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

Its hard and probably impossible, mainly because you can only pass compile-time constants as template values to templates, hence your last example with the int_12_bit x; will never be able to be a template value (and types can't be passed as parameters, of course). I played around a bit with decltype, declval and different templates, but I simply could not get it to take in types and (non-constand expression) values with a single "call". It's really unfortunate decltype doesn't accept types, I wonder why the committee choose to only accept expressions.

Since you mentioned gcc-extensions, there is an extension which can make it work, __typeof__.

I personally have never used this extension, but it seems like it works similar to decltype but it also accepts types directly.

This snipped compiles under gcc x86-64 8.3 for me:

template
struct bits_trait;

template<>
struct bits_trait{};

void f() {
    int x;
    bits_trait<__typeof__(x)>();
    bits_trait<__typeof__(int)>();
}

But this will only compile under gcc.

Edit: Clang seems to support it as well, no luck with MSVC though.