== operator

[Andrei Alexandrescu via Digitalmars-d]

On 1/3/15 5:30 PM, Jonathan Marler wrote:
> I've recently looked at how the '==' operator works with classes.  I was
> disappointed to find that 'a == b' always gets rewritten to:
>
> .object.opEquals(a, b);
>
> The reason for my disappointment is that this results in unnecessary
> overhead. I would think that the compiler would first try to rewrite the
> '==' operator using a type-specific opEquals method, then fall back on
> the generic version if one did not exist.  Is there a reason for this?

TDPL has a detailed explanation of that, including a reference to Java's 
approach. There's less overhead in calling the free function in object 
(it's inlinable and if e.g. the references are equal there's no virtual 
call overhead).

Andrei