CODEWITHSUNDEEP
haskell
user606521
user606521

Reputation: 15434

IO String and IO Data type

I have this:

data SomeData = SomeData Int Int

getDataFromUser :: SomeData
getDataFromUser = do
{
    read (getLine)::SomeData;
}

This doesnt compile : Expected type String Actual type IO String

How I can fix it? I need this data deserialization...

Upvotes: 1

Views: 682

Answers (2)

Luis Casillas
Luis Casillas

Reputation: 30227

You need to read up more on how Haskell IO works and make sure you understand it.

A couple of points on your example. If you want to use read to deserialize to SomeData, you need to provide a Read instance for the type. You can use the default one:

data SomeData = SomeData Int Int deriving (Read)

Second: getLine is an IO action that returns a String, not a String; since read wants a String, this is the cause of your error. This is closer to what you want:

getDataFromUser :: IO SomeData
getDataFromUser = do str <- getLine
                     return (read str)

This can be simplified to the following, but make sure you understand the above example before you worry too much about this:

getDataFromUser :: IO SomeData
getDataFromUser = liftM read getLine

Upvotes: 7

ehird
ehird

Reputation: 40787

You're trying to treat getLine as a String, but it's an IO String — an IO action that, when executed, produces a string. You can execute it and get the resulting value from inside a do block by using <-, but since getDataFromUser does IO, its type has to be IO SomeData:

getDataFromUser :: IO SomeData
getDataFromUser = do
  line <- getLine
  return $ read line

More broadly, I would recommend reading a tutorial on IO in Haskell, like Learn You a Haskell's chapter on IO; it's very different from the IO facilities of most other languages, and it can take some time to get used to how things fit together; it's hard to convey a full understanding with answers to specific questions like this :)

Upvotes: 9

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