Haskell中的OOP样式继承

Question

In C# I can declare the following

class A {
    int Field;
}

class B : A {
    int Field2;
}

static int f(A a) { return a.Field; }
static int f(B b) { return a.Field + b.Field2; }

static void Main(string[] args) {
    A a = new A() { Field = 1 };
    A b = new B() { Field = 1, Field = 2};

    Console.WriteLine(f(a) + f(b));
}

In Haskell I would type out the above as

data A = A { field :: Int } | B { field :: Int, field2 :: Int }

f :: A -> Int
f (A a) = a
f (B a b) = a + b

main :: IO()
main = do putStrLn $ show (f(a) + f(b))
    where a = A 1
          b = B 1 2

What I don't like about the Haskell counterpart is that I have to repeat field twice in the data definition of A (this becomes more tiresome as the number of fields increases that are present in A that need to be in B ). Is there a more concise way in Haskell to write B as a subclass of A (that is somewhat similar to the C# way)?

Answer 1

If A has a lot of fields, one design that might make sense is to have two different datatypes for A and B, where a B contains an A, and then use a typeclass to define f. Like so:

data A = A {field1 :: Int, field2 :: Int, ..., field9999 :: Int}
data B = B {a :: A, field10000 :: Int}

class ABC a where
    f :: a -> Int

instance ABC A where
    f a = field1 a + field101 a

instance ABC B where
    f b = f (a b) + field10000 b

Answer 2

Your data definition is a lateral move, both the A and B constructors are instances of the type A, where what you're looking for is a type B that is-a type A but not just an instance of type A.

Try:

data Foo = Foo {field :: Type}
newtype Bar = Bar Foo

fromBar Bar x = x
toBar x = Bar x

This will allow you to declare all sorts of specialty functions on type Bar that will not apply to Foo, as well as operate on Bar using functions designed for type Foo.

If you want to extend the data parameters of Foo with additional information in Bar, you could use a data declaration and a has-a relationship. This isn't as desirable as a newtype, as newtype is lazily evaluated where the data declaration is strict. Still:

data Foo = Foo {field :: Type}
data Bar = Bar {fooField :: Foo, moreField :: Type}

As sepp2k points out, what you really want to do to define the "nature" of a type is not by how it looks, but by how it works . We do this in Haskell by defining a type class, and having a type instance that class.

Hope this helps.