I'm using an array of pointer to function. I wrote the code like this since some of elements can not be expressed with function template.
extern void zero(); // isr 0 is defined somewhere else
void one() {
// isr 1
}
template <std::size_t N>
void Nth() {
// isr N
}
using func = void (*)();
constexpr func interrupt_vector[256] = {
&zero,
&one,
&Nth<2>,
&Nth<3>,
...
&Nth<254>,
&Nth<255>,
};
I've read about static table generation with variadic template, but those were about initializing the whole array.
How can I simplify the code?
@ Actually It's a part of interrupt vector. Since it should be called directly, I cannot use template specialization such as
template <>
void Nth<0>() {
zero();
}
@@ Edited the code. I think that cout things were confusing.
If you can change to use std::array
then something like this would work.
using func = void (*)();
template<int...>
struct index_sequence { };
template<int From, int N, int... Is>
struct make_index_sequence_from : make_index_sequence_from<From, N - 1, N - 1, Is...> { };
template<int From, int... Is>
struct make_index_sequence_from<From, From, Is...> : index_sequence<Is...> { };
template<int... Is> constexpr
std::array<func, 256> make_interrupt_vector_array(index_sequence<Is...>)
{
return {{zero, one, Nth<Is>...}};
}
constexpr
std::array<func, 256> make_interrupt_vector_array()
{
return make_interrupt_vector_array(make_index_sequence_from<2, 256>());
}
constexpr auto interrupt_vector = make_interrupt_vector_array();
I would recommended that you wrap your function around a class/struct so that you can take advantage of template specialization and declare your function as static inside of the class/struct
#include <iostream>
template <std::size_t N>
struct Nth
{
static void print()
{
std::cout << N << "!!" << std::endl;
}
};
template <>
struct Nth<0>
{
static void print()
{
std::cout << "Zero!!" << std::endl;
}
};
template <>
struct Nth<1>
{
static void print()
{
std::cout << "One!!" << std::endl;
}
};
int main()
{
Nth<0>::print();
Nth<1>::print();
Nth<2>::print();
}
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