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parsing ok((2+3)==5)
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Subject: parsing ok((2+3)==5)
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From: amaha...@cs.indiana.edu (Abhijit Mahabal)
I hunted down the cause of the non-parsing of
ok((2 + 3) == $five, "== (sum on lhs)");
in 03operator.t, but am not yet up to speed in Haskell to fix it.
Below is the location of the problem.
The error is in Parser.hs, in the blocks for parseApply and
parseParamList. parseApply eats parens using maybeDotParens, and then
calls parseParamList, which is also willing enough to gobble more parens.
Thus the following fails to parse: f((2+3),5), but this succeeds:
f(((2+3), 5)).
--abhijit
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