Hi Alex,<br><br>In Haskell, data structures cache, while functions do not.<br><br>"Memoization" is conversion of functions into data structures (and then trivially re-wrapping as functions) so as to exploit the caching property of data structures to get caching functions.<br>
<br> - Conal<br><br><div class="gmail_quote">On Wed, Sep 15, 2010 at 11:03 AM, Alex Rozenshteyn <span dir="ltr"><<a href="mailto:rpglover64@gmail.com">rpglover64@gmail.com</a>></span> wrote:<br><blockquote class="gmail_quote" style="margin: 0pt 0pt 0pt 0.8ex; border-left: 1px solid rgb(204, 204, 204); padding-left: 1ex;">
I feel that there is something that I don't understand completely: I have been told that Haskell does not memoize function call, e.g.<div>> slowFib 50</div><div>will run just as slowly each time it is called. However, I have read that Haskell has call-by-need semantics, which were described as "lazy evaluation with memoization"</div>
<div><br></div><div>I understand that</div><div>> fib50 = slowFib 50</div><div>will take a while to run the first time but be instant each subsequent call; does this count as memoization?</div><div><br></div><div>(I'm trying to understand "Purely Functional Data Structures", hence this question)<br clear="all">
<br>-- <br><div dir="ltr"><div> Alex R</div></div><br>
</div>
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