From: "Booker C.Bense" <bbense+comp.lang.ruby.Dec.06.06@...>
Date: 2006-12-07T04:45:06+09:00
Subject: Re: variable pointers

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In article <iIKdnQyjjPDVj-rYnZ2dnUVZ_vGinZ2d@comcast.com>,
xyz <xyz@xyz.com> wrote:
>
>I have a situation where I'm walking through an unordered sequence of text 
>(splitting at newlines) looking for particular patterns. If a particular pattern 
>matches then I perform some processing unique to that pattern and assign the 
>output to a particular location (the location could be anything, e.g. local 
>scalar, a particular array element, an instance variable, etc.).

You are thinking in "C", this just doesn't work well in
Ruby. Everything in Ruby is an object and every variable is just
a pointer to that object. When you say 

a = b 

in Ruby, it means in C 

a = &b 

and 

a 

always evaluates to 

*a 


>
>WARNING: C syntax approaching. Please restore your seats to their upright and 
>locked positions and apply seat belts.
>
>Using "&" (take address) and "*" (de-reference pointer) from C, conceptually 
>what I want to do is:
>
>   template = [
>                [ patternRegex, processingProc, &storage ]
>                ...
>               ]
>   textSequence.each { |line|
>      template.each { |tuple|
>         *tuple[2] = tuple[1].call(line) if line =~ tuple[0]
>      }
>   }
>
>Obviously this could be achieved with eval() but that is very heavyweight.
>
>Are there other options?
>

Just use 

tuple[2] = tuple[1].call(line) if line =~ tuple[0] 

You can't easily pass storage around in Ruby, like you can in
C. What you pass around is a pointer to an object. The
processingProc should handle the createtion of the object, 
the storage item is just a name for where you store the pointer
to the object. If you are managing storage, rather than objects
you are doing things the hard way in Ruby. 

_ Booker C. Bense 


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