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    <p>Let's start with a code example:</p> <pre><code>foob :: forall a b. (b -&gt; b) -&gt; b -&gt; (a -&gt; b) -&gt; Maybe a -&gt; b foob postProcess onNothin onJust mval = postProcess val where val :: b val = maybe onNothin onJust mval </code></pre> <p>This code doesn't compile (syntax error) in plain Haskell 98. It requires an extension to support the <code>forall</code> keyword.</p> <p>Basically, there are 3 <strong>different</strong> common uses for the <code>forall</code> keyword (or at least so it <em>seems</em>), and each has its own Haskell extension: <code>ScopedTypeVariables</code>, <code>RankNTypes</code>/<code>Rank2Types</code>, <code>ExistentialQuantification</code>.</p> <p>The code above doesn't get a syntax error with either of those enabled, but only type-checks with <code>ScopedTypeVariables</code> enabled.</p> <p><strong>Scoped Type Variables:</strong></p> <p>Scoped type variables helps one specify types for code inside <code>where</code> clauses. It makes the <code>b</code> in <code>val :: b</code> the same one as the <code>b</code> in <code>foob :: forall a b. (b -&gt; b) -&gt; b -&gt; (a -&gt; b) -&gt; Maybe a -&gt; b</code>.</p> <p><em>A confusing point</em>: you may hear that when you omit the <code>forall</code> from a type it is actually still implicitly there. (<a href="https://stackoverflow.com/questions/3071136/what-does-the-forall-keyword-in-haskell-ghc-do/3071365#3071365">from Norman's answer: "normally these languages omit the forall from polymorphic types"</a>). This claim is correct, <strong>but</strong> it refers to the other uses of <code>forall</code>, and not to the <code>ScopedTypeVariables</code> use.</p> <p><strong>Rank-N-Types:</strong></p> <p>Let's start with that <code>mayb :: b -&gt; (a -&gt; b) -&gt; Maybe a -&gt; b</code> is equivalent to <code>mayb :: forall a b. b -&gt; (a -&gt; b) -&gt; Maybe a -&gt; b</code>, <strong>except</strong> for when <code>ScopedTypeVariables</code> is enabled.</p> <p>This means that it works for every <code>a</code> and <code>b</code>.</p> <p>Let's say you want to do something like this.</p> <pre><code>ghci&gt; let putInList x = [x] ghci&gt; liftTup putInList (5, "Blah") ([5], ["Blah"]) </code></pre> <p>What's must be the type of this <code>liftTup</code>? It's <code>liftTup :: (forall x. x -&gt; f x) -&gt; (a, b) -&gt; (f a, f b)</code>. To see why, let's try to code it:</p> <pre><code>ghci&gt; let liftTup liftFunc (a, b) = (liftFunc a, liftFunc b) ghci&gt; liftTup (\x -&gt; [x]) (5, "Hello") No instance for (Num [Char]) ... ghci&gt; -- huh? ghci&gt; :t liftTup liftTup :: (t -&gt; t1) -&gt; (t, t) -&gt; (t1, t1) </code></pre> <p>"Hmm.. why does GHC infer that the tuple must contain two of the same type? Let's tell it they don't have to be"</p> <pre><code>-- test.hs liftTup :: (x -&gt; f x) -&gt; (a, b) -&gt; (f a, f b) liftTup liftFunc (t, v) = (liftFunc t, liftFunc v) ghci&gt; :l test.hs Couldnt match expected type 'x' against inferred type 'b' ... </code></pre> <p>Hmm. so here ghc doesn't let us apply <code>liftFunc</code> on <code>v</code> because <code>v :: b</code> and <code>liftFunc</code> wants an <code>x</code>. We really want our function to get a function that accepts any possible <code>x</code>!</p> <pre><code>{-# LANGUAGE RankNTypes #-} liftTup :: (forall x. x -&gt; f x) -&gt; (a, b) -&gt; (f a, f b) liftTup liftFunc (t, v) = (liftFunc t, liftFunc v) </code></pre> <p>So it's not <code>liftTup</code> that works for all <code>x</code>, it's the function that it gets that does.</p> <p><strong>Existential Quantification:</strong></p> <p>Let's use an example:</p> <pre><code>-- test.hs {-# LANGUAGE ExistentialQuantification #-} data EQList = forall a. EQList [a] eqListLen :: EQList -&gt; Int eqListLen (EQList x) = length x ghci&gt; :l test.hs ghci&gt; eqListLen $ EQList ["Hello", "World"] 2 </code></pre> <p>How is that different from Rank-N-Types?</p> <pre><code>ghci&gt; :set -XRankNTypes ghci&gt; length (["Hello", "World"] :: forall a. [a]) Couldnt match expected type 'a' against inferred type '[Char]' ... </code></pre> <p>With Rank-N-Types, <code>forall a</code> meant that your expression must fit all possible <code>a</code>s. For example:</p> <pre><code>ghci&gt; length ([] :: forall a. [a]) 0 </code></pre> <p>An empty list does work as a list of any type.</p> <p>So with Existential-Quantification, <code>forall</code>s in <code>data</code> definitions mean that, the value contained <strong>can</strong> be of <strong>any</strong> suitable type, not that it <strong>must</strong> be of <strong>all</strong> suitable types.</p>
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    1. POWhat does the `forall` keyword in Haskell/GHC do?
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    1. COAnd an explanation with code now. Thanks! It's 2AM here, almost, so I'll have to read this more fully tomorrow. Thanks for weighing in.
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    2. COOK, I got my six hours and can now decode your answer. :) Between you and Norman I got exactly the kind of answer I was looking for. Thanks.
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    3. COActually, you make `ScopedTypeVariables` seem worse than it is. If you write the type `b -> (a -> b) -> Maybe a -> b` with this extension on it will still be exactly equivalent to `forall a b. b -> (a -> b) -> Maybe a -> b`. However, if you want to refer to *the same* `b` (and not have it implicitly quantified) *then* you need to write the explicitly quantified version. Otherwise, `STV` would be an extremely intrusive extension.
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