Recursive templates don't work as expected with static variables - c++

Question

The code
#include <iostream>
using namespace std;
template<int n> struct Fibo { static int x; };
template<> int Fibo<0>::x = 1;
template<> int Fibo<1>::x = 1;
template<int n> int Fibo<n>::x = Fibo<n-1>::x + Fibo<n-2>::x; //marked line
int main() {
cout << Fibo<5>::x << endl;
cout << Fibo<4>::x << endl;
cout << Fibo<3>::x << endl;
cout << Fibo<2>::x << endl;
cout << Fibo<1>::x << endl;
cout << Fibo<0>::x << endl;
return 0;
}
outputs
0
0
1
2
1
1
in VC++. (According to user M M. it compiles as expected in gcc). When the compiler gets to the marked line with n=5 it doesn't compile that same line again for n=4, but just treats Fibo<4>::x as if it were declared with
template<> int Fibo<4>::x; // x defaults to 0
Why is that? Why does it work as expected when using
template<int n> struct Fibo { enum { x = Fibo<n-1>::x + Fibo<n-2>::x }; };
template<> struct Fibo<0> { enum { x = 1 }; };
template<> struct Fibo<1> { enum { x = 1 }; };
instead, but not with a static variable? And how do you fix the first code (without enum)?

Answer 1

The Standard is very clear on this:
14.7.1 Implicit instantiation [temp.inst]
9 The implicit instantiation of a class template does not cause any
static data members of that class to be implicitly instantiated.
All the calls in main() to your Fibo<n>::x for n > 1, are explicit instantiations, that through the Fibonnaci recursion will implicitly instantiate Fibo<n-1> and Fibo<n-2> but not their members x. This means that at those points, the static members x will be evaluated to their default initialization of 0. For n=1 and n=0, the compiler will see the explicit initialization values of 1. So effectively, you get the following computation
Fibo<5>::x --> Fibo<4>::x + Fibo<3>::x --> 0 + 0 = 0
Fibo<4>::x --> Fibo<3>::x + Fibo<2>::x --> 0 + 0 = 0
Fibo<3>::x --> Fibo<2>::x + Fibo<1>::x --> 0 + 1 = 1
Fibo<2>::x --> Fibo<1>::x + Fibo<0>::x --> 1 + 1 = 2
Fibo<1>::x --> 1
Fibo<0>::x --> 1
You need to instantiate the static member x before evaluating the Fibonacci recursion. You can do this through a static const int or enum member x, or through a function (possibly constexpr in C++11) as shown by #Jarod42.

Answer 2

I'm not sure if the initialization order of the static variables of template<int n> int Fibo<n>::x = Fibo<n-1>::x + Fibo<n-2>::x; is specified...
You may write this:
template <int N> struct Fibo { int operator()() const { static int x = Fibo<N - 1>()() + Fibo<N - 2>()(); return x; } };
template <> struct Fibo<1> { int operator()() const { static int x = 1; return x; } };
template <> struct Fibo<0> { int operator()() const { static int x = 1; return x; } };
The dependencies are respected.
[Edit]
In a case where the value may be modified (according to your comment), you may use similar technique but returning reference:
template <int N> struct Fibo {
private:
int& operator()() { static int x = Fibo<N - 1>()() + Fibo<N - 2>()(); return x; }
public:
int operator()() const { return const_cast<Fibo&>(*this)(); }
// This change Fibo<0> and Fibo<1> and then update value up to Fibo<N>.
int operator(int fibo0, int fibo1) {
int n_1 = Fibo<N - 1>()(fibo1, fibo2);
(*this)() = n_1 + Fibo<N - 2>()();
}
};
template <> struct Fibo<1> {
private:
int& operator()() { static int x = 1; return x; }
public:
int operator()() const { return const_cast<Fibo&>(*this)(); }
void operator(int fibo0, int fibo1) { Fibo<0>()(fibo0); (*this)() = fibo1; }
};
template <> struct Fibo<0> {
private:
int& operator()() { static int x = 1; return x; }
public:
int operator()() const { return const_cast<Fibo&>(*this)(); }
void operator(int fibo0) { (*this)() = fibo0; }
};

Answer 3

The solution presented by #Jarod42 appears overly complicated to me.
Consider instead the simpler code below.
template<int N>
struct fib {
static const int val = fib<N-1>::val + fib<N-2>::val;
};
template<>
struct fib<0> { static const int val = 0;};
template<>
struct fib<1> { static const int val = 1;};
int main() {
std::cout << fib<45>::val << "\n";
return 0;
}