我无法弄清楚这个简单代码的问题在哪里,我认为这是输出到控制台的问题可能是死锁或其他什么的,有人可以帮忙。
#include <iostream>
#include <string>
#include <sstream>
#include <boost/thread.hpp>
using namespace std;
struct IntegrateTask
{
int id;
double from, to, step, result;
IntegrateTask(int id, double from, double to, double step)
{
this -> id;
this -> from = from;
this -> to = to;
this -> step = step;
}
~IntegrateTask()
{}
};
vector<IntegrateTask> * tasks = new vector<IntegrateTask>();
boost::mutex mutlist;
boost::mutex iomutex;
boost::condition_variable condtask;
bool isInterrupted = false;
double foo(double x)
{
return x * x;
}
void integrate(IntegrateTask * task)
{
double result = 0;
double step = task -> step;
for(double i = task -> from ; i != task -> to; i =+ step)
{
result += foo(i) * step;
}
task -> result = result;
}
void integrateThread()
{
boost::thread::id id = boost::this_thread::get_id();
try
{
{
boost::mutex::scoped_lock iolock(iomutex);
cout << "Thread #" << id << " is working!" << endl;
}
while(!isInterrupted)
{
IntegrateTask * currtask = NULL;
{
boost::mutex::scoped_lock lock(mutlist);
while(!isInterrupted && tasks -> empty())
{
condtask.wait(lock);
}
if (!tasks -> empty())
{
currtask = &tasks->back();
tasks->pop_back();
}
}
if (currtask != NULL)
{
integrate(currtask);
boost::mutex::scoped_lock iolock(iomutex);
cout << "Task #" << (currtask->id) << "; result = " << (currtask->result) << endl;
}
}
boost::mutex::scoped_lock iolock(iomutex);
cout << "Thread # " << id << " stoped working normal!" << endl;
}
catch(boost::thread_interrupted)
{
boost::mutex::scoped_lock ioLock(iomutex);
cout << "Thread # " << id << " stoped working by interruption!" << endl;
}
catch(exception & e)
{
boost::mutex::scoped_lock iolock(iomutex);
cout << "Error: " << e.what() << endl;
}
}
int main()
{
cout << "Function for integration: f(x)=x*x" << endl;
cout << "For stopping program press EXIT" << endl;
int thcount = 6;// or boost::thread::hardware_concurrency()
boost::thread_group thgroup;
for (int i = 1; i <= thcount; i++){
thgroup.create_thread(&integrateThread);
}
int id = 0;
while (true)
{
string line;
{
boost::mutex::scoped_lock iolock(iomutex);
cout << "Task #" << ++id << "; left bound, right bound and step: ";
getline(cin, line);
}
if (line.find("e") != string::npos)
{
isInterrupted = true;
condtask.notify_all();
thgroup.interrupt_all();
thgroup.join_all();
return 0;
}
double from, to, step;
istringstream input(line);
input >> from >> to >> step;
IntegrateTask * task = new IntegrateTask(id, from, to, step);
{
boost::mutex::scoped_lock lock(mutlist);
tasks->push_back(*task);
}
condtask.notify_one();
}
}
答案 0 :(得分:1)
我没有尝试遵循你想要达到的目标的逻辑,但有两个问题(至少):
您未在id
的构造函数中使用IntegrateTask
。您通常应该优先使用初始化列表而不是构造函数体中的赋值,并且在函数签名中使用类成员变量名称也是一种灾难,因此我将构造函数更改为:
IntegrateTask(int id_init, double from_init, double to_init, double step_init)
: from(from_init), to(to_init), step(step_init), id(id_init) {}
!=
的{{1}}循环中使用了for
。如果你改为integrate
你的程序不应该挂起,但我不确定这是否会破坏程序的逻辑。