C++ binary constant/literal

Question

I\'m using a well known template to allow binary constants

template< unsigned long long N >
struct binary
{
  enum { value = (N % 10) + 2 * binary<          


        

Answer 1

You can add more non-type template parameters to "simulate" additional bits:

// Utility metafunction used by top_bit.
template 
struct compare {
    enum { value = N1 > N2 ? N1 >> 1 : compare::value };
};

// This is hit when N1 grows beyond the size representable
// in an unsigned long long.  It's value is never actually used.
template
struct compare<0, N2> {
    enum { value = 42 };
};

// Determine the highest 1-bit in an integer.  Returns 0 for N == 0.
template 
struct top_bit {
    enum { value = compare<1, N>::value };
};

template 
struct binary {
    enum {
        value =
            (top_bit::value>::value << 1) * binary::value +
            binary::value
    };
};

template 
struct binary {
    enum { value = (N1 % 10) + 2 * binary::value };
};

template <>
struct binary<0> {
    enum { value = 0 } ;
};

You can use this as before, e.g.:

binary<1001101>::value

But you can also use the following equivalent forms:

binary<100,1101>::value
binary<1001,101>::value
binary<100110,1>::value

Basically, the extra parameter gives you another 20 bits to play with. You could add even more parameters if necessary.

Because the place value of the second number is used to figure out how far to the left the first number needs to be shifted, the second number must begin with a 1. (This is required anyway, since starting it with a 0 would cause the number to be interpreted as an octal number.)