On Montag, 24. Dezember 2012, 06:47:34, jugree@lavabit.com wrote:
Hello.
Could you explain this example(0)? Could you show its step by step execution?
gcd' :: Int -> Int -> Writer (DiffList String) Int gcd' a b | b == 0 = do tell (toDiffList ["Finished with " ++ show a]) return a | otherwise = do result <- gcd' b (a `mod` b) tell (toDiffList [show a ++ " mod " ++ show b ++ " = " ++ show (a `mod` b)]) return result
Why does the above append the log to the beginning of the list?
It doesn't. Note that it first computes the gcd of b and (a `mod` b), logging the steps, and only afterwards "tell"s the original arguments. So the very first thing that is logged is the "Finished with" message. gcd' 3 2 -- nothing logged yet gcd' 2 1 -- nothing logged yet gcd' 1 0 -- start logging tell (["Finished with 1"]++) return 1 -- log is (["F. w. 1"]++) tell (["2 mod 1 = 0"]++) return 1 -- log is ((["F. w. 1"]++) . (["2 mod 1 = 0"]++)) tell (["3 mod 2 = 1"]++) return 1 -- log is (((["F. w. 1"]++) . (["2 mod 1 = 0"]++)) . (["3 mod 2 = 1"]++))
What value will result have in the following?
result <- gcd' 2 (3 `mod` 2)
result will be bound to 1 (the value of gcd 2 1). Basically, a `Writer monoid a` is a pair `(a, monoid)` and the monadic bind `(>>=)` that the do-notation desugars to is (x,log) >>= f = let (y, newLog) = f x in (y, log `mappend` newLog) so do result <- gcd' 2 (3 `mod` 2) tell (["3 mod 2 = 1"]++) return result becomes gcd' 2 1 >>= \result -> (tell (...) >> return result) and substituting gcd' 2 1 with its result: (1,log) >>= \result -> (tell (...) >> return result) ~> let (y, newLog) = (\result -> (tell (...) >> return result)) 1 in (y, log <> newLog) ~> let (y, newLog) = tell (...) >> return 1 in (y, log <> newLog) ~> let (y, newLog) = let (_, told) = tell (...) in (1, mempty <> told) in (y, log <> newLog)
(0) http://learnyouahaskell.com/for-a-few-monads-more#writer