From: Rick DeNatale <rick.denatale@...>
Date: 2007-10-22T21:31:03+09:00
Subject: Re: How does overload work in ruby?

On 10/22/07, Bertram Scharpf <lists@bertram-scharpf.de> wrote:
> Am Montag, 22. Okt 2007, 14:25:49 +0900 schrieb Xavier Noria:

> >   def sum(a, b, c=0)
> >     a + b + c
> >   end
>
> If you want to allow the caller to select the default value
> explicitly you may say
>
>   def sum a, b, c = nil
>     c ||= 0
>     a + b + c
>   end

It seems to me that this is just a slightly less efficient way of
doing the same thing as the code you quoted.

In either case the caller isn't specifying the default value, it's
overriding the default.

In the call sum(1,2)  ruby will evaluate the default expression for c,
in the call sum(1,2,3) it won't since c has a value given by the third
parameter.  Making the default value of c nil, and then doing c ||= 0
justs slows down both cases.


-- 
Rick DeNatale

My blog on Ruby
http://talklikeaduck.denhaven2.com/