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copied!<h3>Explanation of the error message</h3> <p>We have</p> <pre><code>foldr (:) :: [a] -> ([a] -> [a]) (.) :: (c -> d ) -> (b -> c) -> (b -> d) </code></pre> <p>From this, we conclude that the second argument <code>y</code> in the composition</p> <pre><code>foldr (:) . y </code></pre> <p>must result in a list, i.e., be of type </p> <pre><code>e -> [a] </code></pre> <p>In our case, <code>y = (* 2) [] [1..10]</code>. So we have:</p> <pre><code>(* 2) [] [1..10] :: e -> [a] -- for some e and a </code></pre> <p>Thus:</p> <pre><code>(* 2) [] :: [f] -> e -> [a] -- for some e, a and f where (Num f, Enum f) </code></pre> <p>(The constraints on <code>f</code> are because we use numeric literals and range notation in the list <code>[1..10]</code>.) Now, consider that <code>2</code> and <code>(*)</code> are overloaded, and</p> <pre><code>(* 2) :: Num g => g -> g </code></pre> <p>From this we get that <code>[]</code> must have the same type as <code>(* 2) []</code> (and that type must be an instance of <code>Num</code>):</p> <pre><code>[] :: [f] -> e -> [a] -- for some e, a and f where -- (Num f, Enum f, Num ([f] -> e -> [a])) </code></pre> <p>But this is not true, because <code>[]</code> has type <code>[g]</code> for some <code>g</code>. Now let's look at the error message again:</p> <pre><code>Couldn't match expected type `[t0] -> a0 -> [a1]' with actual type `[a2]' In the first argument of `* 2', namely `[]' </code></pre> <p>So modulo renaming of type variables, this is exactly what the error message says.</p> <h3>Addendum</h3> <p>The error points to <code>`[]'</code>, so it's perhaps enlightening to see what the typechecker thinks should go in this position. In GHC 7.8, we'll be able to use the <code>TypeHoles</code> extension for this, but for now, an easy way to do it is to abstract from <code>[]</code>, i.e., replace it by a function argument:</p> <pre><code>Prelude> :t \ x -> foldr (:) . (* 2) x [1..10] \ x -> foldr (:) . (* 2) x [1..10] :: (Enum t, Num ([t] -> a -> [a1]), Num t) => ([t] -> a -> [a1]) -> a -> [a1] -> [a1] </code></pre> <p>From this, you can easily see that this term is actually still type-correct, it only requires <code>x</code> to be of the type</p> <pre><code>[t] -> a -> [a1] </code></pre> <p>with the additional class constraints (which are admittedly unlikely to hold).</p>

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