Overloaded function as argument of variadic template function

Question

I\'m trying to make variadic template function, which takes as arguments overloaded function and its arguments :)

int sumall(int a) { return a; }
int sumall(         


        

Answer 1

(If you're prepared to use variadic macros, then scroll to the end of this answer to see a better answer which make everything fully variadic. But I think that variadic macros are just a g++ extension.)

It can be made to work, if you're prepared to put the name of the function at the end of the parameter list. By putting it later, the compiler can deduce the necessary types from the earlier parameters to doit:

cout << doit(7, 6, sumall) << endl;
cout << doit(10, sumall) << endl;

Here is a demo on ideone.

The downside is that you have to implement one doit for each number of parameters. I've only implemented it for one- and two- parameter functions, but it shouldn't be a problem to extend this:

int sumall(int a) { return a; }
int sumall(int a, int b) { return a+b; }

template<typename A1, typename A2, typename R>
auto doit( A1 a1, A2 a2, R (*f) (A1,A2)) -> R {
    return f(a1, a2);
}
template<typename A1, typename R>
auto doit( A1 a1, R (*f) (A1)) -> R {
    return f(a1);
}

Update: Sometimes, it might appear that you can get away with having f as the first argument. But that's not as robust as putting it at the end. Consider the example where where are two functions that take the same number of arguments, but different types of parameters. e.g.:

int sumall(int a, int b) { return a+b; }
string sumall(string a, string b) { return a+" "+b; }

You need to have the function as the last argument, in order that the template deduction can use the type and number of parameters at the start to deduce the types of the arguments. Here's a demo on ideone of function-arg first and function-arg last.

The only downside with putting the arg at the end is that we can't then use variadic templates - variadic arg packs must be at the end. And you must get the types exactly right - see how I had to use string("hi") instead of simply "hi".

Using variadic macros to have the best of all worlds

By implementing doit as a macro, and using variadic macros (a gcc/g++ extension), it is possible to have a fully variadic solution with the function name appearing first. A demo on ideone.

cout << doit(sumall, 7, 6) << endl;
cout << doit(sumall, 10) << endl;
cout << doit(sumall, string("hi"), string("world")) << endl;

By using decltype and a couple of other simple classes, we can use the args provided to deduce the types of the args and then it can use that to select the right method from the overload set and deduce the return type from that.

template<typename ...Args>
struct OverloadResolved {
        template<typename R>
        static auto static_doit( R (*f) (Args...), Args ... args ) -> R {
                return f(args...);
        }
};

template<typename ...Args>
auto deduce(Args...) -> OverloadResolved<Args...> {
        return OverloadResolved<Args...>();
}

template<typename T>
struct dummy : public T { };

#define doit(f, ...) ( dummy<decltype(deduce( __VA_ARGS__ ))> :: static_doit(f, __VA_ARGS__) )

I'm pretty sure this is a safe use of macros, nothing will be evaluated twice (nothing inside decltype actually executes.

Answer 2

When taking a pointer of a function the compiler needs to know which of the overloads you want to use. There is no way to pass a pointer to an "overload set" and have the compiler decide later. Neither of you examples works with any of the compilers I tried (very recent versions of EDG, gcc, and clang).

I don't think you can do what you want without changing the notation of your call. If you are willing to change the call you can encapsulate the knowledge about the function to be called into a class, e.g.:

struct sumall_t {
    template <typename... T>
    auto operator()(T... args) -> decltype(sumall(args...)) {
        return sumall(args...);
    }
};

This effectively creates a wrapper for an overload set. Since the result type can't be deduced directly and may depend on how the function is called, you'd need to use a different version of doit() as well:

template<typename Func, typename... A>
auto doit( Func f, A... a) ->decltype(f(a...)) {
    return f(a...);
}

This would then be used something like this:

doit(sumall_t(), 1, 2);

Another way to fix this is to mandate specification of the result type: in some way you try to do two things at once: you want to deduce the result type of the function to be called and you want to guide the compiler to choose a specific overload of a result set. However, these are interdependent. If you remove any dependency on deducing any template from the function pointer, you don't need wrap the overload set because you can determine the choice of overloaded function from the first argument to the function. In case you claim that "my compiler can do it if the return type isn't void" I'd say that your compiler is actually wrong in doing this.