Am Freitag, 16. September 2005 16:02 schrieb Adam Wyner:
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Suppose I have two expressions:
emptyListA = null
emptyListB = []
emptyListA is apparently a function from empty lists to Bool.
emptyListA is a function from *arbitrary* lists to Bool.
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The problem is that there is no "show" function for emptyListB, which is just []
emptyListB
ERROR - Cannot find "show" function for: *** Expression : emptyListB *** Of type : [a]
The type [a] means forall a. [a], i.e., you can replace the "a" in [a] with an arbitrary type t and the expression in question has type [t]. This is only true for the empty list. [] has type [Integer], [String] etc. A show function which only accepted empty lists would have type (forall a. [a]) -> String. This is not possible in Haskell 98. You would need explicit universal quantification. What is possible in Haskell is to give show the type [a] -> String. This means forall a. ([a] -> String), i.e., show has every type [t] -> String where t is an arbitrary type. This would allow show to be applied to non-empty lists whose elements itself cannot be shown via the show functions. This has, of course, to be disallowed. The type [a] -> String is too general. The actual type of show in Haskell 98 is Show a => [a] -> String. The element type has to be an instance of Show which means that its values can be shown via the show function.
What I would like, simply is:
emptyListB
[]
You can specialize the type of emptyListB: input: emptyListB :: [Int] output: []
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Best wishes, Wolfgang