Am Mittwoch, 1. Oktober 2008 12:18 schrieb Cetin Sert:
warn :: String → IO Int warn = return 1 << putStrLn -- causes an error
try warn = (return 1 <<) . putStrLn
-- = \msg → return 1 << putStrLn msg -- works just fine -- = \msg → putStrLn msg >> return 1 -- works just fine
(<<) :: Monad m ⇒ m b → m a → m b b << a = a >>= \_ → b
(<<) = flip (>>)
Why do I get this compile-time error?? How can one define << ?
cetin@linux-d312:~/lab/test/qths/p> ghc -fglasgow-exts -O2 -o d64x --make demo2.hs system.hs [1 of 2] Compiling Netman.System ( system.hs, system.o )
system.hs:23:14: No instance for (Num (IO Int)) arising from the literal `1' at system.hs:23:14 Possible fix: add an instance declaration for (Num (IO Int)) In the first argument of `return', namely `1' In the first argument of `(<<)', namely `return 1' In the expression: return 1 << putStrLn
Okay warn = (return 1) << putStrLn putStrLn :: String -> IO () return 1 :: m b (<<) :: m b -> m a -> m b warn :: String -> IO Int so we must have (String -> IO ()) === m a (String -> IO Int) === m b So the monad is ((->) String), a === IO () b === IO Int, hence in return 1 :: String -> IO Int the 1 must have type IO Int. Now 1 is actually fromInteger 1, fromInteger :: (Num a) => Integer -> a, so the compiler looks for the instance Num (IO Int) where ... which it doesn't find.