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  <author_name>wanpark</author_name>
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  <blog_title>こども(てれび)</blog_title>
  <blog_url>https://wanpark.hatenadiary.org/</blog_url>
  <categories>
    <anon>Haskell</anon>
    <anon>SICP</anon>
  </categories>
  <description>そういうことで、問題1.11 の反復的なものを Haskell で書くときはこの前書いた f n = iter 2 1 0 n where iter a b c count | count == 0 = c | otherwise = iter (a + 2 * b + 3 * c) a b (count - 1)では固定スペースで実行できなくて、こうやらないといけない。 f n = iter 2 1 0 n where iter a b c count | count == 0 = c | otherwise = (iter $! (a + 2 * b + 3 * c)) a b (coun…</description>
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  <provider_name>Hatena Blog</provider_name>
  <provider_url>https://hatena.blog</provider_url>
  <published>2006-04-19 09:42:06</published>
  <title>遅延評価と末尾再帰</title>
  <type>rich</type>
  <url>https://wanpark.hatenadiary.org/entry/20060419/1145407326</url>
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