The layout of your code is very important when writing haskell code: Your code : expr = do t <- term do symbol "+" e <- expr return e return (t + e) +++ return t
is equivalent to:
expr = do { t <- term
; do { symbol "+"
; e <- expr
; return e
}
; return (t + e) -- e is not in scope
}
+++ return t -- t is not in scope
Both t and e are not in scope:
* e is in a nested do-block
* the expression 'return t' is outside the do-block
What you probably mean is: expr = do t <- term do symbol "+" e <- expr return (t + e) +++ return t
which is equivalent to:
expr = do { t <- term
; do { symbol "+"
; e <- expr -- (dropped the return e line)
; return (t + e)
}
+++ return t
}
Now t and e are in scope. The parser 'expr' will first recognize a 'term' and then try to recognize a '+' symbol followed by an expression. If that fails it returns 't'.
Cheers,
Arthur
On 15-mrt-05, at 16:28, Nicola Whitehead wrote:
Curiouser and curiouser...
expr :: Parser Int
expr = do t <- term
do symbol "+"
e <- expr
return (t + e)
+++ return t
solves the undefined variable problem but introduces a new 'Last operator in do {...} must be an expression' error, which then disappears if I explicitly return e
expr :: Parser Int
expr = do t <- term
do symbol "+"
e <- expr
return e
return (t + e)
+++ return t
to give me the original undefined variable t error at the line expr = do t <- term . It looks in scope to me... :(
Thanks,
Nik
Dr Nik Freydís Whitehead
University of Akureyri, Iceland
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Having the moral high ground and an FGMP-15 is better.
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