(Uh-oh: better - better define a quality measure!)
- Your code doesn't tell what it's about.
constexpr looks good - "my" environment complains with C++11.
"The array&index manipulation" where "simultaneous assignment" is wanted is hard to read (could be worse: the elements of result[] could be non-interchangeable).
Alas, what I found for "modern C++" is ghastly compared to python's a, b = b, a + b. I appreciate the attempt to avoid avoidable assignments; I'm mildly curious if it makes any difference in the code generated by an optimising compiler.
Is there any better? Well, with output size limited by a constant, there's a tighter limit:
the runtime of your code is in O(1), just as any other.
In a comment, you express concern about the complexity of multiplication. If you accept ("bit-wise") "shift" as a (very) cheap operation, you can take three steps in the Fibonacci sequence at once without an increase in "logic gate complexity":
#include <cstdlib>
#include <tuple>
#include <iostream>
/// Iterates an a,b = b,a+b sequence in steps of three.
//constexpr
static unsigned long tri(int previous, int current, const unsigned int n) {
if (n < 2)
return n ? current : previous;
std::div_t split = std::div(n-2, 3);
while (0 <= --split.rem)
std::tie(previous, current)
= std::make_tuple(current, current+previous);
unsigned long
a = current - previous,
b = current + previous;
while (0 <= --split.quot)
std::tie(a, b) = std::make_tuple(b, (b<<2)+a);
return b;
}
/// Iterates the Fibonacci sequence in steps of three.
unsigned long fibonacci(const unsigned int n) {
return tri(0, 1, n);
}
/// Iterates the Lucas numbers in steps of three.
unsigned long lucas(const unsigned int n) {
return tri(2, 1, n);
}
(For variants using arrays of precomputed elements in stead of "the setup-loop" (and a main()), consult the edit history.)
(b*Phi³ coincidentally can be computed with just two summands (and no other power up to 2³² can).)
nfor which that's better. \$\endgroup\$std::fibonacci(or similar in Boost etc), I think you'd be right. \$\endgroup\$