c++ - 条件变量和 notify_all

Question

在下面的代码中(取自 cpp 引用并添加了额外的 cout)为什么我们看不到 ...finished waiting. 在第一个 cv.notify_all 之后？

#include <iostream>
#include <condition_variable>
#include <thread>
#include <chrono>

std::condition_variable cv;
std::mutex cv_m; // This mutex is used for three purposes:
                 // 1) to synchronize accesses to i
                 // 2) to synchronize accesses to std::cerr
                 // 3) for the condition variable cv
int i = 0;

void waits()
{
    std::unique_lock<std::mutex> lk(cv_m);
    std::cerr << "Waiting... \n";
    cv.wait(lk, []{return i == 1;});
    std::cerr << "...finished waiting. i == 1\n";
}

void signals()
{
    std::this_thread::sleep_for(std::chrono::seconds(1));
    {
        std::lock_guard<std::mutex> lk(cv_m);
        std::cerr << "Notifying...\n";
    }
    cv.notify_all();

    std::cerr << "I should see i here...\n";

    std::this_thread::sleep_for(std::chrono::seconds(1));

    {
        std::lock_guard<std::mutex> lk(cv_m);
        i = 1;
        std::cerr << "Notifying again...\n";
    }
    cv.notify_all();
}

int main()
{
    std::thread t1(waits), t2(waits), t3(waits), t4(signals);
    t1.join(); 
    t2.join(); 
    t3.join();
    t4.join();
}

输出:

Waiting... 
Waiting... 
Waiting... 
Notifying...
I should see i here...
Notifying again...
...finished waiting. i == 1
...finished waiting. i == 1
...finished waiting. i == 1

Answer 1

因为这一行:

cv.wait(lk, []{return i == 1;});

您传递了一个谓词，因此只有当谓词返回 true 时它才会唤醒。在第一个 notify_all() 之后，谓词不满足。