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    <p>Your problem is that the function <code>encode</code> expects "unpacked" lists, but you pass a "packed" list.</p> <p>Adding type signatures helps a lot here, I added a type signature for <code>encode'</code></p> <pre><code>{-# LANGUAGE ScopedTypeVariables #-} pack :: (Eq a) =&gt; [a] -&gt; [[a]] pack [] = [] pack (x:xs) = (x : takeWhile (== x) xs) : (pack $ dropWhile (== x) xs) encode :: forall a. (Eq a) =&gt; [a] -&gt; [(Int, a)] encode [] = [] encode list = (encode' $ head packed) : (encode $ tail packed) where packed = pack list encode' :: [a] -&gt; (Int, a) encode' l = (length l, head l) </code></pre> <p>and the compiler finds the error quickly:</p> <pre><code>[1 of 1] Compiling Main ( test.hs, interpreted ) test.hs:9:42: Couldn't match type `a' with `[a]' `a' is a rigid type variable bound by the type signature for encode :: Eq a =&gt; [a] -&gt; [(Int, a)] at test.hs:8:1 Expected type: [(Int, a)] Actual type: [(Int, [a])] In the second argument of `(:)', namely `(encode $ tail packed)' In the expression: (encode' $ head packed) : (encode $ tail packed) Failed, modules loaded: none. </code></pre> <p>Because that would only work if <code>a</code> was the same as <code>[a]</code> and therefore the same as <code>[[a]]</code> etc. That's an infinite type error. Or simply the difference between a "packed" list (<code>[[a]]</code>) and a "unpacked" list (<code>[a]</code>) in your sample.</p> <p>(For better understanding: "packed" list is a list after applying the <code>pack</code> function ;-) )</p> <p><em>edit: fixed ScopedTypeVariables vs. ExistentialQuantification error, thanks John L</em></p>
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    1. POhaskell - yet another infinite type error
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    1. USJohannes Weiss
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    1. COthis looks... too advanced for me (just starting with haskell).
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    2. COLet me rephrase that: The compiler tries to tell you: An "unpacked" list would be the same as a "packed" list iff `[a]` would be the same as `[[a]]`. That's true! See your definition of `pack`: `pack :: (Eq a) => [a] -> [[a]]`. `pack` transforms `[a]` to `[[a]]`. If that would be the same, "packed" and "unpacked" lists would be the same and your program would work. BUT: Then we'd have `a == [a] == [[a]] == [[[a]]] == [[[[a]]]]` so: an infinite number of "names" for the type `a`. And the compiler told you `infinite type error` :-)
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    3. CONow I understand why the error is called "infinite type error"... By the way, I referred to the ExistentialQuantification thing when I said that your answer looks too advanced. I found one reference in wikibooks and it looks... well... too advanced, for now :)
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