12

我正在尝试一个程序：

#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;
}

编译器 VS11 x64。

建造：

cl /EHsc /Zi async.cpp && async

对我来说，这个程序崩溃了。我怀疑，同时运行的期货数量是有限的。如果我将迭代次数减少到几个订单，它就会起作用。

所以，两个问题：

  1. 在 C++11 中实际运行期货是否有限制？

  2. 为什么这段代码会崩溃？如果我在“async()”之后立即明确地执行“get()”，它必须在下一次迭代之前完成未来，这意味着一次只运行一个未来。

更新

我已将代码简化为：

#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;
}

它仍然对我崩溃。它不会抛出，我已经检查过了。但现在我有一个可见的堆栈跟踪：

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

和线程：

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

我正在使用 VS 11.0.40825.2 PREREL。

4

4 回答 4

4

  1. 显然存在实现限制，就像数组的大小有限制一样。如果启动策略是 lauch::async 并且系统无法启动新线程，则 std::async 可以发出错误信号“resource_unavailable_try_again”。但是您没有收到此错误。

  2. 该程序不应该崩溃，也不适合我（VS11 x64，发布版本，相同的源代码和命令行）。

    我相信即使没有.get()该程序，一次也不会有多个异步操作。您将未来分配给一个局部变量，未来在每次循环迭代时都会被销毁，从而强制异步操作​​在下一次循环迭代中开始另一个操作之前完成。

于 2012-04-24T18:16:33.820 回答
1

  1. 不，标准没有说关于线程支持库的限制（包括线程、未来等）。

  2. 这取决于线程支持库和底层 API 的实现质量。正如您所说f.get()，等待任务编译（此行为需要 C++ 标准）。当库实现可能无法重用底层 API 的线程句柄等资源时，可能会导致系统资源不足和程序崩溃。这是图书馆实施的质量。

于 2012-05-02T07:38:35.967 回答
1

尝试使用 try-catch 将代码包含在 main() 中，并检查它是否引发 std::exception。这可能会给出提示。除此之外，VS 中的 C++11 仍处于测试阶段。

于 2012-04-29T15:28:22.967 回答
0

如果您将 g++ 与 std=c++11 选项一起使用，请确保您拥有最新版本并与 pthread 链接。

例如，在 CodingGround 上，您获得的默认编译（使用编译按钮）是：

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

它失败了"what(): Unknown error -1"。但是您可以手动将-lpthread添加到编译命令中：

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

它会工作得很好。

于 2017-03-17T06:43:21.740 回答