How to write a variadic template recursive function?

Question

I\'m trying to write a variadic template constexpr function which calculates sum of the template parameters given. Here\'s my code:

template<         


        

Answer 1

Just sum it up the usual way.

template<int... Args>
constexpr int f() {
   int sum = 0;
   for(int i : { Args... }) sum += i;
   return sum;
}

Answer 2

Your base case was wrong. You need a case for the empty list, but as the compiler suggests, your second try was not a valid template specialization. One way to define a valid instantiation for zero arguments is to create an overload that accepts an empty list

template<class none = void>
constexpr int f()
{
    return 0;
}
template<int First, int... Rest>
constexpr int f()
{
    return First + f<Rest...>();
}
int main()
{
    f<1, 2, 3>();
    return 0;
}

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EDIT: for completeness sake also my first answer, that @alexeykuzmin0 fixed by adding the conditional:

template<int First=0, int... Rest>
constexpr int f()
{
    return sizeof...(Rest)==0 ? First : First + f<Rest...>();
}

Answer 3

It is quite similar to what @T.C suggested above:

#include<iostream>
#include<numeric>

template<int First, int... Rest>
constexpr int f()
{
    auto types = {Rest...};
    return std::accumulate(std::begin(types), std::end(types),0);   
}

int main()
{
    std::cout <<f<1, 2, 3>();
    return 0;
}

Answer 4

I find it generally easier to move the code from template arguments to function arguments:

constexpr int sum() { return 0; }

template <class T, class... Ts>
constexpr int sum(T value, Ts... rest) {
    return value + sum(rest...);
}

If you really want them as template arguments, you can have your f just call sum by moving them down:

template <int... Is>
constexpr int f() {
    return sum(Is...);
}

It's constexpr, so just using ints is fine.

Answer 5

template<int First, int... Rest>
constexpr int f()
{
    return First + f<Rest...>();
}

template<int First>
constexpr int f()
{
    return First;
}

int main()
{
    f<1, 2, 3>();
    return 0;
}

You get this error:

error C2668: 'f': ambiguous call to overloaded function while trying to resolve f<3,>() call.

This is because a variadic parameter pack can be given 0 arguments, so f<3> could work with template<int First, int... Rest> by "expanding" to template<3, >. However, you also have the specialization of template<int First>, so the compiler does not know which one to choose.

Explicitly stating the first and second template arguments is a completely valid and good solution to this problem.

---

When you try to change the base case to:

template <>
constexpr int f()
{
    return 0;
}

You have a problem because functions cannot be specialized in that way. Classes and structs can be, but not functions.

---

Solution #1: C++17 fold expression with constexpr

template <typename... Is>
constexpr int sum(Is... values) {
    return (0 + ... + values);
}

---

Solution #2: Use a constexpr function

constexpr int sum() {
    return 0;
}

template <typename I, typename... Is>
constexpr int sum(I first, Is... rest) {
    return first + sum(rest...);
}

---

Solution #3: Use Template Metaprogramming

template <int... Is>
struct sum;

template <>
struct sum<>
    : std::integral_constant<int, 0>
{};

template <int First, int... Rest>
struct sum<First, Rest...>
    : std::integral_constant<int,
        First + sum_impl<Rest...>::value>
{};

Answer 6

A more generic solution using std::initializer_list would be:

template <typename... V>                                                                                                                                                                                         
auto sum_all(V... v)
{
  using rettype = typename std::common_type_t<V...>;
  rettype result{};
  (void)std::initializer_list<int>{(result += v, 0)...};
  return result;
}