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    <p>Well, it's not part of the Haskell 2010 standard, so it's not on by default, and is offered as a language extension instead. As for why it's not in the standard, rank-n types are quite a bit harder to implement than the plain rank-1 types standard Haskell has; they're also not needed all that frequently, so the Committee likely decided not to bother with them for reasons of language and implementation simplicity.</p> <p>Of course, that doesn't mean rank-n types aren't useful; they are, very much so, and without them we wouldn't have valuable tools like the <a href="http://hackage.haskell.org/packages/archive/base/latest/doc/html/Control-Monad-ST.html" rel="nofollow noreferrer"><code>ST</code> monad</a> (which offers efficient, local mutable state — like <code>IO</code> where all you can do is use <code>IORef</code>s). But they do add quite a bit of complexity to the language, and can cause strange behaviour when applying seemingly benign code transformations. For instance, some rank-n type checkers will allow <code>runST (do { ... })</code> but reject <code>runST $ do { ... }</code>, even though the two expressions are always equivalent without rank-n types. See <a href="https://stackoverflow.com/questions/8343239/type-error-with-rank-2-types-and-function-composition">this SO question</a> for an example of the unexpected (and sometimes annoying) behaviour it can cause.</p> <p>If, like sepp2k asks, you're instead asking why <code>forall</code> has to be explicitly added to type signatures to get the increased generality, the problem is that <code>(forall x. x -&gt; f x) -&gt; (a, b) -&gt; (f a, f b)</code> is actually a more restrictive type than <code>(x -&gt; f x) -&gt; (a, b) -&gt; (f a, f b)</code>. With the latter, you can pass in any function of the form <code>x -&gt; f x</code> (for any <code>f</code> and <code>x</code>), but with the former, the function you pass in must work for <em>all</em> <code>x</code>. So, for instance, a function of type <code>String -&gt; IO String</code> would be a permissible argument to the second function, but not the first; it'd have to have the type <code>a -&gt; IO a</code> instead. It would be pretty confusing if the latter was automatically transformed into the former! They're two very different types.</p> <p>It might make more sense with the implicit <code>forall</code>s made explicit:</p> <pre><code>forall f x a b. (x -&gt; f x) -&gt; (a, b) -&gt; (f a, f b) forall f a b. (forall x. x -&gt; f x) -&gt; (a, b) -&gt; (f a, f b) </code></pre>
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    1. POWhy doesn't forall (RankNTypes usage) apply by default?
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    1. COI think I need to digest this some more :)
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      1. USPeter Hall
    2. COAlso, for second and higher order polymorphism, there's no such thing as the most general type, `∀ b. (∀ a. a → b) → (b, b)` isn't more general than `∀ c d. (∀ a b. a → b) → (c, d)` nor the other way around. And while type inference is decidable for first and second order polymorphism, it is not for third and higher.
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    3. CO@vitus - can you provide a reference for decidability of second order polymorphism? I'm unfamiliar with it.
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      1. USJohn L
 

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