Question

让我们说我在Java中有这样的东西：

class A {
    int v;
    public A(int v) {
        this.v = v;
    }
    public int a() {
        return v;
    }
}

class B extends A {
    public B(int v) { 
        super(v);
    }
    @Override
    public int a() {
        return super.a() + 5;
    }
}

我怎样才能在Haskell中做类似的事情？我对类型类有一个非常基本的了解。我想在虚拟方法中使用虚拟方法来获得ArrayList<A>个对象（或者只是Haskell中的[A]），这些方法将继承自超类。

Answer 1

有一种＆＃34;模式＆＃34;这样工作：

data Animal m = Animal {
  makeSound :: m ()
}

cat = Animal {
  makeSound = print "Meow!"
}

dog = Animal {
  makeSound = print "Woof!"
}

cow = Animal {
  makeSound = print "Moo!"
}


animals = [cat, cat, dog, cow]

main = mapM_ makeSound animals

使用这种模式，您可以做一些或多或少的疯狂事情，例如：

import Data.Maybe

data I a = I {
  f :: a,
  sup :: Maybe (I a)
}

mkA = I { f = 10,  sup = Nothing }

mkB = let a = mkA in I { 
   f = 5 + f a,
   sup = Just a
 }

ls = [mkA, mkB, fromJust $ sup mkB]

main = mapM_ (print . f) ls

此模式还可用于实现像工厂一样的功能，如：

import Data.IORef

data Storage m a = Storage {
  putData :: String -> m (),
  readData :: m String
}

fileStorage path = Storage {
  putData = \s -> writeFile path s,
  readData = readFile path
}

inMemStorage ref = Storage {
  putData = \s -> writeIORef ref s,
  readData = readIORef ref
}

readStorage storage = readData storage

chooseStorage "file" = return $ fileStorage "tmp.txt"
chooseStorage "mem" = do
  r <- newIORef ""
  return $ inMemStorage r

main = do
  storage <- chooseStorage "mem"
  putData storage "hi there"
  txt <- readData storage
  print txt

基本上，您定义了一个以函数为成员的数据类型，然后定义＆＃34; constructors＆＃34; （它们只是常规函数）返回该数据类型的实例。

Haskell替代虚方法和方法继承

1 个答案: