Why is this variadic function ambiguous? - c++

Question

This is related to my earlier post. I'd like to know why one attempted solution didn't work.
template <typename... T> /* A */
size_t num_args ();
template <>
size_t num_args <> ()
{
return 0;
}
template <typename H, typename... T> /* B */
size_t num_args ()
{
return 1 + num_args <T...> ();
}
If I try to call, say, num_args<int,float>() then the error is that the function call is ambiguous:
A with T={int,float}
B with H=int, T={float}
I don't understand how this is ambiguous -- A is a declaration and B is a definition of the function declared by A. Right?
I'm trying to make this example work and the responses to my earlier question seem to claim that it can never work.
If that's the case, what's the point of variadic free functions? What can they do?

Answer 1

I don't understand how this is ambiguous -- A is a declaration and B
is a definition of the function declared by A. Right?
No. A is a declaration of a function template, and B is a declaration (and definition) of another function template.
The compiler has no way to decide between the two: they both have no arguments, and the template arguments are a match for both.
The one in the middle is an explicit total specialization of the function template declared in A.
If you tried to make B another specialization of A:
template <typename H, typename... T> /* B */
size_t num_args<H, T...>()
{
return 1 + num_args <T...> ();
}
... you'd end up with a partial specialization of a function template, which is not allowed.
You can do this with the usual trick of using a class template with partial specializations and a function template that calls into the class template:
template <typename... T>
class Num_Args;
template <>
struct Num_Args <>
{
static constexpr size_t calculate() {
return 0;
}
};
template <typename H, typename... T>
struct Num_Args <H, T...>
{
static constexpr size_t calculate() {
return 1 + Num_Args<T...>::calculate();
}
};
template <typename... T> /* B */
constexpr size_t num_args ()
{
return Num_Args<T...>::calculate();
}

Answer 2

Apropos the usefulness/uselessness of free variadic function templates: the usual use case for these is to have a variadic function parameter list, in which case a regular overload for the empty case will do just fine:
size_t num_args()
{
return 0;
}
template <typename H, typename... T> /* B */
size_t num_args (H h, T... t)
{
return 1 + num_args(t...);
}
EDIT:
As far as I can see, the following abuse of enable_if ought to work as a solution to your original question:
#include <utility>
// Only select this overload in the empty case
template <typename... T>
typename std::enable_if<(sizeof...(T) == 0), size_t>::type
num_args()
{
return 0;
}
template <typename H, typename... T>
size_t
num_args()
{
return 1 + num_args<T...>();
}
(Edit2: Reversed the order of the overloads to make the code actually compile)

Answer 3

While I really like the std::enable_if<sizeof...(T) == 0, _> hack by JohannesD, I'll still drop the below hack that I don't actually remember where I learned from, which resolves the ambiguity with only free functions without having to use classes:
template <typename One>
size_t num_args() {
return 1;
}
template <typename First, typename Next, typename... Rest>
size_t num_args() {
return 1 + num_args<Next, Rest...>();
}
No partial specialization required, and it totally avoids the zero parameter case by ending the unpacking recursion at one parameter. Also, see https://stackoverflow.com/a/58295007/1525238 for an even nicer c++17 solution that uses the same idea.
The only catch is that it actually doesn't cover the zero template parameter case, which is impossible in the context of free functions AFAIK. If that is required from the consumer's perspective, a partially specialized class must be used, see R. Martinho Fernandes' answer.
As for the utility of variadic template free functions, I find them particularly useful as constexpr type-safe accumulating calculator utilities.