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copied!<p>The first thing to note is the signature of <code>getLines</code>. Since <code>p `isPrefixOf`</code> has type <code>(Eq a) => [a] -> Bool</code>, <code>getLines p</code> has type <code>(Eq a) => [[a]] -> [[a]]</code> (based on the type of <code>filter</code>). Here <code>[a]</code> seems to be <code>String</code>, so <code>getLines p</code> has type <code>[String] -> [String]</code> and <code>p</code> thus seems to have type <code>String</code>. So in fact, <code>getLines</code> has type <code>String -> [String] -> [String]</code>.</p> <p>Lastly, <code>head</code> has (specialized) type <code>[String] -> String</code>, and you're trying to postcompose it with <code>getLines</code>. I would guess you're trying to build a function with type <code>String -> [String] -> String</code>, defined by <code>\p ss -> head (getLines p ss)</code>. That is, however, <em>not</em> what <code>head . getLines</code> is!</p> <p>To see this, consider <code>f :: a -> b -> c</code> and <code>g :: c -> d</code> (where I mean to say that the <code>c</code>s are the same type in both signatures, so I'm not really writing proper Haskell signatures here). Since one often likes to think of <code>f</code> as a "function of two variables", one might err and think of <code>g . f</code> as being the function <code>\x y -> g (f x y)</code> (type <code>a -> b -> d</code>). This is not the case: See for example <a href="https://stackoverflow.com/q/5821089/453616">this question and its answers</a> or the concrete example in <a href="https://stackoverflow.com/a/4341123/453616">this answer.</a> Or even better: Look at <a href="http://hackage.haskell.org/package/base-4.6.0.1/docs/Prelude.html#g:10" rel="nofollow noreferrer">the type of</a> <code>(.)</code> and work out what <code>g . f</code> must be yourself! (Hint: <a href="http://learnyouahaskell.com/higher-order-functions" rel="nofollow noreferrer">Every function takes precisely <em>one</em> argument</a>. What is the type of <code>f</code>'s one argument?)</p>

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