3 ms·
I think gp's point is that bif[n_] == 0 for all n_. A 'smart' optimser would recognise this and so the time to compute would be (a) constant irrespective of th
by spinningslate 2y ago
I think gp's point is that bif[n_] == 0 for all n_. A 'smart' optimser would recognise this and so the time to compute would be (a) constant irrespective of the value of n_, and (b) instantaneous because the function call can be re-rewritten as the constant value 0.
- bonzini 2y agoUnderstood, and my point is that user Someone suggested "bif" as a way to avoid testing bigint performance; but fib[15] does not test bigint performance. So using bif is unnecessary and potentially harmful if it's optimized to zero.