Reputation: 1486
I think it's not possible but I'd like to ask you before give up about it.
I want something like a constexpr increment.
#include <iostream>
constexpr int inc() {
static int inc = 0;
return inc++;
}
class Foo {
static const int Type = inc();
};
class Foo2 {
static const int Type = inc();
};
int main() {
std::cout << "Foo1 " << Foo1::Type << st::endl;
std::cout << "Foo2 " << Foo2::Type << st::endl;
return 0;
}
I want to call it into some classes not manually (I use CRTP for that), to give a different type to each of them, but the type need to be const. There is anyway to achieve something like that in C++? (C++17 + TS)
Upvotes: 9
Views: 1198
Reputation: 1486
So there is the solution by Filip Roseen called the constant-expression counter:
#include <iostream>
template<int N>
struct flag {
friend constexpr int adl_flag (flag<N>);
};
template<int N>
struct writer {
friend constexpr int adl_flag (flag<N>) {
return N;
}
static constexpr int value = N;
};
template<int N, int = adl_flag (flag<N> {})>
int constexpr reader (int, flag<N>) {
return N;
}
template<int N>
int constexpr reader (float, flag<N>, int R = reader (0, flag<N-1> {})) {
return R;
}
int constexpr reader (float, flag<0>) {
return 0;
}
template<int N = 1>
int constexpr next (int R = writer<reader (0, flag<32> {}) + N>::value) {
return R;
}
class Foo {
public:
static const int Type = next();
};
class Foo2 {
public:
static const int Type = next();
};
int main() {
std::cout << "Foo1 " << Foo::Type << std::endl;
std::cout << "Foo2 " << Foo2::Type << std::endl;
return 0;
}
Thanks guys :) But it's too risky for use it in my main library which will be use in every project.
PS: I won't close this right now if there is another answer. Because yes it's ugly.
Upvotes: 3