From mboxrd@z Thu Jan  1 00:00:00 1970
Received: from news.lsuc.on.ca ([142.57.1.101]) by archone.tamu.edu with SMTP id <18886>; Fri, 20 Dec 1991 09:12:03 -0600
Received: by news.lsuc.on.ca (/\==/\ Smail3.1.24.1 #24.4)
	id <m0kulrZ-000CfoC@news.lsuc.on.ca>; Fri, 20 Dec 91 10:10 EST
Received: by golem.uucp (/\==/\ Smail3.1.24.1 #24.13)
	id <m0kujDy-0000MxC@golem.uucp>; Fri, 20 Dec 1991 07:21 -0500
To: Chris Siebenmann <cks@hawkwind.utcs.toronto.edu>
cc: rc@archone.tamu.edu
Subject: Re: how would you do this in rc 
In-reply-to: Your message of "Thu, 19 Dec 91 19:09:33 EST."
             <91Dec19.190941est.2742@hawkwind.utcs.toronto.edu> 
Date: Fri, 20 Dec 1991 06:21:09 -0600
Message-ID: <7071.693231669@golem.UUCP>
From: "David J. Fiander" <david@golem.uucp>

>From: 	Chris Siebenmann <hawkwind.utcs.toronto.edu!cks@rutgers.uucp>
> You have a list l and a string s. What's the most efficient way to not
>merely find out if s exists in l, but to find its index in l if it does?
>Is there an efficient method to do this? If there isn't, should there be? :)
>[probably not; rc is not a general programming language, after all.]

I'd go with the obvious implementation, unless it is a
bottleneck.

	i = 0
	if (~ $s $l) {
		# if s is a member of l, then s matches an elt.
		i = 1 while (!~ $s $l($i)) {
			i = `{expr $i + 1}
		}
	}

At the end of the loop (i == 0) if s is not a member,
otherwise, i == the offset.  If you wanted to avoid the expr
overhead, and use byron's 'break' feature, then you could say

	for (i in `{seq $#l}) ...

- David