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copied!<p>I can't say with too much authority on this subject, but here's what I think may be happening:</p> <p>Consider what the typechecker has to do in each of these cases. <code>(print . runST)</code> has type <code>Show b => (forall s. ST s t) -> IO ()</code>. <code>fourty_two</code> has type <code>ST x Int</code>. </p> <p>The <code>forall</code> here is an existential type qualifier - here it means that the argument passed in has to be <em>universal</em> on <code>s</code>. That is, you must pass in a polymorphic type that supports any value for <code>s</code> whatsoever. If you don't explicitly state <code>forall</code>, Haskell puts it at the outermost level of the type definition. This means that <code>fourty_two :: forall x. ST x Int</code> and <code>(print . runST) :: forall t. Show t => (forall s. ST s t) -> IO ()</code></p> <p>Now, we can match <code>forall x. ST x Int</code> with <code>forall s. ST s t</code> by letting <code>t = Int, x = s</code>. So the direct call case works. What happens if we use <code>$</code>, though?</p> <p><code>$</code> has type <code>($) :: forall a b. (a -> b) -> a -> b</code>. When we resolve <code>a</code> and <code>b</code>, since the type for <code>$</code> doesn't have any explicit type scoping like this, the <code>x</code> argument of <code>fourty_two</code> gets lifted out to the outermost scope in the type for <code>($)</code> - so <code>($) :: forall x t. (a = forall s. ST s t -> b = IO ()) -> (a = ST x t) -> IO ()</code>. At this point, it tries to match <code>a</code> and <code>b</code>, and fails.</p> <p>If you instead write <code>((print . runST) $) fourty_two</code>, then the compiler first resolves the type of <code>((print . runST $)</code>. It resolves the type for ($) to be <code>forall t. (a = forall s. ST s t -> b = IO ()) -> a -> b</code>; note that since the second occurance of <code>a</code> is unconstrained, we don't have that pesky type variable leaking out to the outermost scope! And so the match succeeds, the function is partially applied, and the overall type of the expression is <code>forall t. (forall s. ST s t) -> IO ()</code>, which is right back where we started, and so it succeeds.</p>

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