#include <iostream>
#include <vector>
#include <thread>
#include <mutex>
#include <condition_variable>
#include <atomic>
using namespace std;
mutex m;
condition_variable cov;
bool ready = false;
bool processed = false;
void showNum(int &f_, atomic_bool &alive_)
{
while(alive_)
{
unique_lock<mutex> lk(m);
cov.wait(lk,[]{ return ready;});
f_++;
ready = false;
processed= true;
lk.unlock();
cout<<f_<<endl;
cov.notify_one();
}
}
int main() {
vector<int> va;
for (int i = 0; i < 10; ++i) {
va.push_back(i);
}
int f = 0;
atomic_bool alive{ true };
std::thread t1(showNum,ref(f),ref(alive));
auto sizeofVector = va.size();
for (int j = 0; j < sizeofVector; ++j) {
{
lock_guard<mutex> lk0(m);
f = va.back();
cout<<f<<" ";
ready = true;
}
cov.notify_one();
va.pop_back();
{
unique_lock<mutex> lk(m);
cov.wait(lk,[]{return processed;});
processed = false;
lk.unlock();
}
}
alive = false;
t1.join();
return 0;
}
我只想在多线程中测试条件变量。上面的代码是我的测试代码。
错误是线程t1无法正常连接。我打印alive_
,它总是如此,不能被主线程中的alive = false
设置为false。
我尝试使alive
成为一个全局变量,但仍然是同样的错误。
你能给我一些建议吗?
可以改变
cov.wait(lk,[]{ return ready;});
至
cov.wait(lk,[&alive_]{ return ready || !alive_;});
if (!alive_)
break;
并在alive_=false;
下方添加该行
cov.notify_one();
完整的代码如下
#include <iostream>
#include <vector>
#include <thread>
#include <mutex>
#include <condition_variable>
#include <atomic>
using namespace std;
mutex m;
condition_variable cov;
bool ready = false;
bool processed = false;
void showNum(int &f_, atomic_bool &alive_)
{
while(alive_)
{
unique_lock<mutex> lk(m);
cov.wait(lk,[&alive_]{return ready || !alive_;});
if (!alive_)
break;
f_++;
ready = false;
processed= true;
lk.unlock();
cout<<f_<<endl;
cov.notify_one();
}
}
int main() {
vector<int> va;
for (int i = 0; i < 10; ++i) {
va.push_back(i);
}
int f = 0;
atomic_bool alive{ true };
std::thread t1(showNum,ref(f),ref(alive));
auto sizeofVector = va.size();
for (int j = 0; j < sizeofVector; ++j) {
{
lock_guard<mutex> lk0(m);
f = va.back();
cout<<f<<" ";
ready = true;
}
cov.notify_one();
va.pop_back();
{
unique_lock<mutex> lk(m);
cov.wait(lk,[]{return processed;});
processed = false;
lk.unlock();
}
}
alive = false;
cov.notify_one();
t1.join();
return 0;
}
在t1
,该功能不会经常测试alive
。您已经设计了它,以便每个循环都以等待条件变量开始。然后它会在通知时进入睡眠状态并唤醒。不幸的是,当main将alive设置为false时,t1线程仍处于等待状态。
您可以轻松地观察到:
void showNum(int &f_, atomic_bool &alive_)
{
while(alive_)
{ cout<<"waiting..."<<endl;
unique_lock<mutex> lk(m);
cout<<"waiting more..."<<endl;
cov.wait(lk,[]{ return ready;}); ///<<<<< stuck here
cout<<"go..."<<endl;
f_++;
ready = false;
processed= true;
lk.unlock();
cout<<" sn:"<<f_<<endl;
cov.notify_one();
}
}
只有当main
确保对条件变量进行更多通知时,它才会醒来。只有在这一刻,它才会退出等待状态,经过处理后发现alive
是false
。
为了避免永远陷入困境,您可以更改代码并使用wait_for()
,以便该函数可以检查是否仍然保持alive
超时。
#include <iostream>
#include <vector>
#include <thread>
#include <mutex>
#include <condition_variable>
#include <atomic>
#include <chrono>
using namespace std;
mutex m;
condition_variable cov;
bool ready = false;
bool processed = false;
atomic_bool alive{ true };
void showNum(int &f_, atomic_bool &alive_)
{
while(alive)
{
unique_lock<mutex> lk(m);
cov.wait(lk,[]{ return ready || !alive;});
if(!alive)
break;
f_++;
ready = false;
processed= true;
lk.unlock();
cout<<f_<<endl;
cov.notify_one();
}
}
int main() {
vector<int> va;
for (int i = 0; i < 10; ++i) {
va.push_back(i);
}
int f = 0;
std::thread t1(showNum,ref(f),ref(alive));
auto sizeofVector = va.size();
for (int j = 0; j < sizeofVector; ++j) {
{
lock_guard<mutex> lk0(m);
f = va.back();
cout<<f<<" ";
ready = true;
}
cov.notify_one();
va.pop_back();
{
unique_lock<mutex> lk(m);
cov.wait(lk,[]{return processed;});
processed = false;
lk.unlock();
}
}
alive = false;
cov.notify_one();
t1.join();
return 0;
}
整合建议,我修改我的代码在上面。它按我的预期输出。感谢所有人提供的建议,最好的问候。