CODEWITHSUNDEEP
haskell
darko
darko

Reputation: 321

Adding an additional step to ">>="

I have this which works fine:

forM_ [1..10] $ \x -> myFunc1 x 99 >>= putStrLn . show >> return ()

myFunc1 :: Int -> Int -> IO Bool
myFunc1 .....

I want to do add an additional string to the output:

   forM_ [1..10] $ \x -> myFunc1 x 99 >>= (++) "the result is: " >>= putStrLn . show >> return ()

But that doesn't compile. I've tried different variations but still no success. Your suggestions?

Upvotes: 0

Views: 87

Answers (3)

Nagarjuna Pamu
Nagarjuna Pamu

Reputation: 14825

Replace ++ with add

add x y = return (x ++ show (y))


forM_ [1..10] $ \x -> myFunc1 x 99 >>= add "the result is: " >>= putStrLn . show >> return ()

reason: The return type of ++ is not IO type

Upvotes: 0

ErikR
ErikR

Reputation: 52029

The first thing that sticks out is that the expression:

(++) "the result is: "

is not an IO-action - it's just a pure function String -> String, and that's one reason why you code isn't type checking.

To turn it into an IO-action you can compose it with return:

return . ( (++) "the result is: " )
  :: String -> IO String

Now you can use it with >>=.

However, that's not where the real problem is...

To prepend "the result is: " before the show you have to insert it in the putStrLn call:

... >>= putStrLn . (++ "the result is: ") . show

(Note that there is no need to >> return () since putStrLn already returns ()).

Honestly, this is a lot simpler using do-notation:

forM_ [1..10] $ \x -> do
  s <- myFunc1 x 99
  putStrLn $ "the result is: " ++ show s

Upvotes: 7

Guerric Chupin
Guerric Chupin

Reputation: 461

You can compose show with the concatenation function like so:

forM_ [1..10] $ \x -> myFunc1 x 99 >>= putStrLn . ("the result is: " ++) . show >> return ()

Upvotes: 1

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