Re: Dropping C++03 support (again)

From: Gavin Lambert (boost_at_[hidden])
Date: 2023-02-20 05:35:15

Mere moments ago, quoth I:
> A common type does not need to represent all values of both types; only 
> enough to determine equivalence.  This usually means the "common type" 
> is the smaller type, because a failure to convert the larger type to the 
> smaller type inherently implies non-equivalence.

Applying this to the optional<T> example used previously, the "common 
type" is T, and any failure to convert optional<T> to T can just return 
false, and this is perfectly correct.

More generally, the "common type" would be the *intersection* between 
the two types (not the union, as you stated); any failure to convert 
either argument type to the common type can just return false, because 
if one value cannot be represented in the other type then they cannot 
possibly be equivalent.

Assuming it doesn't fastpath out, you now have two values of the same 
common type that can use that type's homogenous operator==.  It should 
not matter which order it compares these in, unless commutability has 
been violated.

(For product types, you might need to repeat this for additional pairs 
of members, of course, but it can short-circuit as soon as there is a 
failure.)