On Sat, 25 Jan 2003 15:06:54 +0200, "Vesa Karvonen" <[EMAIL PROTECTED]> wrote:
>Gennaro Prota: >>Am I missing something? > >Yes. I have, in fact, previously explained how to use choose_n correctly. >Please read more carefully. If something doesn't appear to make sense, >reconsider your basic assumptions. Only after you have carefully >reconsidered your assumptions, should you challenge the assertions of >others. Gulp! Sounds like a scold :-) Ok my fault. I admit that I should have understood that the initial value was chosen as you say. In retrospect, it's obvious that the choice as initial value of the n such that 2^n < MAX <= 2^ (2n) - 1 is the right way to go, since it ensures MAX / 2^n <= 2^n - 1/2^n < 2^n and thus that at each step the part that "remains to be computed with the subsequent steps" of the logarithm is at most n-1 (which can be expressed as sum of n/2, n/4 ... 1) > <excellent code snipped> I'll do some benchmarks against the current implementation in static_log2.hpp and post them here. I haven't tested your code yet but now I'm convinced that it works :-) Thanks! Genny. _______________________________________________ Unsubscribe & other changes: http://lists.boost.org/mailman/listinfo.cgi/boost
