如何将 lambda 作为模板类的成员函数的参数

Question

I have C++ class which is template. it have member function which should take any lamda as parameter;

basically this is what I wanna do:-

#include <QFuture>
#include <QFutureWatcher>

template <class T>
class EFuture  {
private:
    QFuture<T> future;
    QFutureWatcher<T> watcher;

public:
    explicit EFuture(QFuture<T> &future);

    void onFinished(void (*f)() );
};


template <class T>
EFuture<T>::EFuture(QFuture<T> &future ): future(future)
{  }

template<class T>
void EFuture<T>::onFinished(void (*f)()){

    QObject::connect(watcher,&QFutureWatcher<T>::finished,f);
    watcher.setFuture(future);
}

This have serious restriction as I can't capture anything in lambda which I am passing. where I try to do something like this:-

future->onFinished([someobject](){
   ...
});

I get following error:-

connectionworker.cpp:106:24: error: no viable conversion from '(lambda at /home/noone/Development/Exatation/Exever/src/connectionworker.cpp:106:24)' to 'void (*)()'
efuture.h:17:28: note: passing argument to parameter 'f' here

Answer 1

Only non-capturing and non-generic lambda expressions are convertible to a function pointer. Itself, any lambda expression -- both captureless and capturing ones -- has its own type, known only to the compiler. In such a case, there are two alternatives:

• Use a function template that can deduce the type of the lambda expression:

   template <typename F> void onFinished(F f);

• Use a type-erasing technique, eg, std::function<void()> :

   #include <functional> void onFinished(std::function<void()> f);