Non-type variadic function templates in C++11

Question

I saw a blog post which used non-type variadic templates (currently not supported by gcc, only by clang).

template 
stru         


        

Answer 1

Luc Danton's solution doesn't behave correctly with parameters which are not of type int, but can be implicitly converted to an int. Here's one which does:

template<typename T, typename U> struct first_type { typedef T type; };
template<typename T> int max_checked(T n) { return n; }
template<typename T1, typename T2, typename ...Ts>
int max_checked(T1 n1, T2 n2, Ts ...ns)
{
  int maxRest = max_checked(n2, ns...);
  return n1 > maxRest ? n1 : maxRest;
}
template<typename ...T> auto max(T &&...t) ->
  decltype(max_checked<typename first_type<int, T>::type...>(t...))
{
  return max_checked<typename first_type<int, T>::type...>(t...);
}

struct S { operator int() { return 3; } };
int v = max(1, 2.0, S()); // v = 3.

Here, max forwards all arguments unchanged to max_checked, which takes the same number of arguments of type int (provided by performing a pack-expansion on the first_type template). The decltype(...) return type is used to apply SFINAE if any argument can't be converted to int.

Answer 2

Here are two ways of defining a variadic function template only accepting int parameters. The first one generates a hard-error when instantiated, the second uses SFINAE:

template<typename... T>
struct and_: std::true_type {};

template<typename First, typename... Rest>
struct and_
: std::integral_constant<
    bool
    , First::value && and_<Rest...>::value
> {};

template<typename... T>
void
foo(T... t)
{
    static_assert(
        and_<std::is_same<T, int>...>::value
        , "Invalid parameter was passed" );
    // ...
}

template<
    typename... T
    , typename = typename std::enable_if<
        and_<std::is_same<T, int>...>::value
    >::type
>
void
foo(T... t)
{
    // ...
}

As you can see, non-type template parameters aren't used here.

Answer 3

This will print out all elements, get max could be implemented similarly

template <int N>
void foo(){
  cout << N << endl;
}

template <int N, int M, int ... Rest>
void foo(){
  cout << N << endl;
  foo<M, Rest...>();
}


int main(){
  foo<1, 5, 7>();

  return 0;
}

Answer 4

You are simply confusing type names and non-type names. What you want simply doesn’t work.

You can probably use variadic non-type templates in functions, but not as (non-template) arguments:

template <int N, int... Rest>
int max()
{
    int tmp = max<Rest...>();
    return N < tmp ? tmp : N;
}

std::cout << max<3, 1, 4, 2, 5, 0>() << std::endl;

… although I haven’t tested this and I’m not sure how this should work given that you need to have a partial specialisation as the base case. You could solve this by dispatching to a partially specialised struct:

template <int N, int... Rest>
struct max_t {
    static int const value = max_t<Rest...>::value > N ? max_t<Rest...>::value : N;
};

template <int N>
struct max_t<N> {
    static int const value = N;
};


template <int... NS>
int max()
{
    return max_t<NS...>::value;
}

This will work.

Answer 5

Here's how you could achieve variadic args in your max example such that it can take any number of arithmetic arguments:

template<typename T, typename ... Ts>
struct are_arithmetic{
    enum {
        value = std::is_arithmetic<T>::value && are_arithmetic<Ts...>::value
    };
};

template<typename T>
struct are_arithmetic<T>{
    enum {
        value = std::is_arithmetic<T>::value
    };
};

template<typename Arg, typename = typename std::enable_if<std::is_arithmetic<Arg>::value>::type>
Arg max(Arg arg){
    return arg;
}

template<typename Arg, typename Arg1, typename ... Args, typename = typename std::enable_if<are_arithmetic<Arg, Arg1, Args...>::value>::type>
auto max(Arg arg, Arg1 arg1, Args ... args){
    auto max_rest = max(arg1, args...);
    return arg > max_rest ? arg : max_rest;
}

int main(){
    auto res = max(1.0, 2, 3.0f, 5, 7l);
}

This is good because it can take any numeric type and will return the maximum number by its original type rather than just int, too.