Haskell conversion between types

Question

Again stuck on something probably theoretical. There are many libraries in Haskell, i'd like to use less as possible. If I have a type like this:

data Note = Note { _noteID :: Int
                 , _noteTitle :: String
                 , _noteBody :: String
                 , _noteSubmit :: String
                 } deriving Show

And use that to create a list of [Note {noteID=1...}, Note {noteID=2...}, ] et cetera. I now have a list of type Note . Now I want to write it to a file using writeFile . Probably it ghc will not allow it considering writeFile has type FilePath -> String -> IO () . But I also want to avoid deconstructing (writeFile) and constructing (readFile) the types all the time, assuming I will not leave the Haskell 'realm'. Is there a way to do that, without using special libs? Again: thanks a lot. Books on Haskell are good, but StackOverflow is the glue between the books and the real world.

Answer 1

If you're looking for a "quick fix", for a one-off script or something like that, you can derive Read in addition to Show , and then you'll be able to use show to convert to String and read to convert back, for example:

data D = D { x :: Int, y :: Bool }
    deriving (Show, Read)

d1 = D 42 True

s = show d1
-- s == "D {x = 42, y = True}"

d2 :: D
d2 = read s
-- d2 == d1

However, please, please don't put this in production code. First, you're implicitly relying on how the record is coded, and there are no checks to protect from subtle changes. Second, the read function is partial - that is, it will crash if it can't parse the input. And finally, if you persist your data this way, you'll be stuck with this record format and can never change it.

For a production-quality solution, I'm sorry, but you'll have to come up with an explicit, documented serialization format. No way around it - in any language.