在C++ 11

Question

我想一個程序編號異步/期貨：

#include <iostream> 
#include <thread> 
#include <future> 

int foo() { 
    return 0; 
} 

int main(int argc, char* argv[]) { 
    for (auto i = 0L; i < 10000; ++i) { 
    auto f = std::async(foo); 
    f.get(); 
    } 
    return 0; 
} 

編譯VS11 64。

體形：

cl /EHsc /Zi async.cpp && async 

對於我這個程序崩潰。我懷疑，同時實際運行的期貨數量是有限的。如果我將迭代次數減少到幾個訂單，它就可以工作。

於是，兩個問題：

1. 是否有實際運行期貨在C++ 11的限制？

2. 爲什麼此代碼會崩潰？如果我明確地在「async（）」之後立即執行「get（）」，它必須在下一次迭代之前完成未來，這意味着一次只能運行一個未來。

UPDATE

我已經簡化了代碼：

#include <future> 

int main(int argc, char* argv[]) { 
    for (auto i = 0L; i < 1000000; ++i) { 
    auto f = std::async([](){ return 0; }); 
    f.get(); 
    } 
    return 0; 
} 

它仍然崩潰了我。它不扔，我檢查了。但現在我有一個可見的堆棧跟蹤：

async.exe!_Mtx_unlock(_Mtx_internal_imp_t * * mtx) Line 229 C++ 
async.exe!std::_Mtx_unlockX(_Mtx_internal_imp_t * * _Mtx) Line 84 C++ 
async.exe!std::_Mutex_base::unlock() Line 47 C++ 
async.exe!std::unique_lock<std::mutex>::~unique_lock<std::mutex>() Line 284 C++ 
async.exe!std::_Associated_state<int>::_Set_value(int && _Val, bool _At_thread_exit) Line 358 C++ 
async.exe!std::_Packaged_state<int __cdecl(void)>::_Call_immediate() Line 569 C++ 
async.exe!std::_Async_state<int>::`std::U_Nil::ain::ain'::`3'::<lambda_A200A86DFF9A63A1>::operator()() Line 700 C++ 
async.exe!??$_ApplyX@X@?$_Callable_obj@V<lambda_A200A86DFF9A63A1>@?2???$?0V?$_Bind@$0A@XV<lambda_23AC5A2FBB53FD4D>@?5?main@U_Nil@std@@U23@U23@U23@U23@U23@U23@@std@@@?$_Async_state@H@std@@QEAA@$$QEAV?$_Bind@$0A@XV<lambda_23AC5A2FBB53FD4D>@?5?main@U_Nil@std@@U23@U23@U23@U23@U23@U23@@3@@Z@$0A@@std@@QEAAXXZ() Line 420 C++ 
async.exe!?_Do_call@?$_Func_impl@U?$_Callable_obj@V<lambda_A200A86DFF9A63A1>@?2???$?0V?$_Bind@$0A@XV<lambda_23AC5A2FBB53FD4D>@?5?main@U_Nil@std@@U23@U23@U23@U23@U23@U23@@std@@@?$_Async_state@H@std@@QEAA@$$QEAV?$_Bind@$0A@XV<lambda_23AC5A2FBB53FD4D>@?5?main@U_Nil@std@@U23@U23@U23@U23@U23@U23@@3@@Z@$0A@@std@@V?$allocator@V?$_Func_class@XU_Nil@std@@U12@U12@U12@U12@U12@U12@@std@@@2@XU_Nil@2@U42@U42@U42@U42@U42@U42@@std@@UEAAXXZ() Line 217 C++ 
async.exe!std::_Func_class<void,std::_Nil,std::_Nil,std::_Nil,std::_Nil,std::_Nil,std::_Nil,std::_Nil>::operator()() Line 486 C++ 
async.exe!`Concurrency::details::_MakeVoidToUnitFunc'::`3'::<lambda_25D33530A43E1C90>::operator()() Line 1056 C++ 
async.exe!std::_Callable_obj<`Concurrency::details::_MakeVoidToUnitFunc'::`3'::<lambda_25D33530A43E1C90>,0>::_ApplyX<Concurrency::details::_Unit_type>() Line 420 C++ 
async.exe!std::_Func_impl<std::_Callable_obj<`Concurrency::details::_MakeVoidToUnitFunc'::`3'::<lambda_25D33530A43E1C90>,0>,std::allocator<std::_Func_class<Concurrency::details::_Unit_type,std::_Nil,std::_Nil,std::_Nil,std::_Nil,std::_Nil,std::_Nil,std::_Nil> >,Concurrency::details::_Unit_type,std::_Nil,std::_Nil,std::_Nil,std::_Nil,std::_Nil,std::_Nil,std::_Nil>::_Do_call() Line 217 C++ 
async.exe!std::_Func_class<Concurrency::details::_Unit_type,std::_Nil,std::_Nil,std::_Nil,std::_Nil,std::_Nil,std::_Nil,std::_Nil>::operator()() Line 486 C++ 
async.exe!`Concurrency::details::_Task_impl<Concurrency::details::_Unit_type>::_ScheduleTask'::`3'::<lambda_7D9BCD859405C05B>::operator()() Line 325 C++ 
async.exe!Concurrency::details::_PPLTaskHandle<`Concurrency::details::_Task_impl<Concurrency::details::_Unit_type>::_ScheduleTask'::`3'::<lambda_7D9BCD859405C05B> >::operator()() Line 72 C++ 
async.exe!Concurrency::details::_UnrealizedChore::_InvokeBridge<Concurrency::details::_PPLTaskHandle<`Concurrency::details::_Task_impl<Concurrency::details::_Unit_type>::_ScheduleTask'::`3'::<lambda_7D9BCD859405C05B> > >(Concurrency::details::_PPLTaskHandle<`Concurrency::details::_Task_impl<Concurrency::details::_Unit_type>::_ScheduleTask'::`3'::<lambda_7D9BCD859405C05B> > * _PChore) Line 4190 C++ 
async.exe!Concurrency::details::_UnrealizedChore::_UnstructuredChoreWrapper(Concurrency::details::_UnrealizedChore * pChore) Line 275 C++ 
async.exe!Concurrency::details::_PPLTaskChore::_DeletingChoreWrapper(Concurrency::details::_UnrealizedChore * pChore) Line 78 C++ 
async.exe!Concurrency::details::InternalContextBase::ExecuteChoreInline(Concurrency::details::WorkItem * pWork) Line 1600 C++ 
async.exe!Concurrency::details::InternalContextBase::Dispatch(Concurrency::DispatchState * pDispatchState) Line 1704 C++ 
async.exe!Concurrency::details::FreeThreadProxy::Dispatch() Line 191 C++ 
async.exe!Concurrency::details::ThreadProxy::ThreadProxyMain(void * lpParameter) Line 173 C++ 
kernel32.dll!0000000076df652d() Unknown 
ntdll.dll!0000000076f2c521() Unknown 

