On Dec 8, 2007 3:12 PM, Ilya Tsindlekht
On Sat, Dec 08, 2007 at 02:59:16PM -0200, Felipe Lessa wrote:
Prelude> do {1 <- return 3; return' "ok"} Nothing Prelude> return 3 >>= \1 -> return' "ok" *** Exception: <interactive>:1:13-30: Non-exhaustive patterns in lambda What seems confusing to you?
\1 -> foo is the same as \x -> case x of {1 -> foo;}
When this function is evaluated with parameter different from 1, Haskell fails to find matching pattern for x and exception occurs.
The problem is that with the do notation it doesn't raise an exception. In the example you quoted, do {1 <- return 3; (return "ok" :: Maybe String)} == Nothing =/= _|_ The report says (http://www.haskell.org/onlinereport/exps.html#sect3.14) that the do above is translated to let ok 1 = return 3 >> (return "ok" :: Maybe String); ok _ = fail "..." in return 3 >>= ok --> let ok 1 = Just 3 >> Just "ok"; ok _ = Nothing in Just 3 >>= ok --> case Just 3 of {Nothing -> Nothing; Just x -> case x of {1 -> Just 3 >> Just "ok"; _ -> Nothing}} --> case 3 of {1 -> Just 3 >> Just "ok"; _ -> Nothing}} --> Nothing and not as return 3 >>= \1 -> (return "ok" :: Maybe String) --> _|_ as I initially thought. -- Felipe.