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 semapho

Answer 1

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.