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POhaskell -- not understanding why this associated type example needs more inference
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Id
7382171
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7382228
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1
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CreationDate
2011-09-12T00:04:05.503
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2011-09-12T00:27:34.503
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2011-09-12T00:27:34.503
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text
Body
Consider the following code, <pre><code>data MyBaseExpr α where ConstE :: Show α => α -> MyBaseExpr α class Monad => MyMonadCls where type ExprTyp :: * -> * var :: String -> ExprTyp α -> (ExprTyp α) expTypArg :: forall α. MyMonadCls => ExprTyp α -> α expTypArg a = undefined -- dummy type which will be used as an instance newtype A α = A ( α) </code></pre> Then, if one tries to write an instance using the expTypeArg function, <pre><code>instance forall . (Monad , Monad (A )) => MyMonadCls (A ) where type ExprTyp (A ) = MyBaseExpr var nme init@(expTypArg -> typb) = return init </code></pre> the compiler complains <pre><code>Couldn't match type `ExprTyp 0' with `MyBaseExpr' Expected type: ExprTyp (A ) α Actual type: ExprTyp 0 α </code></pre> But, if one adds some scoped type expressions, <pre><code>instance forall . (Monad , Monad (A )) => MyMonadCls (A ) where type ExprTyp (A ) = MyBaseExpr var nme init@((expTypArg :: MyMonadCls (A ) => ExprTyp (A ) α -> (A α)) -> typb) = return init </code></pre> then it resolves fine. What's the problem resolving <code>ExprTyp == MyBaseExpr</code> for <code>expTypArg</code>? <h2>edit</h2> Thanks very much, Daniel! Here's a way to take some of the verbage out, after noticing that only the type of <code></code> need be enforced. <pre><code>ignore_comp :: α -> β -> β ignore_comp a b = b instance forall . (Monad , Monad (A )) => MyMonadCls (A ) where type ExprTyp (A ) = MyBaseExpr var nme init@(expTypArg -> typb) = typb `ignore_comp` return init </code></pre>
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<haskell><typeclass><associated-types>
Title
haskell -- not understanding why this associated type example needs more inference
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