[Haskell-cafe] Functional vs Imperative
thomasspriggs at hotmail.com
Tue Sep 13 11:06:41 EDT 2005
>From: Thomas Davie <tom.davie at gmail.com>
>To: dhaemon at gmail.com
>CC: Haskell-Cafe at haskell.org
>Subject: Re: [Haskell-cafe] Functional vs Imperative
>Date: Tue, 13 Sep 2005 14:55:14 +0100
>On 13 Sep 2005, at 14:45, Dhaemon wrote:
>>I'm quite interested in haskell, but there is something I don't
>>understand(intuitively). I've been crawling the web for an answer, but
>>nothing talks to me...
>>So I was hoping I could find some help here:
>>"How is evaluating an expression different from performing action?"
>>I'm puzzled... Doesn't it amount to the same thing? Maybe I have a wrong
>>definition of "evaluating"(determine the value of an expression)?
>>Examples would be appreciated.
>>Also, just for kicks, may I had this: I read the code of some
>>haskell-made programs and was astonished. Yes! It was clean and all, but
>>there were "do"s everywhere... Why use a function language if you use it
>>as an imperative one?(i.e. most of the apps in http://
>The difference is all about referential transparency -- in short, a
>function given the same inputs will always give the same result. This is
>not the same as in imperative languages, where functions/ methods/actions
>can have 'side-effects' that change the behavior of the rest of the
>Take this example:
>#define square(x) ((x) * (x))
>#define inc(x) ((x)++)
>int myFunc (int *x)
> return square(inc(*x));
>the C preprocessor will re-write the return line to:
>return ((((x)++)) * (((x)++)));
Shouldn't that be:
return ((((*x)++)) * (((*x)++)));
>this will be performed in sequence, so, x will be incremented (changing
>the value of x), and that result will be multiplied by x incremented
>so if we run myFunc(&y), where y is 5, what we get is 5 incremented to 6,
>and them multiplied by 6 incremented to 7. So the result of the function
>is 42 (when you might reasonably expect 36), and y is incremented by 2,
>when you might reasonably expect it to be incremented by 1.
>square x = x * x
>inc = (+1)
>myFunc = square . inc
>and we now call myFunc 5, we get this evaluation:
>myFunc 5 is reduced to (square . inc) 5
>(square . inc) 5 is reduced to square (inc 5)
>square (inc 5) is reduced to square ((+1) 5)
>square ((+1) 5) is reduced to square 6
>square 6 is reduced to 6 * 6
>6 * 6 is reduced to 36
>If you want to study these reductions on a few more examples, you might
>want to download the Hat tracer, and use hat-anim to display reductions
>step by step.
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