Emulating the special properties of sizeof with constexpr

Question

In C++ sizeof is somewhat unique in that it\'s legal to write this:

int x;
sizeof(x); // a variable

As well as simply:

<         


        

Answer 1

Building upon the quite magical answer from n.m., with just tiny bit of massage, it seems it is possible to have bitsof mimic sizeof.

#include 
#include 
#include 

template 
struct bits_traits {
  enum { value = sizeof(T) * CHAR_BIT };
};

struct int_12_bit {
  enum { bits = 12 };
};

template <>
struct bits_traits {
  enum { value = int_12_bit::bits };
};

#define bits_traits_of(k) decltype(bits_traits_of_left+(k)+bits_traits_of_right)

struct bits_traits_of_left_t {
    template 
    bits_traits operator+(const T&);
} bits_traits_of_left;

struct bits_traits_of_right_t {
    template 
    friend T operator+(const T&, bits_traits_of_right_t);

    bits_traits_of_right_t operator+();

    template 
    operator T() const;

} bits_traits_of_right;

#define bitsof(x) bits_traits_of(x)::value

int main() {
  using std::size_t;
  size_t a = bitsof(int);
  size_t b = bitsof(int_12_bit);
  std::cout <<"a="<< a <<", b="<< b << std::endl;

  int_12_bit x;
  size_t c = bitsof(x);
  std::cout <<"c="<< c << std::endl;
}

The only thing that I changed, other than adding in definitions for bits_traits, is to redefine bitsof so that it returns the bits_traits::value rather than the bits_traits type.

$ ./a.out 
a=32, b=12
c=12

I'm just writing this up to verify that it can work. All credits should go to n.m.'s answer.