How can I pass multiple function overloads as a single argument?

Question

This is another case of trying to get rid of a macro. Consider

void f_(int i) { printf("f_(int)\t%d\n", i); }
void f_(int i, double x) { printf("f_(int, double)\t%d, %5.3f\n", i, x); }
void g_(int i) { printf("g_(int)\t%d\n", i); }
void g_(int i, double x) { printf("g_(int, double)\t%d, %5.3f\n", i, x); }

(Imagine f_() gets data form a .foo file or uses hard-coded "dummy" values, g_() does the same for .bar .) There might be a function to decide which overload to call:

void f(int i, double d) { (i > 0) ? f_(i, d) : f_(i); }

with the same logic duplicated for g_() :

void g(int i, double x) { (i > 0) ? g_(i, x) : g_(i); }

Getting rid of that duplicated code is easy with a macro:

#define h(i, x, func_) (i > 0) ? func_(i, x) : func_(i);
// ...
h(-1, 314.1, f_);
h(99, 314.1, f_);
h(-1, 314.1, g_);
h(99, 314.1, g_);

But of course we'd rather not use macros in C++. The "obvious" template

template<typename T>
void h2(int i, double x, T t)
{
   (i > 0) ? t(i, x) : t(i);
}
// ...
h2(-1, 314.1, f_);

fails because the compiler can't figure out what overload of f_() to use.

How can I replace the functionality of the macro h ?

Answer 1

You can use a variadic lambda and have the lambda call the function you wish to call.

template<typename T>
void h2(int i, double x, T t)
{
   i > 0 ? t(i, x) : t(i);
}

int main()
{
    h2(-1, 314.1, [](auto... args){ f_(args...); });
    //                              ^^ change this to g_ if you want to use g_ instead of f_
}

Answer 2

If you don't mind changing the definition's of f_ and g_ , you can do something like

template<class... Ts> struct overloaded : Ts... { using Ts::operator()...; };
template<class... Ts> overloaded(Ts...) -> overloaded<Ts...>;

auto f_ = overloaded(
  [](int i) { printf("f_(int)\t%d\n", i); },
  [](int i, double x) { printf("f_(int, double)\t%d, %5.3f\n", i, x); }
);

auto g_ = overloaded(
  [](int i) { printf("g_(int)\t%d\n", i); },
  [](int i, double x) { printf("g_(int, double)\t%d, %5.3f\n", i, x); }
);

Then your h template is exactly what you want

template<typename T>
void h(int i, double x, T t)
{
   i > 0 ? t(i, x) : t(i);
}

The overloaded template shamelessly copied from this example for std::visit .

To make it work for C++11 , you have to adapt the overloaded template and add a helper function to deduce the type parameters

template<class... Ts> struct overloads;
template<> struct overloads<>{};
template<class T, class... Ts> struct overloads<T, Ts...> : T, overloads<Ts...> 
{ 
    overloads(T t, Ts... ts) : T(t), overloads<Ts...>(ts...) {}
    using T::operator(); 
};

template<class... Ts> overloads<Ts...> overloaded(Ts&&... ts) 
{ return overloads<Ts...>(std::forward<Ts>(ts)...); }

Answer 3

You cannot do it directly, because the type infered for T would need to include the parameter types, but you can do something along the line of

#include <iostream>

struct overloader {
    void foo(int) { std::cout << "int\n";}
    void foo(double) { std::cout << "double\n";}
};

template <typename T>
void bar(int x,double y,bool opt, T t){
    if (opt) { t.foo(x); }
    else { t.foo(y); }
}

int main() {
    overloader ol;
    bar(1,1,true,ol);
    bar(1,1,false,ol);
}

prints:

int

double