Re: function to determine if an integer is divisible by 2

From:

Carl Barron <cbarron413@adelphia.net>

Newsgroups:

comp.lang.c++.moderated

Date:

Sat, 10 Feb 2007 12:41:01 CST

Message-ID:

<100220071102244960%cbarron413@adelphia.net>

In article <1171102541.055615.311310@v45g2000cwv.googlegroups.com>,
James Kanze <james.kanze@gmail.com> wrote:

I think it's an interesting exercise. It's pretty
straightforward if you use dynamic initialization for the
table, but then, you've written the algorithm twice. What
I'd like to see is some sort of meta-programming trick to
generate a staticly initialized C-style table.

I assume your table is not INT_MAX in size:) then it requires a look
up. Since numeric_limits<int>::digits is probably <= 128 a linear
search is all we need. If your compiler can fathom 128 levels of
template recursion then here is one table lookup solution.
#include <limits>

template <int N> struct power_two_table_entry
{
    static bool exec(int n)
    {
       return n == N ? true:
          power_two_table_entry<N/2>::exec(n);
    }
};

template <> struct power_two_table_entry<0>
{
    static bool exec(int) {return false;}
};

bool is_power2(int x)
{
    return x < 1 ?
       false :

power_two_table_entry<std::numeric_limits<int>::digits-1>::exec(x);
}

not much 'trickery':) I still prefer
inline bool is_power_two(int x) {return x>0 && x & (x-1) == 0;}

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