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From: john@basser.oz (John Mackin)
Newsgroups: comp.sources.d
Subject: Re: PD version of "gmatch"?
Summary: it's not trivial to do it with a regular expression matcher
Message-ID: <1555@basser.oz>
Date: 23 Oct 88 06:39:52 GMT
References: <24641@tut.cis.ohio-state.edu>
Reply-To: john@basser.oz (John Mackin)
Organization: Dept of Comp Sci, Uni of Sydney, Australia
Lines: 32

In article <24641@tut.cis.ohio-state.edu>,
	jgreely@oz.cis.ohio-state.edu (J Greely) writes:

> In article <3286@ttidca.TTI.COM> Steve Alter (alter@ttidca.tti.com) writes:

> >and the function returns non-zero if the string fits the pattern.  The
> >pattern may contain '*', '?', or '[a-z]' style constructs.  You've all
> >seen this before;
> >I'm just clarifying this to tell it apart from
> >regular expressions, which are entirely different beasts.
> 
> Only in syntax.  And it's easy to transform between the two.
> 
> Use the regex package (Berkeley calls them re_comp and re_exec).  In
> your pattern string, replace '?' with '.', '*' with '.*', and '[a-z]'
> with itself.

Unfortunately, this doesn't work.  You have neglected to consider
the fact that the pattern given to gmatch may contain characters
that are metacharacters to regex but not gmatch: .[]^$ spring to
mind instantly.  If it was just .^$ you could easily add \ in
front of them, but character classes complicate it to where it's
barely worth doing in my opinion, and when you start to consider
the semantics of \ itself in the gmatch pattern, it's certainly
no harder to just write gmatch than it is to translate a gmatch
pattern into a regular expression.

John Mackin, Basser Department of Computer Science,
	     University of Sydney, Sydney, Australia

john@basser.oz.AU (john%basser.oz.AU@UUNET.UU.NET)
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