I can't have my int class wrapper acting like a primitive int in template specialization.
I've prepared this code to explain my issue in detail:
#include <iostream>
#include <stdlib.h>
class Integer
{
int _v;
public:
constexpr explicit Integer(int v) : _v(v) {}
constexpr Integer next() const { return Integer(_v + 1); }
constexpr operator int() const { return _v;}
};
static constexpr auto integer1 = Integer(1);
static constexpr auto integer2a = Integer(2);
static constexpr auto integer2b = integer1.next();
template <const Integer& i>
void foo_Integer()
{
static auto foo_id = rand();
std::cout << foo_id << std::endl;
}
static constexpr auto int1 = 1;
static constexpr auto int2a = 2;
static constexpr auto int2b = int1 + 1;
template <int i>
void foo_int()
{
static auto foo_id = rand();
std::cout << foo_id << std::endl;
}
int main()
{
foo_int<int1>();
foo_int<int2a>();
foo_int<int2b>(); // same template specialization as above -> :)
foo_Integer<integer1>();
foo_Integer<integer2a>();
foo_Integer<integer2b>(); // different template specialization -> :(
}
As you can see running the code
foo_int<int2a>();
foo_int<int2b>();
use the same template specialization, while
foo_Integer<integer2a>();
foo_Integer<integer2b>();
use different template specializations.
This is, of course, correct, from the compiler point of view, since the template accepts a const Integer&, but I hope there are other better approaches to workaround the issue.
You can easily make Integer a structural type (C++20). Then its values are valid template parameters.
class Integer
{
public:
int _v;
constexpr explicit Integer(int v) : _v(v) {}
constexpr Integer next() const { return Integer(_v + 1); }
constexpr operator int() const { return _v;}
};
template <Integer i>
void foo_Integer()
{
static auto foo_id = rand();
std::cout << foo_id << std::endl;
}
Live
And the values would be equivalent, even if they come from objects with different identities.
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