> PRS: (>>=) :: Parser a -> Parser b -> Parser b
> p >>= f = \inp ->
>case p inp of
> [] -> []
> [(v, out)] -> parse (f v) out
You probably want:
(>>=) :: Parser a -> (a -> Parser b) -> Parser b
p >>= f = \inp -> case parse p inp of
[] -> []
> Hi
> (>>=) :: Parser a -> Parser b -> Parser b
> p >>= f = \inp ->
>case p inp of
> [] -> []
> [(v, out)] -> parse (f v) out
> based on a lot of guesswork, after the mess created by the OCR, I
> managed to get the above example to work syntactically but is it
> semantically correct?
Hi
(>>=) :: Parser a -> Parser b -> Parser b
p >>= f = \inp ->
case p inp of
[] -> []
[(v, out)] -> parse (f v) out
based on a lot of guesswork, after the mess created by the OCR, I
managed to get the above example to work syntactically but is it
semantically correct?
Thanks, Paul
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