutive == start)
        {
            std::unique_lock<std::mutex> locker(mu);
            std::cout << "Thread " << start << " -> " << x << std::endl;
            x += consecutive;
            change++;
            //to counter overflow
            change %= consecutive;
        }
    }
}

int main()
{
    //change num = 2 for printing odd and even
    const int num = 7;
    const int endValue = 1000;
    std::thread threads[num];
    //Create each consecutive threads
    for (int i = 0; i < num; i++)
    {
        threads[i] = std::thread(printConsecutiveNumbers, i, endValue, num);
    }

    //Joins all thread to the main thread
    for (int i = 0; i < num; i++)
    {
        threads[i].join();
    }

    return 0;
}

Answer 12

This code will work . I have tested it on visual studio 2017

#include "stdafx.h"
#include <iostream>
#include <mutex>
#include <thread>
#include <condition_variable>
using namespace std;
mutex m;
condition_variable cv;
int num = 1;


void oddThread() {​​​​​​​
    
    for (; num < 10;) {​​​​​​​
        unique_lock<mutex> lg(m);
        cv.wait(lg, [] {​​​​​​​return (num % 2 ==1); }​​​​​​​);
        cout << "From odd Thread " << num << endl;
        num++;
        lg.unlock();
        cv.notify_one();
    }​​​​​​​
}​​​​​​​
void evenThread() {​​​​​​​
    for (; num < 100;) {​​​​​​​
        unique_lock<mutex> lg(m);
        cv.wait(lg, [] {​​​​​​​return (num % 2 == 0); }​​​​​​​);
        cout << "From even Thread " << num << endl;
        num++;
        lg.unlock();
        cv.notify_one();
    }​​​​​​​
}​​​​​​​


int main() {​​​​​​​
    
    thread t1{​​​​​​​ oddThread}​​​​​​​; //odd function thread
    thread t2{​​​​​​​ evenThread}​​​​​​​;
    t1.join();
    t2.join();
    cin.get();
    return 0;
}​​​​​​​

output

from oddThread: 1 from evenThread: 2 from oddThread: 3 from evenThread: 4 from oddThread: 5 from evenThread: 6 from oddThread: 7 from evenThread: 8 from oddThread: 9 from evenThread: 10

Answer 13

#include <iostream>
#include <thread>
#include <mutex>
#include <condition_variable>
using namespace std;
std::mutex m;
std::condition_variable cv;
int counter =1;

void printEven()
{
    std::unique_lock<std::mutex> lk(m);
    while(1)
    {
        if(counter > 10)
            break;
        if(counter %2 != 0)
        {
            cv.wait (lk);
        }
        else
        {
            cout << "counter : " << counter << endl;
            counter++;
            //lk.unlock();
            cv.notify_one();
        }
    
    
    }
}


void printOdd()
{
    std::unique_lock<std::mutex> lk(m);
    
    while(1)
    {
        if(counter > 9)
        break;
        if(counter %2 == 0)
        {
            cv.wait (lk);
        }
        else
        {
            cout << "counter : " << counter << endl;
            counter++;
            //lk.unlock();
            cv.notify_one();
        }
    
    
    }
}

int main()
{
    std::thread t1(printEven);
    std::thread t2(printOdd);
    t1.join();
    t2.join();
    cout << "Main Ends" << endl;
}

Answer 14

i have used anonymous func (lambda) to do this and clubbed cond variable and mutex.

