constexpr base case not considered during compilation - c++

Question

I was trying out the constexpr functions and stumbled upon the below example of implementing Fibonacci numbers
There is no logical difference between fibon2 and fibon1 but I still get the compilation error of exceeding template initializations for the fibon1.
What i'm missing here?
#include <cstdint>
#include <iostream>
template <int32_t x>
constexpr int32_t fibon2() {
if constexpr (x == 0)
return 1;
else if constexpr (x == 1)
return 1;
else if constexpr (x == 2)
return 1;
else
return fibon2<x - 1>() + fibon2<x - 2>();
}
template <int32_t x>
constexpr int32_t fibon1() {
if constexpr (x == 0) return 1;
if constexpr (x == 1) return 1;
if constexpr (x == 2) return 1;
return fibon1<x - 1>() + fibon1<x - 2>();
}
int32_t fibon3(int32_t x) {
if (x == 1) return 1;
if (x == 2) return 1;
return fibon3(x - 1) + fibon3(x - 2);
}
int main() {
std::cout << fibon3(2) << std::endl;
std::cout << fibon2<2>() << std::endl;
std::cout << fibon1<2>() << std::endl;
return 0;
}

Answer 1

The return in fibon1 is not under constexpr and thus has to be compiled regardless of the template argument provided.
Clang show this nicely: https://godbolt.org/z/577f15Kv1
If you put the return inside its own if constexpr then your fibon1 will compile:
template <int32_t x>
constexpr int32_t fibon1() {
if constexpr (x == 0) return 1;
if constexpr (x == 1) return 1;
if constexpr (x == 2) return 1;
if constexpr (x > 2)
return fibon1<x - 1>() + fibon1<x - 2>();
}

Answer 2

The last return in fibon2 is in a constexpr-if even though there's only an else there. It's the same as if you'd made it:
else if constexpr (x == 2)
return 1;
else if constexpr(true) // <- like this
return fibon2<x - 1>() + fibon2<x - 2>();
That's not the case in fibon1 which is why it fails. If you want a freestanding if constexpr for it, then
template <int32_t x>
constexpr int32_t fibon1() {
if constexpr (x == 0) return 1; // shouldn't this be 0 ?
if constexpr (x == 1) return 1;
if constexpr (x == 2) return 1; // this doesn't seem correct if fibon1<0> => 1
// added constexpr-if:
if constexpr (x < 0 || x > 2) return fibon1<x - 1>() + fibon1<x - 2>();
}
Note that the above will fail if you supply a negative value as a template parameter as it will never reach one of your terminating conditions. Either the recursion will be to deep or you'll get a signed integer overflow. If you instead want symmetry around 0, you can simply negate the template parameter and return value in case the template parameter is negative:
template <int32_t x>
constexpr int32_t fibon1() {
if constexpr (x < 0)
return -fibon1<-x>(); // like this
else if constexpr (x == 0 || x == 1) // assuming you want 0 for input 0
return x;
else
return fibon1<x - 1>() + fibon1<x - 2>();
}