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[英]passing std::future to a boost::thread vs a std::thread
[英]How do I stop std/boost::thread copying my object rather than passing references?
從這個示例項目的輸出中,我看到我的對象的三個副本在我只期望一個時創建。 並且只想要一個。 我該怎么解決這個問題?
在我的真實代碼中,ThreadedThing是一個更大/更重的類,它位於一個線程池中,我寧願只有我真正需要的那么多。 但我寫了一個小演示應用程序(下面)來演示這種行為。 我從增強線程樣本中復制了基本代碼,所以我希望它能正常工作,所以我擔心這是一個C ++新手問題。
我之前編寫過多線程代碼,但是在Delphi而不是C ++中。 對於這個代碼,valgrind說沒有泄漏,這一切都很好,但仍然有三個對象被創建,我寧願有一個。
#include <iostream>
#include <boost/thread.hpp>
using namespace std;
class ThreadedThing {
public:
ThreadedThing() {
cout << "ThreadedThing created" << endl;
}
ThreadedThing(const ThreadedThing &orig) {
cout << "ThreadedThing copy created" << endl;
}
~ThreadedThing() {
cout << "ThreadedThing destroyed" << endl;
}
void operator()() {
cout << "ThreadedThing running" << endl;
sleep(2);
}
};
int main() {
std::vector < shared_ptr < boost::thread >> threads;
cout << "Started" << endl;
ThreadedThing thing;
std::shared_ptr<boost::thread> thread(new boost::thread(thing));
threads.push_back(thread);
for (std::vector < std::shared_ptr < boost::thread >> ::iterator it = threads.begin(); it != threads.end(); ++it) {
(*it)->join();
cout << "joined" << endl;
}
cout << "stopped" << endl;
return 0;
}
/* output
Started
ThreadedThing created
ThreadedThing copy created
ThreadedThing copy created
ThreadedThing destroyed
ThreadedThing running
ThreadedThing destroyed
joined
stopped
ThreadedThing destroyed
*/
在無法訪問C ++ 11時,只需在C ++ 11中使用std::ref
或使用boost::ref
。
有充分證據表明, boost::thread
的構造函數與std::thread
,不接受引用作為參數。 要解決此問題,您可以編寫一個包含引用的小shell對象,並在復制時保留它。
我就是這樣做的:
/* container to allow passing an object reference to std::thread()
* without this std::thread() would run on a copy
*/
template<typename T>
struct holder {
holder(T& payload) : payload(payload) {}
void operator()() { payload(); }
T& payload;
};
然后像這樣使用它:
BackgroundTaskQueue queue;
// start worker thread
holder<BackgroundTaskQueue> h(queue);
queuethread = new std::thread(h);
// h is not needed anymore
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