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Dilawar Singh
Dilawar Singh

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An example of `boost::fiber` library

Fiber is just a thread implemented in user space.

Fibers are easier to reason about than threads. And have much cheaper context switching cost. Fibers are well suited for handling concurrent IO operations where a processor mostly wait for data, and threads usually have pretty big context switching cost. So multiple fibers running in a single thread is an effective solution.

Here is a simple program I wrote to explore fibers. You can find the full example below.

playground/fiber.cpp at a5fba9f21ff121249c71bff75ee8964c1016aa3f · dilawar/playground

This program has two functions: print_a prints a and print_b prints b and then launches a thread that prints B (in detached mode).

void print_a()
{
    cout << "a";
    boost::this_fiber::yield();
}

void print_b()
{
    cout << "b";
    std::thread j([]() { printf("B"); });
    j.detach();
    boost::this_fiber::yield();
}
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Following is the main function. We created a shared variable i initialized to 0. We create two detached fibers. The first one keeps calling print_a till i < 20. Similarly, the second one loops on print_b till i < 20. Both increment i by 1. When i = 20, both fibers should be able to join.

int main()
{
    int i = 0;
    boost::fibers::fiber([&]() {
        do {
            print_a();
            i++;
        }
        while (i < 20);
    }).detach();

    boost::fibers::fiber([&]() {
        do {
            i++;
            print_b();
        } while (i < 20);
    }).detach();

    printf("X");
    return 0;
}
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Let’s guess the output of this program. It is most likely to be the same as if std::threads were used instead of fiber.

X is printed first? Yes. Note that detach() is called on each fiber so neither of their functions is called. They are put in the background. Control passes to the fiber manager at return 0; when it asks the fibers to join. In fact, you can put more computations after the printf("X"); statement, and it would be computed before any fiber is called.

As soon as we try to return from the main, the fiber manager is asked to join the fibers. The first fiber awakes, a is printed, and the fiber yields the control to the manager. The fiber manager then wakes up the second fiber (who was waiting in the queue) that prints b and also launch a thread in the background that prints B. We can not be sure if B will be printed immediately after the b (it is a std::thread). print_b yields the control and goes to sleep. The fiber manager wakes up first fiber again that calls print_a again, and a is printed, and so on. Note that i is incremented every time either of the fibers is called.

When i hits 20, both fibers terminate and joined, and the main function return 0;. So we have print_a called ten times, and print_b is also called ten times. In the output, we should have ten as, ten bs, and 10 Bs. B may not strictly follow b, but b must come after a.

Here are a few runs of the program. Note that the location of B is not deterministic.

  • XababBabBabBababBBabBabBabBabBB
  • XababBabBabBabBabBabBabaBbBabBB
  • XababBabBabBabBabBabBabBabBabBB
  • XababBabBabBabBabBabBabBabBabBB
  • XababBabBabBBababBabBabBabBabBB
  • XababBabBabBabBabBabBabBabBabBB
  • XababBabBabBababBBabBabBababBBB
  • XababBabBabBababBBabBabBabBabBB
  • XababBabBabBababBBabBabBabBabBB
  • XababBabBabBabBabBababBBabBabBB
  • XababBabBabBabBabBabBabBabBabBB

References

  1. A great talk by Nat on Boost fibers https://youtu.be/e-NUmyBou8Q

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