CODEWITHSUNDEEP
haskell
Olumide
Olumide

Reputation: 5829

Understanding Haskell Map.lookup (Example in LYH)

I'd appreciate help understanding the origin of the extra map in Map.lookup lockerNumber map, from the following example taken from chapter 8 of Learn You a Haskell for Great Good

lockerLookup :: Int -> LockerMap -> Either String Code  
lockerLookup lockerNumber map =   
    case Map.lookup lockerNumber map of   
        Nothing -> Left $ "Locker number " ++ show lockerNumber ++ " doesn't exist!"  
        Just (state, code) -> if state /= Taken   
                                then Right code  
                                else Left $ "Locker " ++ show lockerNumber ++ " is already taken!"

Upvotes: 1

Views: 3985

Answers (2)

chi
chi

Reputation: 116174

Briefly put, here map is a variable of type LockerMap, which I assume to be of the form Data.Map.Map someKeyType someValueType. It represents a (finite) association between elements of someKeyType and those of someValueType. This is sometimes also called a (finite) mapping, (finite) function, dictionary, hash, etc.

The expression Map.lookup x map simply accesses the map with key value x. The result is Nothing is there is no association for x in the map, and Just v if x is associated with value v.

Upvotes: 1

Zeta
Zeta

Reputation: 105915

There's no "additional" map (as in map :: (a -> b) -> [a] -> [b]) in that code. It gets more clear if we rename map to m:

lockerLookup :: Int -> LockerMap -> Either String Code  
lockerLookup lockerNumber m =   
    case Map.lookup lockerNumber m of   
      -- snip

Now you just need to remember that Map.lookup has type Ord k => k -> Map k a -> Maybe a and therefore you need to supply a Map. Luckily, the second argument (m) is a LockerMap, which is the same as Map Int (LockerState, Code).

Upvotes: 2

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