尝试使用condition_variables在线程之间切换

Question

我有两个主题。两者都被初始化，但第一个线程首先运行，而第二个线程等待第一个线程（call_once(flagEven1, [&](){cond2.wait(lock, [&](){return zeroEvenOddFlag == 1; }); });）。在满足某个条件后，第一个线程唤醒第二个线程并进入休眠状态。

                zeroEvenOddFlag = 0;
                cond1.notify_all();
                cond2.wait(lock, [&](){return zeroEvenOddFlag == 1; });

此过程交替进行，直到任务完成。现在，线程被锁定，我不知道发生了什么。线程在一起应该处理输入字符串并产生两个零和非零数字的字符串。我通过添加一个在线程（zeroEvenOddFlag）之间共享的整数来纠正虚假唤醒。

#include <string>
#include <thread>
#include <future>
#include <cstdio>
#include <iostream>
#include <queue>
#include <condition_variable>

using namespace std;

mutex mu;
condition_variable cond1;
condition_variable cond2;

once_flag flagZero;
once_flag flagEven;
once_flag flagEven1;
int zeroEvenOddFlag = 0;

string printerFunc(queue<char>& input, int zeroEvenOdd, once_flag& flag){
    string output = "";
    function<void(string&)> f;
    unique_lock<mutex> lock(mu);
    call_once(flag, [&](){
        if (zeroEvenOdd == 1){
            f = [&](string& output){
                int i = input.front() - '0';
                if (i == 0){ 
                output += to_string(i); 
                input.pop();
                }
                else {
                    zeroEvenOddFlag = 1;
                    cond2.notify_all();
                    cond1.wait(lock, [&](){return zeroEvenOddFlag == 0; });
                }
            };
        }
        else if (zeroEvenOdd == 2){
            call_once(flagEven1, [&](){cond2.wait(lock, [&](){return zeroEvenOddFlag == 1; }); });
            f = [&](string& output){
                int i = input.front() - '0';
                if (i != 0){ 
                    output += to_string(i); 
                    input.pop();
                }
                else {
                    zeroEvenOddFlag = 0;
                    cond1.notify_all();
                    cond2.wait(lock, [&](){return zeroEvenOddFlag == 1; });
                }
            };
        }
    });
    while(input.size()!=0){
        f(output);
    }
    return output;
}

int _tmain(int argc, _TCHAR* argv[])
{
     string input = "0102030";
     queue<char> inputQueue;
     for(char c : input){
         inputQueue.push(c);
     }
     auto zeros = async(launch::async, printerFunc, ref(inputQueue), 1, ref(flagZero));
     auto evens = async(launch::async, printerFunc, ref(inputQueue), 2, ref(flagEven));
     string zerosString = zeros.get();
     string evensString = evens.get();
     cout << "these are the zeros " << zerosString << endl;
     cout << "these are the evens " << evensString << endl;

     this_thread::sleep_for(chrono::seconds(10));
    return 0;
}

Answer 1

您需要始终检查虚假的唤醒条件。 cond2.wait(lock, [&](){return zeroEvenOddFlag == 1; });之类的使用有助于检查。此外，您在循环结束时遇到问题。哪个线程最终拥有互斥锁必须在它消耗队列的最后一位时释放它，以便其他线程可以释放完成。最终代码应如下所示：

#include <string>
#include <thread>
#include <future>
#include <cstdio>
#include <iostream>
#include <queue>
#include <condition_variable>

using namespace std;

mutex mu;
condition_variable cond1;
condition_variable cond2;

once_flag flagZero;
once_flag flagEven;
once_flag flagEven1;
int zeroEvenOddFlag = 0;

string printerFunc(queue<char>& input, int zeroEvenOdd, once_flag& flag){
    string output = "";
    function<void(string&)> f;
    unique_lock<mutex> lock(mu);
    call_once(flag, [&](){
        if (zeroEvenOdd == 1){
            f = [&](string& output){
                    int i = input.front() - '0';
                    if (i == 0){
                        output += to_string(i);
                        input.pop();
                        if (input.size() == 0){
                            zeroEvenOddFlag = 1;
                            cond2.notify_all();
                        }
                    }
                    else {
                        zeroEvenOddFlag = 1;
                        cond2.notify_all();
                        cond1.wait(lock, [&](){return zeroEvenOddFlag == 0; });
                    }
            };
        }
        else if (zeroEvenOdd == 2){
            call_once(flagEven1, [&](){cond2.wait(lock, [&](){return zeroEvenOddFlag == 1; }); });
            f = [&](string& output){
                    int i = input.front() - '0';
                    if (i != 0){
                        output += to_string(i);
                        input.pop();
                        if (input.size() == 0){
                            zeroEvenOddFlag = 0;
                            cond1.notify_all();
                        }
                    }
                    else {
                        zeroEvenOddFlag = 0;
                        cond1.notify_all();
                        cond2.wait(lock, [&](){return zeroEvenOddFlag == 1; });
                    }
            };
        }
    });
    while(input.size()!=0){
        f(output);
    }
    return output;
}

int _tmain(int argc, _TCHAR* argv[])
{
     string input = "0102030";
     queue<char> inputQueue;
     for(char c : input){
         inputQueue.push(c);
     }
     auto zeros = async(launch::async, printerFunc, ref(inputQueue), 1, ref(flagZero));
     auto evens = async(launch::async, printerFunc, ref(inputQueue), 2, ref(flagEven));
     string zerosString = zeros.get();
     string evensString = evens.get();
     cout << "these are the zeros " << zerosString << endl;
     cout << "these are the evens " << evensString << endl;

     this_thread::sleep_for(chrono::seconds(10));
    return 0;
}