如何更改此notify_one以便它选择一个随机线程

How to change this notify_one so that it chooses a random thread?

本文关键字：一个 线程 随机 选择 notify 何更改 one      更新时间：2023-10-16

提前感谢您的帮助。试图制作一个可以创建6个线程的程序，然后每2秒随机选择一个线程并打印其编号。我显然做错了什么，因为它只是不断地打印0-1-2-3-4-5。代码如下。主要问题是，我应该怎么做才能使随机线程解锁？

#include <thread>
#include <memory>
#include <chrono>
#include <condition_variable>
std::condition_variable* cv = new std::condition_variable();
std::mutex cv_m;
void threadFunc(std::shared_ptr<bool> flag2, int id)
{
while (true)
{
std::unique_lock<std::mutex> lock(cv_m);
cv->wait(lock);
if (true)
if (*flag2) std::cout << "Thread" << " " << id << std::endl;
}
}
int main() {
std::shared_ptr<bool> f2 = std::make_shared<bool>(false);
std::thread threads[6];
for (int i = 0; i < 6; i++)
threads[i] = std::thread(threadFunc, f2, i);
*f2 = true;
while (true)
{
cv->notify_one();
std::this_thread::sleep_for(std::chrono::seconds(2));
}
return 0;
}

您可以为每个线程使用一个条件变量，每个线程一开始应该为false，然后将一个随机条件变量更改为true并通知所有线程，这将使一个随机线程(拥有该条件变量的线程(苏醒这是完整的解决方案

#include <iostream>
#include <thread>
#include <mutex>
#include <condition_variable>
#include <unistd.h>
#include "UserInterruptHandler.h"
using namespace std;
UserInterruptHandler h;
condition_variable conditionalVariable;
mutex mtx;
bool flag = true;
void myMethod(int id, bool *canWork) {
unique_lock<mutex> ul(mtx);
while (flag) {
conditionalVariable.wait(ul,[=]{return *canWork;});
if(!flag)
break;
cout << "thread " << id << endl;
}
cout << "thread " << id << " exits.." << endl;
}
int main() {
cout << "input thread count" << endl;
int n;
cin >> n;
thread myThreads[n];
bool *canWork = new bool[n];
for (int i = 0; i < n; i++) {
canWork[i] = false;
myThreads[i] = thread(myMethod, i + 1, &canWork[i]);
}

while (!h.checkInterruption()) {
int i = rand() % n;
canWork[i] = true;
conditionalVariable.notify_all();
canWork[i] = false;
usleep(1000);
}
flag = false;
int i = 0;
for (thread &th:myThreads) {
canWork[i++] = true;
conditionalVariable.notify_all();
if (th.joinable())
th.join();
}
}

注意，这里我使用头UserInterruptHandler.h来处理CTR+C事件，以优雅地结束所有线程

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