(恐ろしく不自然な例を言い訳してください)"where"句に式ジェネリック型を与えるにはどうすればよいですか?私が何をしたいか
は、WHERE句でタイプを指定します:
somemap :: (a -> b) -> [a] -> [b]
somemap f xs = ys
where
some = take 5 xs :: [a]
ys = map f some :: [b]
しかし、これはエラーが発生します。
*Main> :load file.hs
[1 of 1] Compiling Main (file.hs, interpreted)
fil.hs:15:18:
Couldn't match expected type `a1' against inferred type `a'
`a1' is a rigid type variable bound by
an expression type signature at file.hs:15:25
`a' is a rigid type variable bound by
the type signature for `somemap' at file.hs:12:12
Expected type: [a1]
Inferred type: [a]
In the second argument of `take', namely `xs'
In the expression: take 5 xs :: [a]
file.hs:16:13:
Couldn't match expected type `b1' against inferred type `b'
`b1' is a rigid type variable bound by
an expression type signature at file.hs:16:24
`b' is a rigid type variable bound by
the type signature for `somemap' at file.hs:12:17
In the first argument of `map', namely `f'
In the expression: map f some :: [b]
In the definition of `ys': ys = map f some :: [b]
Failed, modules loaded: none.
具体的な型を指定する場合、a
とにはInt
を代入しますb
について、何の問題:
somemap :: (Int -> Bool) -> [Int] -> [Bool]
somemap f xs = ys
where
some = take 5 xs :: [Int]
ys = map f some :: [Bool]
だから私の質問はありません:どのように私は、WHERE句でジェネリック型と型制約を指定するのですか? where
句インサイド
http://stackoverflow.com/questions/7408911/haskell-magical-code-whats-going-on-hereのコードは、このような「醜いヘルパー機能」から生じる可能性のある混乱の良い例です。 –