Java支持Currying吗?

我想知道是否有任何方法可以在Java中拉动它。我认为如果没有对闭包的原生支持,这是不可能的。


答案 1

Java 8(2014年3月18日发布)确实支持咖喱。missingfaktor 在答案中发布的示例 Java 代码可以重写为:

import java.util.function.*;
import static java.lang.System.out;

// Tested with JDK 1.8.0-ea-b75
public class CurryingAndPartialFunctionApplication
{
   public static void main(String[] args)
   {
      IntBinaryOperator simpleAdd = (a, b) -> a + b;
      IntFunction<IntUnaryOperator> curriedAdd = a -> b -> a + b;

      // Demonstrating simple add:
      out.println(simpleAdd.applyAsInt(4, 5));

      // Demonstrating curried add:
      out.println(curriedAdd.apply(4).applyAsInt(5));

      // Curried version lets you perform partial application:
      IntUnaryOperator adder5 = curriedAdd.apply(5);
      out.println(adder5.applyAsInt(4));
      out.println(adder5.applyAsInt(6));
   }
}

...这相当不错。就个人而言,有了Java 8,我认为没有理由使用另一种JVM语言,如Scala或Clojure。当然,它们提供了其他语言功能,但这还不足以证明过渡成本和较弱的IDE/工具/库支持IMO是合理的。


答案 2

在Java中,咖喱和部分应用程序是绝对可能的,但是所需的代码量可能会让您失望。


一些代码来演示Java中的currying和部分应用程序:

interface Function1<A, B> {
  public B apply(final A a);
}

interface Function2<A, B, C> {
  public C apply(final A a, final B b);
}

class Main {
  public static Function2<Integer, Integer, Integer> simpleAdd = 
    new Function2<Integer, Integer, Integer>() {
      public Integer apply(final Integer a, final Integer b) {
        return a + b;
      }
    };  

  public static Function1<Integer, Function1<Integer, Integer>> curriedAdd = 
    new Function1<Integer, Function1<Integer, Integer>>() {
      public Function1<Integer, Integer> apply(final Integer a) {
        return new Function1<Integer, Integer>() {
          public Integer apply(final Integer b) {
            return a + b;
          }
        };
      }
    };

  public static void main(String[] args) {
    // Demonstrating simple `add`
    System.out.println(simpleAdd.apply(4, 5));

    // Demonstrating curried `add`
    System.out.println(curriedAdd.apply(4).apply(5));

    // Curried version lets you perform partial application 
    // as demonstrated below.
    Function1<Integer, Integer> adder5 = curriedAdd.apply(5);
    System.out.println(adder5.apply(4));
    System.out.println(adder5.apply(6));
  }
}

FWIW是Haskell相当于上面的Java代码:

simpleAdd :: (Int, Int) -> Int
simpleAdd (a, b) = a + b

curriedAdd :: Int -> Int -> Int
curriedAdd a b = a + b

main = do
  -- Demonstrating simpleAdd
  print $ simpleAdd (5, 4)

  -- Demonstrating curriedAdd
  print $ curriedAdd 5 4

  -- Demostrating partial application
  let adder5 = curriedAdd 5 in do
    print $ adder5 6
    print $ adder5 9