On Monday, 6 August 2012 at 15:18:22 UTC, Minas Mina wrote:
That's what I tried to calculate: static x = fib(40);ulong fib(ulong n) { if(n < 2) return n; else return fib(n-1) + fib(n-2); } Maybe because it has a lot of recursion?
That algorithm makes O(2^n) calls to fib. I think templates get only expanded once for every set of parameters, so you get memoization build in and thus it's faster in this case.
