Consider this simple synchronization problem. I have two threads, A and B, that each execute 2 steps. I want step 1a to be performed before step 2b.
| Thread A | Thread B |
|---|---|
| Step 1a | Step 1b |
| Step 2a | Step 2b |
I have some options for how to implement this.
std::condition_variable + std::mutex + bool
This is the solution proposed by this stack overflow answer and this leetcode discussion page.
Thread B will wait on the condition variable, and Thread A will notify the condition variable. The mutex is required because it is the argument of the condition_variable's wait.
#include <iostream>
#include <thread>
#include <condition_variable>
std::condition_variable step_1a;
std::mutex a_mutex_I_guess;
bool step_1a_done = false;
void Step_1a() {
std::cout << "step 1a" << "\n";
}
void Step_2a() {
std::cout << "step 2a" << "\n";
}
void Step_1b() {
std::cout << "step 1b" << "\n";
}
void Step_2b() {
std::cout << "step 2b" << "\n";
}
void A() {
//std::unique_lock<std::mutex> lck{ a_mutex_I_guess }; unnecessary
Step_1a();
step_1a_done = true;
//lck.unlock(); unnecessary
step_1a.notify_one();
Step_2a();
}
void B() {
Step_1b();
std::unique_lock<std::mutex> lck{ a_mutex_I_guess };
step_1a.wait(lck, []() { return step_1a_done; });
Step_2b();
}
int main() {
std::thread thread_A{ A };
std::thread thread_B{ B };
thread_A.join();
thread_B.join();
}
To me, this seems like overkill. std::condition_variables are designed to handle multiple waiting threads. std::mutex is intended to protect shared data, not to be fodder for wait. On top of all of that, I needed bool step_1a_done to actually keep track of whether or not step_1a had completed.
As a measure of their complexity, the mutex, condition_variable, and bool together require 153 (80 + 72 + 1) bytes of memory on my machine.
std::binary_semaphore
Alternatively, I can use a binary semaphore. Semantically, the binary semaphore isn't meant for one-time-use. However, it gets the job done with simpler tools than the previous option.
#include <iostream>
#include <thread>
#include <semaphore>
std::binary_semaphore step_1a_sem{ 0 };
void Step_1a() {
std::cout << "step 1a" << "\n";
}
void Step_2a() {
std::cout << "step 2a" << "\n";
}
void Step_1b() {
std::cout << "step 1b" << "\n";
}
void Step_2b() {
std::cout << "step 2b" << "\n";
}
void A() {
//std::unique_lock<std::mutex> lck{ a_mutex_I_guess }; unnecessary
Step_1a();
step_1a_sem.release();
Step_2a();
}
void B() {
Step_1b();
step_1a_sem.acquire();
Step_2b();
}
int main() {
std::thread thread_A{ A };
std::thread thread_B{ B };
thread_A.join();
thread_B.join();
}
step_1a_sem requires only 1 byte of memory.
Question
My assessment is that binary_semaphore is better. However, even better would be a "one_time_semaphore" that documents (or enforces) in my code that release should only be called once. Are there C++ concurrency primitives that are a better fit for this thread synchronization problem?
EDIT: std::promise<void>
@Daniel Langr has pointed out that std::promise<void> also works. While this seems like the exact use case of std::promise<void>, things appear significantly more complicated under the hood than with a binary_semaphore. The memory requirement is 24 bytes.
#include <iostream>
#include <thread>
#include <future>
std::promise<void> step_1a_done;
void Step_1a() {
std::cout << "step 1a" << "\n";
}
void Step_2a() {
std::cout << "step 2a" << "\n";
}
void Step_1b() {
std::cout << "step 1b" << "\n";
}
void Step_2b() {
std::cout << "step 2b" << "\n";
}
void A() {
Step_1a();
step_1a_done.set_value();
Step_2a();
}
void B() {
Step_1b();
step_1a_done.get_future().wait();
Step_2b();
}
int main() {
std::thread thread_A{ A };
std::thread thread_B{ B };
thread_A.join();
thread_B.join();
}
