Specialize a method template with type parameter with a non-type one

Question

I'm trying to implement a class template with static member functions that can handle different types. My code looks like this (very simplified example):

#include <array>

struct Foo {
    template<typename T>
    static constexpr size_t foo(T v);
};

template<>
constexpr size_t Foo::foo(double v) {
    return sizeof(v);
}

template<int count>
constexpr size_t Foo::foo(const std::array<int, count>& v) {
    return sizeof(int) * v.size();
}

However, when I try to compile it, I get the following error:

<source>:16:23: error: out-of-line definition of 'foo' does not match any declaration in 'Foo'
   16 | constexpr size_t Foo::foo(const std::array<int, count>& v) {
      |                       ^~~
1 error generated.
Compiler returned: 1

How can I specialise a template with another template?

Answer 1

There is no partial specialzation for function templates, but you can partially specialize a class template:

#include <array>
#include <cstddef>

template <typename T> struct Foo_impl;

template <> struct Foo_impl<double> {
    constexpr static size_t value = sizeof(double);
};

template <size_t count> struct Foo_impl<std::array<int,count>> {
    constexpr static size_t value = count * sizeof(int);
};
 
struct Foo {
    template<typename T>
    static constexpr size_t foo(T v) { 
        return Foo_impl<T>::value;
    }
};

Also note that you must use the right type, size_t not int.