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Subject: Re: Common Lisp style question
From: Peter Frings <peter.fri...@gmail.com>
In-Reply-To: <aef4fc95-5edc-43af-820f-5a8955869...@n10g2000yqf.googlegroups.com>
Date: Sat, 4 Jun 2011 13:33:32 +0200
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References: <48AFAE21-1CD0-44BE-AEB4-CDA13E998...@kenworthy.info> <f27accd4-810d-4d56-87da-e37e7565b...@l18g2000yql.googlegroups.com> <aef4fc95-5edc-43af-820f-5a8955869...@n10g2000yqf.googlegroups.com>
To: land-of-lisp@googlegroups.com,
 Purity Control <cr...@craigferry.net>,
 Edward Kenworthy <edw...@kenworthy.info>
X-Mailer: Apple Mail (2.1084)

I prefer the first, with the (let).=20

As Craig said, it's clearer for the programmer and the ones that need to =
maintain it afterwards. Let the compiler figure out how to compile it in =
the most efficient way, and only optimize when it's indeed a verified =
bottleneck. In Knuth's words: "premature optimization is the root of all =
evil". (About optimization: often, using a better algorithm gains you =
more that tweaking a few bits here and there.)

So, go for the most understandable code.

As for the question whether it's too imperative, I would answer no. =
While the language allows you to modify those variables,  you don't: you =
use them as constants. It's not because the language allows mutation, =
that your code is imperative. It's the code. (Which would've been my =
answer to the *ear-muffs* discussion a couple of days ago as well.)

A similar question is when using (nreverse) when returning a newly =
constructed list (common idiom in recursion with accumulators). It's not =
strictly functional, since you modify a variable. But for all practical =
purposes, it is functional (and it's also a bit of premature =
optimization :-) ).

So, adhere to the *ideas* of FP and avoid `extremism'. Else learn =
Haskell where even (random) is strict :-)

Cheers,
Peter.