How to compare aliases of simliar types via the similar type

[miemoe]

I would like to use a Length as a kind of base type for a Width and a Length type.
Following the Tips section I tried:

// have similar Length types
template <typename Tag>
using Length = strong::type<
    double, Tag,
    strong::regular,
    strong::ordered // ,
    // strong::implicitly_convertible_to<double> // make it usable where a double is needed.
    >;

using Width  = Length<struct width_tag>;
using Height = Length<struct height_tag>;

bool higherThanWide(Height h, Width w)
{
    // Height and With are Lengths so we should be able to compare them.
    if ((Length)h > (Length)w) return true;    // (1)
    return false;
}

However the comparison (1) does not compile.

If I add strong::implicitly_convertible_to then

if (h > w) return true; 

works.

I guess, Im looking for a way to use a Height or Width as a Length?

[rollbear]

If this is still an issue for you, there are two simple solutions.

1. Mildly modify your example to use value_of(), like:

   bool higherThanWide(Height h, Width w)
   {
       // Height and With are Lengths so we should be able to compare them.
       if (value_of(h) > value_of(w)) return true;
       return false;
   }
   

2. Make freestanding comparison operators templated on the tag type, although I am not so sure that this is something that you want to do:

   template <typename Tag1, typename Tag2>
   constexpr auto operator<=>(const Length<Tag1>& lh, const Length<Tag2>& rh)
   {
       return value_of(lh) <=> value_of(rh);
   }
   

   Or individually add all <, <=, >=, >, == and != if you're not on C++20 or later.