我不知道如果我真的明白爲什麼std::condition_variable
需要額外std::mutex
作爲參數?它應該不是由自己鎖定?的std :: condition_variable爲什麼它需要一個std ::互斥
#include <iostream>
#include <condition_variable>
#include <thread>
#include <chrono>
std::condition_variable cv;
std::mutex cv_m;
int i = 0;
bool done = false;
void waits()
{
std::unique_lock<std::mutex> lk(cv_m);
std::cout << "Waiting... \n";
cv.wait(lk, []{return i == 1;});
std::cout << "...finished waiting. i == 1\n";
done = true;
}
void signals()
{
std::this_thread::sleep_for(std::chrono::seconds(1));
std::cout << "Notifying falsely...\n";
cv.notify_one(); // waiting thread is notified with i == 0.
// cv.wait wakes up, checks i, and goes back to waiting
std::unique_lock<std::mutex> lk(cv_m);
i = 1;
while (!done)
{
std::cout << "Notifying true change...\n";
lk.unlock();
cv.notify_one(); // waiting thread is notified with i == 1, cv.wait returns
std::this_thread::sleep_for(std::chrono::seconds(1));
lk.lock();
}
}
int main()
{
std::thread t1(waits), t2(signals);
t1.join();
t2.join();
}
二次,在例如它們解鎖互斥第一(signals
方法)。他們爲什麼這樣做? Shoulden't他們首先鎖定,然後通知後解鎖?
好的,謝謝你的解釋。但是,我可以問爲什麼這不是直接投入到'std :: condition_variable'的實現? – Pascal
就是這樣。這就是條件變量的等待函數將鎖作爲參數的原因! –