和線程：

Unflagged  1864 0 Worker Thread ntdll.dll thread ntdll.dll!0000000076f518ca Normal 
Unflagged  10964 0 Main Thread Main Thread async.exe!do_signal Normal 
Unflagged  7436 0 Worker Thread ntdll.dll thread ntdll.dll!0000000076f52c1a Normal 
Unflagged  10232 0 Worker Thread async.exe!Concurrency::details::ThreadProxy::ThreadProxyMain async.exe!Concurrency::details::ThreadProxy::SuspendExecution Normal 
Unflagged > 10624 0 Worker Thread async.exe!Concurrency::details::ThreadProxy::ThreadProxyMain async.exe!_Mtx_unlock Normal 
Unflagged  4756 0 Worker Thread async.exe!Concurrency::details::ThreadProxy::ThreadProxyMain async.exe!Concurrency::details::ThreadProxy::SuspendExecution Normal 
Unflagged  11100 0 Worker Thread async.exe!Concurrency::details::ThreadProxy::ThreadProxyMain async.exe!Concurrency::details::InternalContextBase::WaitForWork Normal 
Unflagged  6440 0 Worker Thread async.exe!Concurrency::details::ThreadProxy::ThreadProxyMain async.exe!std::vector<std::pair<void (__cdecl*)(void * __ptr64),void * __ptr64>,std::allocator<std::pair<void (__cdecl*)(void * __ptr64),void * __ptr64> > >::_Tidy Normal 

我使用VS 11.0.40825.2 PREREL。

Answer 1

1. 顯然存在實施限制，就像數組的大小有限制一樣。如果啓動策略是lauch :: async並且系統無法啓動新線程，則std :: async可以發出錯誤'resource_unavailable_try_again'。但是你沒有得到這個錯誤。

2. 程序不應該崩潰，不適合我（VS11 64，發佈版本，相同的源和命令行）。

   我相信，即使沒有.get()程序不會有一個以上的異步操作會馬上。將未來分配給局部變量，將在每次循環迭代中銷燬未來，迫使異步操作在下次循環迭代中啓動另一個循環之前完成。

Answer 2

嘗試用try-catch將你的代碼放在main（）中，並檢查它是否拋出std :: exception。這可能會提示。除此之外，VS中的C++ 11仍然是測試版。

Answer 3

1. 不，標準沒有對線程支持庫（包括線程，未來等）的限制。

2. 這取決於線程支持庫的實施質量和底層API。正如你所說的f.get()等待任務壓縮（此行爲需要C++標準）。當庫實現不能重用像底層API的線程句柄這樣的資源時，它可能會導致缺乏系統資源和程序崩潰。這是圖書館實施的質量。

Answer 4

如果您使用g ++與std = C++ 11選項，請確保您有最新版本並鏈接到pthread。

例如，在CodingGround，你得到（與編譯按鈕），默認的編譯是：

g++ -std=c++11 -o main *.cpp 

和它失敗「是什麼（）：未知錯誤-1」。但是，您可以手動添加-lpthread到編輯命令：

g++ -std=c++11 -o main -lpthread *.cpp 

，它會工作得很好。