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    <p>Hm. Let's forget about types for a moment.</p> <p>Let's say you have two functions:</p> <pre><code>import qualified Data.IntMap as IM a :: Int -&gt; Float a x = fromInteger (x * x) / 2 l :: Int -&gt; String l x = fromMaybe "" $ IM.lookup x im where im = IM.fromList -- etc... </code></pre> <p>Say there exists some value <code>n :: Int</code> that you care about. Given only the value of <code>a n</code>, how do you find the value of <code>l n</code>? You don't, of course. </p> <p>How is this relevant? Well, the type of <code>myLength</code> is <code>A a -&gt; Int</code>, where <code>A a</code> is the result of applying the "type function" <code>A</code> to some type <code>a</code>. However, <code>myLength</code> being part of a type class, the class parameter <code>a</code> is used to select which implementation of <code>myLength</code> to use. So, given a value of <em>some specific type <code>B</code></em>, applying <code>myLength</code> to it gives a type of <code>B -&gt; Int</code>, where <code>B ~ A a</code> and you need to know the <code>a</code> in order to look up the implementation of <code>myLength</code>. Given only the value of <code>A a</code>, how do you find the value of <code>a</code>? You don't, of course.</p> <p>You could reasonably object that in your code here, the function <code>A</code> <em>is</em> invertible, unlike the <code>a</code> function in my earlier example. This is true, but the compiler can't do anything with that because of the <em>open world</em> assumption where type classes are involved; your module could, in theory, be imported by another module that defines its own instance, e.g.:</p> <pre><code>instance C Bool where type A Bool = [String] </code></pre> <p>Silly? Yes. Valid code? Also yes.</p> <p>In many cases, the use of constructors in Haskell serves to create <em>trivially injective functions</em>: The constructor introduces a new entity that is defined only and uniquely by the arguments it's given, making it simple to recover the original values. This is precisely the difference between the two versions of your code; the data family makes the type function invertible by defining a new, distinct type for each argument.</p>
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    1. CO@cammccann Thanks for the explanation. I guess I don't understand why , for example, the type of the "myLength" in "C Int" doesn't resolve to "[Int] -> Int" by virtue of being defined as part of the "C" typeclass.
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    2. CO@Deech: Well, it does; for `C Int`, `myLength` gets the type `[Int] -> Int`. The problem is that, in an expression like `myLength ([1, 2] :: [Int])`, there's no way to get from `[Int]` back to `A Int`, just like there's no way to get from `12.5` back to `5` (rather than `-5`) in my function `a`.
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    3. CO@cammccann I just changed "let a1 = [1,2,3]" to "let a1 = ([1,2,3] :: A Int)" and similarly to "a2" but I get the same errors. Since I have explicitly given the compiler "a1"'s type shouldn't it now correctly resolve?
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