equivariant functions

[Andrei Alexandrescu]

Steven Schveighoffer wrote:
> These are similar, if you consider that const is a 'base type' of its 
> mutable or invariant version.  But there is one caveat that is hard to 
> explain using this terminology.  const is a type modifier, not a type.  So 
> it's not really a base type of a type, and it can be applied to any type in 
> existance.  Not the same as specifying a base type.

Base type and supertype are not to be confused. By base type I 
understand you mean you actually type class Super{} and class 
Sub:Super{}. The subtyping relation is more general and applies to any 
two types Super and Sub that satisfy a number of constraints.

That const(T) is a supertype of both T and invariant(T) is essential and 
not happenstance. It enshrines the fact that you can always pass a T or 
an invariant(T) into a function. The fact that there is no difference in 
layout and that const is not assignable also means that arrays of 
const(T) are supertypes of both arrays of T and invariant(T), with 
important consequences.



Andrei