Here's my implementation; less elegant than @sepp2k's one, it follows a different approach, actually counting the bits and providing both the position of the MSB and the number of significant bits.
#include <iostream>
#include <limits>
// Number: the number to be examined; Bit: parameter used internally to specify the bit to
// examine (the work starts from the leftmost bit)
template<unsigned int Number, unsigned int Bit=std::numeric_limits<unsigned int>::digits-1>
class BitCounter
{
public:
// Most Significant Bit number; if it's the current, fine, otherwise
// delegate the work to another template that will examine the next bit
static const int MSB=(Number&(1<<Bit))?Bit:BitCounter<Number,Bit-1>::MSB;
// Count of significant bits - actually, MSB+1
static const int Count=MSB+1;
};
// Corner case: we reached the first bit
template<unsigned int Number>
class BitCounter<Number,0>
{
public:
// If it's 1, the MSB is the bit 0 (the rightmost), while if it's 0 it's "one before";
// this is somewhat arbitrary to make Count get 0 for 0 and 1 for 1, you may want to
// change it just to 0
static const int MSB=Number==0?-1:0;
static const int Count=MSB+1;
};
int main()
{
std::cout<<BitCounter<255>::Count<<" "
<<BitCounter<7>::Count<<" "
<<BitCounter<1024>::Count<<" "
<<BitCounter<26>::Count<<" "
<<BitCounter<1>::Count<<" "
<<BitCounter<0>::Count<<std::endl;
return 0;
}
Sample output:
matteo@teoubuntu:~/cpp$ g++ tpl_bitcount.cpp -Wall -Wextra -ansi -pedantic -O3 -o tpl_bitcount.x && ./tpl_bitcount.x
8 3 11 5 1 0