Re: [Haskell-cafe] Functional vs Imperative

Small point,

From: Thomas Davie To: dhaemon@gmail.com CC: Haskell-Cafe@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:

...

Hello, 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:// haskell.org/practice.html)

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 program.

Take this example:

C program: #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 again.

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.

Haskell program:

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.

Bob

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