Haskell数据类型到Java(OO)

Question

我正在尝试将简单的Haskell数据类型和函数转换为OO.但我很困惑..

具有以下Haskell类型用于算术计算:

data Expr = Lit Int | 
      Add Expr Expr |
   deriving Show

--Turn the expr to a nice string
showExpr :: Expr -> String 
showExpr (Lit n) = show n
showExpr (Add e1 e2) = "(" ++ showExpr e1 ++ "+" ++ showExpr e2 ++ ")"

现在我正在尝试转换..

public interface Expr {
  String showExpr(String n);
} 

// Base case
public class Lit implements Expr {
  String n;

  public Lit(String n) {
    this.n = n;
  }

  @Override
  public String ShowExpr() {
    return n;
  }
}

public class Add implements Expr {
  Expr a;
  Expr b;

  public Add(Expr aExpr, Expr bExpr) {
    this.a = aExpr;
    this.b = bExpr;
  }

  public String ShowExpr() {
    return "(" + a.ShowExpr() + "+" + b.ShowExpr() + ")";
  }

  public static void main(String[] args) {
    Lit firstLit  = new Lit("2");
    Lit secLit = new Lit("3");
    Add add = new Add(firstLit,secLit);
    System.out.println(add.ShowExpr());
  }
}

这将导致"(2 + 3)",这是正确的答案.

但是......我不确定..这是考虑它并在OO中建模的正确方法吗？

是不是Haskell数据类型的良好表示？

Answer 1

让我们尽可能地复制代码.

以下是Haskell数据结构具有的一些属性:

1. 它有类型Expr和两个构造函数Lit和Add
2. 您无法在"外部"添加或删除构造函数

因此,如果我们希望这些属性在Java版本中保持正确,您应该这样做:

public abstract class Expr {
    // So that you can't add more subclasses outside this block
    private Expr() {}

    // Simulate pattern matching:
    // (This CAN be done with instanceof, but that's ugly and not OO)
    public boolean isLit() {
        return false;
    }
    public boolean isAdd() {
        return false;
    }
    public Lit asLit() {
        throw new UnsupportedOperationException("This is not a Lit");
    }
    public Add asAdd() {
        throw new UnsupportedOperationException("This is not an Add");
    }

    public static class Lit extends Expr {
        public final int n;
        public Lit(int n) {
            this.n = n;
        }
        @Override
        public boolean isLit() {
            return true;
        }
        @Override
        public Lit asLit() {
            return this;
        }
    }

    public static class Add extends Expr {
        public final Expr a, b;
        public Lit(Expr a, Expr b) {
            this.a = a;
            this.b = b;
        }
        @Override
        public boolean isAdd() {
            return true;
        }
        @Override
        public Add asAdd() {
            return this;
        }
    }
}

现在,转换showExpr:

public static String showExpr(final Expr expr) {
    if(expr.isLit()) {
        return Integer.toString(expr.asLit().n);
    } else if(expr.isAdd()) {
        return "(" + expr.asAdd().a + "+" + expr.asAdd().b + ")";
    }
}

您可以将showExpr静态方法放在Expr类中.我不会把它变成一个实例方法,因为它偏离了Haskell版本.