There are two main ways to get lists with different types.
Method one is the most common.
data MyTypes = MyInt Int | MyString String
--Then in your code you deal with it like so:
blah = map myTypeFunc [MyInt 1, MyString "asdf"]
where
myTypeFunc :: MyTypes -> IO ()
myTypeFunc (MyInt i) = putStrLn $ show i
myTypeFunc (MyString s) = putStrLn s
You can even use parametric types to make it more general
data MyTypes a b c = Type1 a | Type2 b | Type3 c
But you are still going to have to deal with each type explicitly
every function that handles them.
The other way is to use type classes:
class MyClass where
somefunction :: Bool -> a
instance MyClass Int where
somefunction True = 1
somefunction False = 0
instance MyClass String where
somefunction True = "true"
somefunction False = "false"
Then you make functions that ultimately make use of only the class's
functions. Since it is the only property on every element of the list
that is guaranteed to be there, it is all you can use.
myList :: MyClass a => [a]
myList = [somefunction True, somefunction False]
There are cases where one is the better option and cases where the
other is best.
Adapting this to your custom lists, you'd make a datatype like:
data HeteroList = Null | Element MyType HeteroList
On Mon, Jul 11, 2011 at 1:00 AM, Christopher Howard
I'm trying to understand parametric polymorphism and polymorphic data types. I especially want to go beyond simply using the defined polymorphic data types (lists and so forth) and see how I can make my own useful ones.
As my first stab at it, it seemed like I should be able to create my own heterogeneous "list" data type -- i.e., a "list" data type that can contain elements of different types. (like, [3,'a',True], for example)
But I'm a little stuck. My first try was like so:
data HeteroList a b = Null | Element a (HeteroList a b) deriving (Show)
...but this of course did not work, because all elements end up having to be the same type.
Then I tried
data HeteroList a b = Null | Element a (HeteroList b a) deriving (Show)
...but this doesn't work because every other other element has to be the same type:
Element 'a' (Element 1 (Element 'a' (Element 2 Null)))
...I could go on and embarrass myself some more, but since I'm likely widely off-base I'll just ask if somebody can point me in the right direction.
-- frigidcode.com theologia.indicium.us
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