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

using namespace std; 

int main() {
    int count = 1;
    mutex mtx;
    condition_variable condition;
    const int ITERATIONS = 20; // iterations

    //prints odd numbers
    thread t1([&]() {
        while (count < ITERATIONS)
        {
            unique_lock <mutex> lock(mtx);
            condition.wait(lock, [&]() {
                return count % 2 != 0;
                });

            cout << "thread1 prints: " << count << endl;
            count++;
            lock.unlock();
            condition.notify_all();
        }
    });
    
    thread t2([&]
        {
            while (count < ITERATIONS)
            {
                unique_lock <mutex> lock(mtx);
                condition.wait(lock, [&]() {
                    return count % 2 == 0;
                    });
                cout << "thread2 prints: " << count << endl;
                count++;
                lock.unlock();
                condition.notify_all();
            }
        });

    t1.join();
    t2.join();
 }

Answer 15

#include <bits/stdc++.h>
#include <stdlib.h>
#include <unistd.h>
#include <thread>
#include <mutex>
#include <condition_variable>
using namespace std;

mutex m;
condition_variable cv;
unique_lock<mutex> lck(m);

void *printeven(void *arg)
{
    int *n = (int *)arg;
    while (*n <= 100)
    {
        cv.wait(lck);
        *n = *((int *)arg);
        cout << this_thread::get_id() << " : " << *n << endl;
        *n = *n + 1;
        cv.notify_one();
    }
    exit(0);
}

void *printodd(void *arg)
{
    int *n = (int *)arg;
    while (*n <= 100)
    {
        *n = *((int *)arg);
        cout << this_thread::get_id() << " : " << *n << endl;
        *n = *n + 1;
        cv.notify_one();
        cv.wait(lck);
    }
    exit(0);
}

int main()
{
    int num = 1;
    pthread_t p1 = 1;
    pthread_t p2 = 2;

    pthread_create(&p1, NULL, printodd, &num);
    pthread_create(&p2, NULL, printeven, &num);

    pthread_join(p1, NULL);
    pthread_join(p2, NULL);
    return 0;
}

Answer 16

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

std::mutex mtx;
std::condition_variable even_to_odd;
std::condition_variable odd_to_even;
unsigned int count = 1;
constexpr int MAX_COUNT = 20;

void even(int n) {
    for (int i = n; i <= MAX_COUNT; i++) {
        std::unique_lock<std::mutex> lock(mtx);
        if (i %2 == 0) {
            odd_to_even.wait(lock);
            std::cout << "Even: " << i << std::endl;
            even_to_odd.notify_one();
        }
    }
}

void odd(int n) {
    for (int i = n; i <= MAX_COUNT; i++) {
        std::unique_lock<std::mutex> lock(mtx);
        if (i == 1) {
            std::cout << "Odd : " << i << std::endl;
            odd_to_even.notify_one();
        }
        else if (i % 2 != 0) {
            even_to_odd.wait(lock);
            std::cout << "Odd : " << i << std::endl;
            odd_to_even.notify_one();
        }
    }
}

int main() {
    std::thread thread1(even,count);
    std::thread thread2(odd,count);

    thread1.join();
    thread2.join();
    return 0;
}

Answer 17

Please see below working code (VS2005)

#include <windows.h>
#include <stdlib.h>

#include <iostream>
#include <process.h>

#define MAX 100
int shared_value = 0;

CRITICAL_SECTION cs;

unsigned _stdcall even_thread_cs(void *p)
{

    for( int i = 0 ; i < MAX ; i++ )
    {
        EnterCriticalSection(&cs);

        if( shared_value % 2 == 0 )
        {
            printf("\n%d", i);
        }


        LeaveCriticalSection(&cs);

    }
    return 0;
}

unsigned _stdcall odd_thread_cs(void *p)
{
    for( int i = 0 ; i < MAX ; i++ )
    {
        EnterCriticalSection(&cs);

        if( shared_value % 2 != 0 )
        {
            printf("\n%d", i);
        }

        LeaveCriticalSection(&cs);  

    }

    return 0;
}


int main(int argc, char* argv[])
{
     InitializeCriticalSection(&cs);

    _beginthreadex(NULL, NULL, even_thread_cs, 0,0, 0);
    _beginthreadex(NULL, NULL, odd_thread_cs, 0,0, 0);

    getchar();
    return 0;
}

Here, using shared variable shared_value, we are synchronizing the even_thread_cs and odd_thread_cs. Note that sleep is not used.