2 ms·
Recursive fibonacci benchmark - you are doing something different in which case you could just evaluate the closed form φⁿ / √5 and be even faster.
by danbruc 8y ago
Recursive fibonacci benchmark - you are doing something different in which case you could just evaluate the closed form φⁿ / √5 and be even faster.
- super_mario 8y agoI'm perfectly aware it's meant to be recursive, but it's also a completely pointless test. It's not tail recursive, so you are measuring function call overhead in various languages. But various languages have options to perform much faster anyway, but solution is of course going to be language specific.