I tried {-# LANGUAGE ExtendedDefaultRules #-} and it didn't work. I
tried to put it both in Stack.hs and in TestStack.hs
Then I tried to be specific about the type in the test.
Both this code:
(pop EmptyStack :: Stack[Char]) ≡ (Nothing, EmptyStack)
and this:
(pop EmptyStack :: Stack[Char]) ≡ (Nothing :: Maybe[Char],
EmptyStack :: Stack[Char])
...give me:
test/TestStack.hs:20:12:
Couldn't match expected type `Stack [Char]'
with actual type `(Maybe a0, Stack a0)'
In the return type of a call of `pop'
In the first argument of `(==)', namely
`(pop EmptyStack :: Stack [Char])'
In the second argument of `it', namely
`((pop EmptyStack :: Stack [Char]) == (Nothing, EmptyStack))'
On Sat, Jul 30, 2011 at 11:37 PM, Daniel Fischer
On Saturday 30 July 2011, 22:19:28, Ovidiu Deac wrote:
I'm playing with Haskell so I wrote a stack module (see the code below). I have a problem with the pop function which returns a tuple (Nothing, EmptyStack) if called with an EmptyStack.
I kind of understand that the compiler cannot cannot figure out what type to use for a. But how could I tell the compiler that if the list is empty I don't care about that type?
You can't really, the compiler needs a specific type to know which Eq instance to use.
If you had a Num constraint, the defaulting rules (language report, http://www.haskell.org/onlinereport/haskell2010/haskellch4.html#x10-790004.3... ) would let the compiler pick a type (normally Integer, unless you have a default declaration that says otherwise). I'm not sure what GHC's ExtendedDefaultRules extension does, but there's a good chance that a
{-# LANGUAGE ExtendedDefaultRules #-}
pragma at the top will make it compile (and let GHC pick () for the type).
Another option is that you choose a type and write your condition
(pop (EmptyStack :: Stack [Char]) == (Nothing, EmptyStack))
(which is portable, hence preferable).