有关QtConcurrent :: run中的异常的通知

Question

我知道实际上不处理在其他线程中引发的异常是没有意义的，但是有什么方法可以通知我至少发生了异常？例如类似

#include <QtConcurrentRun>

#include <iostream>
#include <stdexcept>

void MyFunction()
{
//  std::cout << "MyFunction()" << std::endl;
  throw std::runtime_error("Test exception.");
}

int main()
{
  try
  {
    QtConcurrent::run(MyFunction);
  }
  catch(...)
  {
    std::cout << "Exception caught!" << std::endl;
  }

}

即使发生异常，也会安静地退出。当异常来自某个调用堆栈深处的异常时，这有时会造成混乱。

- - - - - - 编辑 - - - - - - -

我试图写一个像UmNyobe建议的包装器，但是我对函数指针做错了吗？

#include <QtConcurrentRun>
#include <QFutureWatcher>
#include <QObject>

#include <iostream>
#include <stdexcept>

void MyFunction()
{
//  std::cout << "MyFunction()" << std::endl;
  throw std::runtime_error("Test exception.");
}

template<typename TFirstParam, typename... TParams>
bool ExceptionWrapper(TFirstParam firstParam, TParams&& ...params)
{
  // Here 'firstParam' should be a function pointer, and 'params' are the arguments
  // that should be passed to the function
  try
  {
    firstParam(params...);
  }
  catch(...)
  {
    std::cout << "Exception caught!" << std::endl;
    return false; // failure
  }

  return true; // success
}

struct MyClass : public QObject
{
  Q_OBJECT

  MyClass()
  {
    connect(&this->FutureWatcher, SIGNAL(finished()), this, SLOT(slot_finished()));
  }

  void DoSomething()
  {
    void (*myFunctionPointer)() = MyFunction;
    bool (*functionPointer)(decltype(myFunctionPointer)) = ExceptionWrapper;

    QFuture<bool> future = QtConcurrent::run(functionPointer);
    this->FutureWatcher.setFuture(future);
  }

  QFutureWatcher<void> FutureWatcher;

  void slot_finished()
  {
    std::cout << "Finished" << std::endl;
    if(!this->FutureWatcher.result())
    {
      std::cout << "There was an error!" << std::endl;
    }
  }
};

#include "ExceptionWrapper.moc"

int main()
{
  MyClass myClass = new MyClass;
  myClass->DoSomething();
}

我得到的错误是在这条线上：

QFuture<bool> future = QtConcurrent::run(functionPointer);

error: no matching function for call to 'run(bool (*&)(void (*)()))'

Answer 1

我知道实际上不处理在其他线程中引发的异常是没有意义的，但是有什么方法可以通知我至少发生了异常？

您可以使用从返回的Future来处理它QtConcurrent::run。有关详情，请参见此页面。当您将来收集时，所有未处理的异常都将被重新抛出。您可以创建一个简单的包装器类以捕获异常并在接收线程中对其进行检查。

#include <QtGui>
#include <iostream>
#include <stdexcept>

class MyException : public QtConcurrent::Exception
{
public:
    MyException(std::exception& err) : e(err) {}
    void raise() const { throw *this; }
    Exception* clone() const { return new MyException(*this); }
    std::exception error() const { return e; }
private:
    std::exception e;
};

// first concurrent function
int addFive(int n)
{
    try
    {
        throw std::runtime_error("kablammo!");
        //throw -1;
        return n + 5;
    }
    catch (std::exception& e)
    {
        throw MyException(e);
    }

}

// second concurrent function    
void myVoidFunction()
{
    try
    {
        throw std::runtime_error("oops!");
        //throw -1;
    }
    catch (std::exception& e)
    {
        throw MyException(e);
    }
}

int main(int argc, char* argv[])
{
    QApplication app(argc, argv);

    QFuture<int> f1 = QtConcurrent::run(addFive, 50);
    try
    {
        int r = f1.result();
        std::cout << "result = " << r << std::endl;
    }
    catch (MyException& me)
    {
        std::cout << me.error().what() << std::endl;
    }
    catch (QtConcurrent::UnhandledException&)
    {
        std::cout << "unhandled exception in addFive\n";
    }

    QFuture<void> f2 = QtConcurrent::run(myVoidFunction);
    try
    {
        // result() not available for QFuture<void>, use waitForFinished() to
        // block until it's done.
        f2.waitForFinished();
        std::cout << "myVoidFunction finished\n";
    }
    catch (MyException& me)
    {
        std::cout << me.error().what() << std::endl;
    }
    catch (QtConcurrent::UnhandledException&)
    {
        std::cout << "unhandled exception in myVoidFunction\n";
    }

    QWidget w;
    w.show();

    return app.exec();
}