如何在c ++中使用/创建unique_lock？

Question

Please, can anybody explain how to use and create an unique_lock in c++? It should be used both to get mutual exclusion to any procedure of the monitor and to be able to perform wait() on the condition variable...I'm not understanding from the documentation how I am supposed to create it. Is necessary a mutex? Here is a pseudo-code:

/* compile with g++, flags -std=c++0x -lpthread */

#include <condition_variable>
#include <mutex>
#include <thread>
#include <iostream>
#include <string.h>
#include <unistd.h>

class monitorTh {

private:

    std::mutex m;
    std::condition_variable waitP;
    std::condition_variable waitC;
    char element[32];
    std::unique_lock::unique_lock l;

public:
    void produce(char* elemProd) {
        l.lock();
        if (/*already_present_element*/) {
            waitP.wait(l);
        }
        else {/*produce element*/}
        l.unlock();
    }

    void consume() {
        /*something specular*/
    }
};

int main(int argc, char* argv[]) {

    monitorTh* monitor = new monitorTh();
    char prodotto[32] = "oggetto";

    std::thread producer([&]() {
        monitor->produce(prodotto);
    });

    std::thread consumer([&]() {
        monitor->consume();
    });

    producer.join();
    consumer.join();
}

Answer 1

std::unique_lock use the RAII pattern.

When you want to lock a mutex, you create a local variable of type std::unique_lock passing the mutex as parameter. When the unique_lock is constructed it will lock the mutex, and it gets destructed it will unlock the mutex. More importantly: If a exceptions is thrown, the std::unique_lock destructer will be called and so the mutex will be unlocked.

Example:

#include<mutex>
int some_shared_var=0;

int func() {
    int a = 3;
    { //Critical section
        std::unique_lock<std::mutex> lock(my_mutex);
        some_shared_var += a;
    } //End of critical section
}        

Answer 2

std::unique_lock<std::mutex> holds a lock on a separate std::mutex object. You associate the lock object with the mutex by passing it in the constructor. Unless you specify otherwise, the mutex will be immediately locked. If the lock object holds the lock when it is destroyed then the destructor will release the lock. Typically, the std::unique_lock<std::mutex> object will thus be a local variable, declared at the point where you wish to acquire the lock.

In your case, the produce() function could be written like this:

void produce(char* elemProd) {
    std::unique_lock<std::mutex> lk(m); // lock the mutex
    while (/*already_present_element*/) { // condition variable waits may wake spuriously
        waitP.wait(lk);
    }
    {/*produce element*/}
    // lk releases the lock when it is destroyed
}

Note that I have replaced the if with a while to account for spurious wakes from the wait() call.

Answer 3

A more detailed sample code using condition variables:

#include<mutex>
std::mutex(mu); //Global variable or place within class
std::condition_variable condition; //A signal that can be used to communicate between functions

auto MyFunction()->void
{
  std::unique_lock<mutex> lock(mu);
  //Do Stuff
  lock.unlock(); //Unlock the mutex
  condition.notify_one(); //Notify MyOtherFunction that this is done
}

auto MyOtherFunction()->void
{
   std::unique_lock<mutex> lock(mu);
   condition.wait(lock) //Wait for MyFunction to finish, a lambda can be passed also to protects against spurious wake up e.g (lock,[](){return *some condition*})
   lock.unlock();
}

Answer 4

In this case, I think all you need to do is:

m.lock();
// Critical section code
m.unlock();