Re: binary format of the number.

Tarmo Kuuse a ?crit :


Sorry, ignore my previous post. I thought you wanted to output it in a
stream in bin format.

With binary defined as follow:

template<typename T=unsigned,int nbbit=8>
struct binary
{
     //initialize octet with binary form long
     binary(long v)
     {
         T exp=1;
         val=0;
         //ensure v in range 0 11111111
         v%=11111112;
         //build value
         while(v)
         {
             ldiv_t d=ldiv(v,10); //ldiv from stdlib.h C99
             v=d.quot;
             if(d.rem)val|=exp;
             exp*=2;
         }
     }

     //initialize octet with pre-computed value
     struct value
     {
         T val;
         value(T v):val(v){}
     };
     explicit binary(const value& v):val(v.val){}

     operator T()const{return val;}

     T val;
};

//concatenate binaries
template<typename T,int nbbit_lhs,int nbbit_rhs>
binary<T,nbbit_lhs+nbbit_rhs> operator<<(
         const binary<T,nbbit_lhs>& lhs,
         const binary<T,nbbit_rhs>& rhs)
{
     typedef binary<T,nbbit_lhs+nbbit_rhs> binary_type;
     return binary_type(
             typename binary_type::value(
                 (lhs.val<<nbbit_rhs)|rhs.val
                 )
             );
}

Example:
std::cout<<(binary<>(1)<<binary<>(101))<<std::endl;
int cons=binary<int>(10000000)binary<int>(0);

You can improve of the template expression such as:
  * binary(int val,int number) to build representation of number with
only 0s or 1s in a given number at given position.
  * using fills fill<int,24>(1) -> 111...11 24 bits

And so on.

--
Michael