From: Felipe Contreras <felipe.contreras@...>
Date: 2007-10-11T00:14:24+09:00
Subject: Re: meta variable substitution

On 10/10/07, Robert Dober <robert.dober@gmail.com> wrote:
> On 10/10/07, Felipe Contreras <felipe.contreras@gmail.com> wrote:
> > Hi,
> >
> > I'm creating a meta script that can generate both a shell script and a
> > Makefile. I want the script to be human readable, so I don't want to
> > substitute all the variables, so $(top_srcdir) =
> > "/my/obsenely/huge/directory/name" doesn't get substituted in for
> > example $(srcdir) = $(top_srcdir)/foo.
> >
> > The problem is that bash doesn't use the same syntax for variable substitution.
> >
> > I guess the easiest thing to do is just s/$()/${}/ but I was wondering
> > if there's a better way to do that.
> >
> > Since I know Rubyists can always find better ways to write code I
> > decided to ask here.
> >
> > Is there a better way to implement meta variable substitution?
> >
> Are you parsing the files? I had the impression that you are
> generating them. In that case I would just generate the variable names
> rather late and somehow factorize
> the code into

Yeah, I'm generating them.

> def var_name name
>    "${#{name}}"
> end
>
> def var_name name
>   "$(#{name})"
> end
>
> if this polymorphic approach is not suitable for your design an if
> statement shall do.

A polymorphic approach is fine, but I want this in my code:

foo="<var_start>var_name<var_end>/bar"

So you are suggesting something like:

foo="#{pseudo_var_name}/bar"

Right?

-- 
Felipe Contreras