Clemens Hintze <c.hintze / gmx.net> writes:

> 
>    (c = (a.downcase <=> b.downcase)) == 0 ? a <=> b : c
> 

I had that one, and also

    2*(a.downcase <=> b.downcase) + (a <=> b)

But I still feel there has to be a _neat_ solution with no additional
variables and no redundant comparisons.


Regards



Dave