functor generation from member function pointer type

问题

I am trying to simplify (via make_fn()) the generation of functors that preprocess parameters (via wrap()) for member functions of arity n.
Generating the functors is basically working, but until now only by explicitly specifying the parameter types for the member function.
Now i'd like to generate the correct functor from the member function type it handles:

struct X {};

template<class C, typename T1, bool (C::*F)(T1)>
inline // there are more for T1..TN
bool wrap(C* c, X x) 
{
    return (c->*F)(process<T1>(x));
}

template<class C, typename T1, bool (C::*F)(T1)> 
inline // there are more for T1..TN
boost::function<bool (C*, X)> make_fn(F f) // <- problem here, F is not a type
{
    return boost::bind(&wrap<C, T1, F>, _1, _2);
}

With this however, vc++ and g++ don't see F as a type for the parameter of make_fn(). I must miss something obvious here and am feeling somewhat blind.

The idea was that it should work like this:

struct A 
{
    bool f1(bool) { return true; }
};

void test()
{
    A a;
    X x;
    make_fn(&A::f1)(&a, x);
}

Any ideas on how to make that work?

Background:
I have a fixed interface which, when simplified, looks like this:

bool invoke(C* c, const char* const functionName, int argCount, X* args);

X is a variant type which i have to convert to certain backend types (int, std::string, ...).
To handle these calls i have a map of functors that are looked up by name and map these calls to member functions of some instance.
The intention of the wrapping is to avoid manual conversions and instead generate functors which do the conversion for me or throw. I have this working with a macro based solution, but that solution requires to specify the types and the parameter count explicitly.
Via function overload resolution i hope to generate the correct converting functor implicitly from the member function signature.

回答1:

It appears to me that you are attempting to turn a pointer passed to a function into a non-type template argument, which I'm afraid is not going to work (see comments to your question).

What you could do, is to store the function pointer in a function object. The following appears to compile:

#include <boost/bind.hpp>
#include <boost/function.hpp>

struct X {};

template <class T>
bool process(X) { return true; }


template <class C, class T1, class Func>
struct wrap1
{
    typedef bool result_type;
    Func f;

    wrap1(Func f): f(f) {}

    bool operator()(C* c, X x)
    {
        return (c->*f)(process<T1>(x));
    }
};

template<class C, typename T1>
inline // there are more for T1..TN
boost::function<bool (C*, X)> make_fn(bool (C::*f)(T1))
{
    return boost::bind(wrap1<C, T1, bool (C::*)(T1)>(f), _1, _2);
}


struct A
{
    bool f1(bool) { return true; }
};

void test()
{
    A a;
    X x;
    make_fn(&A::f1)(&a, x);
}

However, I'm not sure if that is any good and how you would create the rest of the wrappers. For the latter you might just get a compiler that supports variadic templates. :)

来源：https://stackoverflow.com/questions/1548686/functor-generation-from-member-function-pointer-type

标签

c++

templates

functor