Ambiguous occurrence in Haskell with “show”

问题

I'm new in functional programming and I'm trying to create and show a Stack with Haskell. I'd like my program to show me the Stack I'm building with it. This is my code:

module Stack (Stack, empty, push, pop, top, isEmpty) where
    data Stack a       = EmptyStack | Stk a (Stack a)
    push x s           = Stk x s
    top (Stk x s)      = x
    pop (Stk _ s)      = s
    empty              = EmptyStack
    isEmpty EmptyStack = True
    isEmpty (Stk x s)  = False`

    instance Show a => Show (Stack a) where
    show EmptyStack = "|"
    show (Stk a b) = (show a) ++ " <- " ++ (show b)

With "show (push 1 empty)" I'd expect an answer (more or less) like: " 1 <- | " But I'm not able to compile the code. When I try it shows the following error:

[1 of 1] Compiling Stack            ( Stack.hs, interpreted )

Stack.hs:12:27:
    Ambiguous occurrence ‘show’
    It could refer to either ‘Stack.show’, defined at Stack.hs:11:9
                      or ‘Prelude.show’,
                         imported from ‘Prelude’ at Stack.hs:1:8-12
                         (and originally defined in ‘GHC.Show’)

Stack.hs:12:47:
    Ambiguous occurrence ‘show’
    It could refer to either ‘Stack.show’, defined at Stack.hs:11:9
                      or ‘Prelude.show’,
                         imported from ‘Prelude’ at Stack.hs:1:8-12
                         (and originally defined in ‘GHC.Show’)
Failed, modules loaded: none.

I understand the error where program could confuse the "show" from Prelude with an "show" defined by be, but I cannot see that error in my code. Besides, some mates have the same code, and the program works well.

There is something I have to change or I have missed?

Thanks!

回答1:

So the first problem is that you've got a ` character in the code you pasted for us. Your second problem is that you don't need to indent all of the lines in the module; most Haskell modules that I see will not indent the body of the module. Your third problem is that you do not need parentheses around show a and show b: precedence in Haskell is really simple; parentheses always take top priority, followed by function application (left-associative or "greedy nom", a function always gobbles up the first thing that it sees in front of it), followed by operators in their defined precedence, followed by special syntactic forms like \a ->, let, do, where. Those are generally aesthetic concerns but you probably still care.

Your final problem is here:

instance Show a => Show (Stack a) where
show EmptyStack = "|"
show (Stk a b) = (show a) ++ " <- " ++ (show b)

You want Haskell to turn this into the single statement:

instance Show a => Show (Stack a) where show tmpvar = case tmpvar of { EmptyStack -> "|"; Stk a b -> show a ++ " <- " ++ show b }

However Haskell has instead turned this into two separate lines:

instance Show a => Show (Stack a) where {} 

show tmpvar = case tmpvar of { EmptyStack -> "|"; Stk a b -> show a ++ " <- " ++ show b }

So the multiple-definition is correctly converted into a case-dispatch, but it's not put within the curly braces above! So, you can omit curly braces {} by using whitespace to indent lines. After where, Haskell does not see any explicit {} so it starts looking for indented lines, and it sees 0 of them, so it converts the clause to where {} (thanks @chi).

Without being in curly braces, either because of indentation or not, that new line defines a different function, Stack.show, distinct from the imported Prelude.show that belongs to the Show typeclass. The problem is that it also references a function called show, which is now ambiguous: is this a recursive call for a function with infinite type show :: Stack (Stack (Stack ...)) -> String or a dispatching call for a function with finite type show :: (Show a) => Stack a -> String? Before it even tries to figure out those types it says "stop it, I don't know what you mean, please clarify."

Probably what you intended was instead:

instance Show a => Show (Stack a) where
    show EmptyStack = "|"
    show (Stk a b) = show a ++ " <- " ++ show b

This indentation clues the Haskell compiler to accepti the two following statements into the where clause.

来源：https://stackoverflow.com/questions/34635468/ambiguous-occurrence-in-haskell-with-show

标签

haskell

ambiguous-call