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Wrong type for the `pParens` parser combinator?
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{-# LANGUAGE UnicodeSyntax #-} | |
-- From : Swiestra (2009). Combinator Parsing: A Short Tutorial | |
-- Tested with : GHC 7.8.4 | |
------------------------------------------------------------------------------ | |
infix 7 <$> | |
infixl 7 <$ | |
infixl 5 <*>, <* | |
newtype Parser s t = P ([s] → [(t, [s])]) | |
-- unP ∷ Parser s t → [s] → [(t, [s])] | |
-- unP (P p) = p | |
pSym ∷ Eq s ⇒ s → Parser s s | |
pSym a = P (\inp → case inp of | |
(s : ss) | s == a → [(s, ss)] | |
| otherwise → [] | |
_ → [] | |
) | |
pReturn ∷ a → Parser s a | |
pReturn a = P (\inp → [(a, inp)]) | |
(<*>) ∷ Parser s (b → a) → Parser s b → Parser s a | |
P p1 <*> P p2 = P (\inp → [ (v1 v2, ss2) | (v1, ss1) ← p1 inp | |
, (v2, ss2) ← p2 ss1 | |
] | |
) | |
(<*) ∷ Parser s a → Parser s b → Parser s a | |
p <* q = const <$> p <*> q | |
(<$>) ∷ (b → a) → Parser s b → Parser s a | |
f <$> p = pReturn f <*> p | |
(<$) ∷ a → Parser s b → Parser s a | |
f <$ p = const <$> pReturn f <*> p | |
pParens ∷ Parser s a → Parser s a | |
pParens p = id <$ pSym '(' <*> p <* pSym ')' | |
-- Error: | |
-- /tmp/ParensType.hs:45:24: | |
-- Couldn't match type `s' with `Char' | |
-- `s' is a rigid type variable bound by | |
-- the type signature for pParens :: Parser s a -> Parser s a | |
-- at /tmp/ParensType.hs:45:1 | |
-- In the first argument of `pSym', namely '(' | |
-- In the second argument of `(<$)', namely pSym '(' | |
-- In the first argument of `(<*>)', namely id <$ pSym '(' |
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