It does compute the result of a function application more than once if you ask for it more than once, and that's why we need memoization.<br><br><div>Dan</div><div><br><div class="gmail_quote">On Tue, Dec 15, 2009 at 10:32 PM, michael rice <span dir="ltr"><<a href="mailto:nowgate@yahoo.com">nowgate@yahoo.com</a>></span> wrote:<br>
<blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex;"><table cellspacing="0" cellpadding="0" border="0"><tbody><tr><td valign="top" style="font:inherit">I'm looking here at the Fibonacci stuff:<br>
<br><a href="http://www.haskell.org/haskellwiki/Memoization" target="_blank">http://www.haskell.org/haskellwiki/Memoization</a><br><br>Since (I've read) Haskell never computes the value<br>of a function more than once, I don't understand the<br>
need for memoization.<br><br>Enlighten me.<br><br>Michael<br><br></td></tr></tbody></table><br>
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