Dining Philosophers Problem I

Dining philosophers in C++ by Andre Adrian

The main() function establishes random numbers in line 42. We set the random number generator seed value to the number of seconds since 1st January 1970. We define our fork resources in line 44. Then we start four threads beginning in line 46. To avoid premature thread termination, we join the threads beginning in line 51. The thread function phil() has a forever loop. The while(true) statement is always true, therefore the thread will never terminate. The problem description says, “he thinks for some random duration”. First, we calculate a random duration with the function myrand(), see line 20 and line 6. The function myrand() produces a pseudo-random return value in the range of [min, max). For program trace, we log the philosopher’s number, his current state of “he thinks,” and the duration to the console. The sleep_for() statement lets the scheduler put the thread for the duration into the waiting state. In a “real” program application, source code uses up time instead of sleep_for().After the philosopher’s thread thinking time ends, he “gets his first fork”. See line 24. We use a function lock() to perform the “gets fork” thing. Currently, the function lock() is very simple because we don’t know better. We just set the fork resource to the value 1. See line 10. After the philosopher thread obtained his first fork, he proudly announces the new state with a.”got ma” console output. Now the thread “gets his second fork”. See line 28. The corresponding console output is “got mb“. The next state is “he eats“. Again, we determine the duration, produce a console output, and occupy the thread with a sleep_for(). See line 31. After the state.”he eats” the philosopher puts down his forks. See lines 35 and 14. The unlock() function is again really simple and sets the resource back to 0.

Please compile the program without compiler optimization. We will see the reason later. The console output of our program looks promising:

Have we already solved the dining philosophers’ problem? Well, the program output is not detailed enough to answer this question.

Multiple Resource Use with Logging

We should add some more logging. Currently, the function lock() does not check if the fork is available before the resource is used. The improved version of lock() in program dp_2.cpp is:

void lock(int& m) {
  if (m) {
    std::cout<<"\t\t\t\t\t\tERROR lock\n";
  }
  m=1;
}

We see “ERROR lock” console output. This output tells us that two philosophers use the same resource simultaneously. What can we do?

Erroneous Busy Waiting without Resource Hierarchy

We can change the if statement in lock() into a while statement. This while statement produces a spinlock. A spinlock is a fancy word for busy waiting. While the fork resource is in use, the thread is busy waiting for a change from the state in use to the state available. At this very moment, we set the fork resource again to state in use. In the program, dp_3.cpp we have:

void lock(int& m) {
  while (m)
    ; // busy waiting
  m=1;
}

int main() {
  std::cout<<"dp_4\n";
  srand(time(nullptr));

  int m1{0}, m2{0}, m3{0}, m4{0};

  std::thread t1([&] {phil(1, m1, m2);});
  std::thread t2([&] {phil(2, m2, m3);});
  std::thread t3([&] {phil(3, m3, m4);});
  std::thread t4([&] {phil(4, m1, m4);});

  t1.join();
  t2.join();
  t3.join();
  t4.join();
}

It is always the same philosopher thread that eats. Is it possible that the setting of compiler optimization can change the behavior of a program? Yes, it is possible. The philosopher threads read from memory the value of fork resource. The compiler optimization optimizes some of these memory reads away. Everything has a price!

Still Erroneous Busy Waiting with Resource Hierarchy

The programming language C++ has the atomic template to define an atomic type. The behavior is well-defined if one thread writes to an atomic object while another reads from it. In the file dp_5.cpp we use atomic<int> for the fork resources. See lines 11, 17, 21, and 47. We include <atomic> in line 5:

// dp_5.cpp
#include <iostream>
#include <thread>
#include <chrono>
#include <atomic>

int myrand(int min, int max) {
  return rand()%(max-min)+min;
}

void lock(std::atomic<int>& m) {
  while (m)
    ; // busy waiting
  m=1;
}

void unlock(std::atomic<int>& m) {
  m=0;
}

void phil(int ph, std::atomic<int>& ma, std::atomic<int>& mb) {
  while(true) {
    int duration=myrand(1000, 2000);
    std::cout<<ph<<" thinks "<<duration<<"ms\n";
    std::this_thread::sleep_for(std::chrono::milliseconds(duration));

    lock(ma);
    std::cout<<"\t\t"<<ph<<" got ma\n";
    std::this_thread::sleep_for(std::chrono::milliseconds(1000));

    lock(mb);
    std::cout<<"\t\t"<<ph<<" got mb\n";

    duration=myrand(1000, 2000);
    std::cout<<"\t\t\t\t"<<ph<<" eats "<<duration<<"ms\n";
    std::this_thread::sleep_for(std::chrono::milliseconds(duration));

    unlock(mb);
    unlock(ma);
  }
}

int main() {
  std::cout<<"dp_5\n";
  srand(time(nullptr));

  std::atomic<int> m1{0}, m2{0}, m3{0}, m4{0};

  std::thread t1([&] {phil(1, m1, m2);});
  std::thread t2([&] {phil(2, m2, m3);});
  std::thread t3([&] {phil(3, m3, m4);});
  std::thread t4([&] {phil(4, m1, m4);});

  t1.join();
  t2.join();
  t3.join();
  t4.join();
}

What’s next?

The next installment of this dining philosopher problem solves the tiny chance of misbehavior. So far, none of the programs have been correct.