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(Update: Apparently my original answer wasn't very helpful for the OP... See below the second HR for an appendix.) <hr> In situations like this, what you really want to do is to launch <code>ghci</code> and use <code>:t</code> to find out the type of various expressions. E.g. <pre><code>:t foldr map -- answer: foldr map :: [b] -> [b -> b] -> [b] </code></pre> If you need to define a name first, use <code>let</code> (which would not be needed in a Haskell source file): <pre><code>let mys = undefined :: a -> a :t map mys -- answer: [a] -> [a] </code></pre> Note the use of <code>undefined</code> with an explicit type signature. It's perfectly ok for finding out the type of expressions of various forms and might even be useful in actual code as a placeholder at early planning stages. The <code>Num a =></code> is a type class constraint on <code>a</code>; see e.g. <a href="http://www.haskell.org/tutorial/classes.html" rel="nofollow noreferrer">Type Classes and Overloading</a> from "A Gentle Introduction to Haskell, Version 98" or <a href="http://book.realworldhaskell.org/read/using-typeclasses.html" rel="nofollow noreferrer">Chapter 6. Using Typeclasses</a> from "Real World Haskell" for more information. (Basically, it does what you think it does. :-)) <hr> The above should be useful for verifying your answers (and the resources on type classes are good). As for how to solve problems of this kind yourself: Type inference is an application of what is called "the unification algorithm". Google for "unification" for a number of resources; if you add a name of a programming language to your query, you'll likely find example implementations. As for how to approach the examples at hand... a. <code>map mys</code>: <code>map</code> takes a function of type <code>a -> b</code> and returns a function of type <code>[a] -> [b]</code>. In general, <code>a</code> can be different to <code>b</code>, but <code>mys</code> is of type <code>a -> a</code>, so the returned function will be of type <code>[a] -> [a]</code>. Here's some hand-unification: <pre><code>(a -> b) -> [a] -> [b]` (a -> a)` ^-- at this point we unify a with b; when propagated to the return type, this produces [a] -> [a] </code></pre> b. <code>mys map</code>: <code>mys</code> is a function which takes an object of some type and returns an object of the same type. In particular, if you pass it an argument of type <code>(a -> b) -> [a] -> [b]</code>, that will be the type of the return value. Incidentally, there is only one "interesting" (not <code>undefined</code>) function whose type signature is <code>a -> a</code> (without type class constraints), namely <code>id</code>. See Philip Wadler's paper "Theorems for free!" (which you can download from <a href="http://homepages.inf.ed.ac.uk/wadler/topics/parametricity.html" rel="nofollow noreferrer">this page</a>) for an extended discussion. c. <code>foldr map</code>: Firstly, note that the <code>a</code>s and <code>b</code>s in <code>foldr</code>'s signature having nothing to do with those in <code>map</code>'s signature. It might be convenient to rename <code>map</code>'s <code>a</code> to <code>c</code> and <code>b</code> to <code>d</code> and use the signature <code>map :: (c -> d) -> [c] -> [d]</code> to make this more readily apparent below. Also note that <code>(a -> b -> b)</code> is just a simpler way of writing <code>(a -> (b -> b))</code>. More hand-unification (explanation follows below): <pre><code>(a -> (b -> b)) (c -> d) -> [c] -> [d] </code></pre> What happens here is that <code>foldr</code> takes a function argument of type <code>(a -> (b -> b))</code>. If you pass in <code>map</code> as that argument, the <code>a</code> gets unified with <code>c -> d</code>, the <code>b</code> with <code>[c]</code> and then again with <code>[d]</code>, which means <code>c</code> equals <code>d</code>. The return value of <code>foldr</code> has type <code>b -> [a] -> b</code>; substituting the more specific types obtained in the previous paragraph we get <pre><code>[c] -> [c -> c] -> [c] ^-- c equals d, right? </code></pre> The <code>c</code> comes from our changed signature for <code>map</code>; with the original <code>b</code>, this becomes <pre><code>[b] -> [b -> b] -> [b] </code></pre> d. <code>foldr map . mys</code>: This is actually <code>(foldr map) . mys</code>, not <code>foldr (map . mys)</code> -- function application ("the invisible operator") binds the strongest! Combining a. and c. from above to solve this one is left as an exercise to the reader. ;-)
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