Calling a zero argument template function pointer with variadic template argument?

Question

Here is a code snippet from a post at Functional C++ blog, describing how a generalized function evaluation can be implemented.

My question is how can you declare template function pointer f like R(C::*f)() with no arguments and still be able to call it with Args…?

// functions, functors, lambdas, etc.
template<
    class F, class... Args,
    class = typename std::enable_if<!std::is_member_function_pointer<F>::value>::type,
    class = typename std::enable_if<!std::is_member_object_pointer<F>::value>::type
    >
auto eval(F&& f, Args&&... args) -> decltype(f(std::forward<Args>(args)...))
{
    return f(std::forward<Args>(args)...);
}

// const member function
template<class R, class C, class... Args>
auto eval(R(C::*f)() const, const C& c, Args&&... args) -> R
{
    return (c.*f)(std::forward<Args>(args)...);
}

template<class R, class C, class... Args>
auto eval(R(C::*f)() const, C& c, Args&&... args) -> R
{
    return (c.*f)(std::forward<Args>(args)...);
}

// non-const member function
template<class R, class C, class... Args>
auto eval(R(C::*f)(), C& c, Args&&... args) -> R
{
    return (c.*f)(std::forward<Args>(args)...);
}

// member object
template<class R, class C>
auto eval(R(C::*m), const C& c) -> const R&
{
    return c.*m;
}

template<class R, class C>
auto eval(R(C::*m), C& c) -> R&
{
    return c.*m;
}

struct Bloop
{
    int a = 10;
    int operator()(){return a;}
    int operator()(int n){return a+n;}
    int triple(){return a*3;}
};

int add_one(int n)
{
    return n+1;
}

int main()
{
    Bloop bloop;

    // free function
    std::cout << eval(add_one,0) << "\n";

    // lambda function
    std::cout << eval([](int n){return n+1;},1) << "\n";

    // functor
    std::cout << eval(bloop) << "\n";
    std::cout << eval(bloop,4) << "\n";

    // member function
    std::cout << eval(&Bloop::triple,bloop) << "\n";

    // member object
    eval(&Bloop::a,bloop)++; // increment a by reference
    std::cout << eval(&Bloop::a,bloop) << "\n";

    return 0;
}

For instance, when I try:

struct Bloop
{
    int a = 10;
    int operator()(){return a;}
    int operator()(int n){return a+n;}
    int triple(){return a*3;}
    int foo(int n) {return n;}
};

template <typename R, typename C, typename... Args>
void eval (R(C::*func)(), C& c, Args... args) {
    (c.*func)(args...);
}

int main()
{
    Bloop bloop;

    eval(&Bloop::foo, bloop, 5);

    return 0;
}

I get this error:

main.cpp: In function 'int main()':
main.cpp:27:31: error: no matching function for call to 'eval(int (Bloop::*)(int), Bloop&, int)'
     eval(&Bloop::foo, bloop, 5);
                               ^
main.cpp:27:31: note: candidate is:
main.cpp:19:6: note: template<class R, class C, class ... Args> void eval(R (C::*)(), C&, Args ...)
 void eval (R(C::*func)(), C& c, Args... args) {
      ^
main.cpp:19:6: note:   template argument deduction/substitution failed:
main.cpp:27:31: note:   candidate expects 1 argument, 2 provided
     eval(&Bloop::foo, bloop, 5);
                               ^

And if I declare func like R(C::*func)(int) , it compiles.

Answer 1

The code in the blog article is incorrect (or at least incomplete); it only works for no-argument functions. You could write eval more correctly like this:

template<class R, class C, class... T, class... Args>
auto eval(R(C::*f)(T...), C& c, Args&&... args) -> R
{
    return (c.*f)(std::forward<Args>(args)...);
}

Note the T... parameter pack for arguments to the pointer to member function type. This is a distinct type pack from Args&&... because the two packs could be deduced differently.

Answer 2

This code could be made simultaneously simpler and more generic by avoiding the analysis of the pointer-to-member-function and pointer-to-member-data types and simply accepting anything for which the calls are well-defined:

#define RETURNS(...) \
  -> decltype(__VA_ARGS__) { \
    return (__VA_ARGS__); \
  }

// Function object type
template<class F, class... Args>
auto eval(F&& f, Args&&... args)
RETURNS(std::forward<F>(f)(std::forward<Args>(args)...))

// pointer to member function, object reference
template<class PMF, class C, class... Args>
auto eval(PMF&& pmf, C&& c, Args&&... args)
RETURNS((std::forward<C>(c).*std::forward<PMF>(pmf))(std::forward<Args>(args)...))

// pointer to member data, object reference
template<class PMD, class C>
auto eval(PMD&& pmd, C&& c)
RETURNS(std::forward<C>(c).*std::forward<PMD>(pmd))

while we're at it, we may as well support the omitted cases of pointer-to-members with object pointers in addition to object references for completeness, especially given that the sample code requires them to evaluate eval(&Bloop::a,&bloop)++ :

// pointer to member data, object pointer
template<class PMD, class P>
auto eval(PMD&& pmd, P&& p)
RETURNS((*std::forward<P>(p)).*std::forward<PMD>(pmd))

// pointer to member function, object pointer
template<class PMF, class P, class... Args>
auto eval(PMF&& pmf, P&& p, Args&&... args)
RETURNS(((*std::forward<P>(p)).*std::forward<PMF>(pmf))(std::forward<Args>(args)...))

DEMO

(Ok, maybe "simpler" was a poor choice of word. "More concise" or "terse" would probably be more accurate.)