Usin case expression to wrap result in Haskell

Question

Suppose I have some type

data Result = Fail | Success deriving (Show)

expressing result of some operation and another arbitrary type

data MyData

and some function which processes the data

someAction :: MyData -> Result
someAction = if ... then Success else Fail

I need a wrapper function formatAction which calls someAction with MyData and formats the result of processing using case statement. Something like that:

formatAction :: MyData -> String
formatAction = let res = someAction in
  case res of 
    Success -> "OK"
    Fail -> "Fail"

It's not a valid code, but I can't undestand how to use result of the function with 'case' expression. Any idea?

Actually I meant a piece of code like that:

data Result = Fail | Success

data SomeData = SDS | SDF 

doSomeWork' :: SomeData -> Result
doSomeWork' SDS = Success
doSomeWork' _ = Fail

processData :: SomeData -> String
processData = let res = doSomeWork' in
    case res of 
        Success -> "OK"
        Fail -> "Fail"

It doesn't compile with the message:

D:\Work\Haskell\revRange.hs:16:17: Couldn't match expected type SomeData -> Result' with actual typeResult' In the pattern: Success In a case alternative: Success -> "OK" In the expression: case res of { Success -> "OK" Fail -> "Fail" } D:\Work\Haskell\revRange.hs:16:28: Couldn't match expected type SomeData -> String' with actual type[Char]' In the expression: "OK" In a case alternative: Success -> "OK" In the expression: case res of { Success -> "OK" Fail -> "Fail" } D:\Work\Haskell\revRange.hs:17:17: Couldn't match expected type SomeData -> Result' with actual typeResult' In the pattern: Fail In a case alternative: Fail -> "Fail" In the expression: case res of { Success -> "OK" Fail -> "Fail" } D:\Work\Haskell\revRange.hs:17:25: Couldn't match expected type SomeData -> String' with actual type[Char]' In the expression: "Fail" In a case alternative: Fail -> "Fail" In the expression: case res of { Success -> "OK" Fail -> "Fail" }

Just can't catch how to match the result to the patterns.

If you are not going to use the derived Show instance, you can implement it yourself (Success mapped to "OK", of source) and define formatAction = show . someAction. — zakyggaps
OK. It's not about 'show'. This code doesn't compile because of: Couldn't match expected type MyData -> String' with actual type [Char]' In the expression: "OK" In a case alternative: Success -> "OK" In the expression: case res of { Success -> "OK" Fail -> "Fail" } — BarbedWire
The problem is that processData does not take any arguments, and does not forward them to doSomeWork'. Hence, res is a function, not a SomeData value, and you can't pattern match on that. Simon H. below did it correctly. — chi
May be you are right. I've tried so many variants and probably decided to use pointless style only..:)) — BarbedWire

Simon H Simon H · Accepted Answer · 2016-03-18T16:16:37

You may need to add the function parameter

formatAction :: MyData -> String
formatAction d = 
  case someAction d of 
    Success -> "OK"
    Fail -> "Fail"

Usin case expression to wrap result in Haskell

1 Answers