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

You can do something like this:

#include 

#define bitsof(k) decltype(bitsof_left+(k)+bitsof_right)

template 
struct bits_traits { /* whatever you want here */ };

struct bitsof_left_t {
    template 
    bits_traits operator+(const T&);
} bitsof_left;

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

    bitsof_right_t operator+();

    template 
    operator T() const;

} bitsof_right;

int main()
{
    using foo = bitsof(42);
    using bar = bitsof(int);

    static_assert(std::is_same>::value);
    static_assert(std::is_same>::value);
}

It works like this.

a + (42) + b is parsed as (a + (42)) + b), then overloaded binary operator+ at either side kicks in. In my example the operators are only declared, not defined, but since it's unevaluated context, it doesn't matter.

a + (int) + b is parsed as a + ((int) (+ b)). Here we employ the overloaded unary + at the right side, then overloaded cast operator, then overloaded binary + at the left side.