Haskell - guard inside case statement

Question

I am going through Learn you a haskell book, and in Chapter 8 there is a snippet of code which looks like this

data LockerState = Taken | Free deriving (Eq, Show         


        

Answer 1

There are two places guards are allowed: function definitions and case expressions. In both contexts, guards appear after a pattern and before the body, so you use = in functions and -> in case branches, as usual:

divide x y
  | y == 0 = Nothing
  --------
  | otherwise = Just (x / y)
  -----------

positively mx = case mx of
  Just x | x > 0 -> Just x
         -------
  _ -> Nothing

Guards are simply constraints for patterns, so Just x matches any non-Nothing value, but Just x | x > 0 only matches a Just whose wrapped value is also positive.

I suppose the definitive reference is the Haskell Report, specifically §3.13 Case Expressions and §4.4.3 Function and Pattern Bindings, which describe the syntax of guards and specify where they’re allowed.

In your code, you want:

Just (state, code)
  | state == Taken -> Left "LockerNumber already taken!"
  | otherwise -> Right code

This is also expressible with patterns alone:

Just (Taken, _) -> Left "LockerNumber already taken!"
Just (_, code) -> Right code