Question

我正在尝试在Java 1.8中实现Church Numerals。我的第一次尝试是：

import java.util.function.UnaryOperator;

@FunctionalInterface
public interface ChurchNumeral {
  public static ChurchNumeral valueOf(int n) {
    if (n < 0) {
      throw new IllegalArgumentException("Argument n must be non-negative.");
    }
    if (n == 0) {
      return (f, arg) -> arg;
    }
    return (f, arg) -> f(valueOf(n-1).apply(f, arg));
  }

  <T> T apply(UnaryOperator<T> f, T arg);
}

由于函数方法具有类型参数，因此失败。（具体来说，带有lambda表达式的行会给出错误：“Illegal lambda expression：类型ChurchNumeral的方法应用是通用的”。）

根据有关泛型与功能接口使用的相关问题的答案，我尝试参数化该类：

import java.util.function.UnaryOperator;

@FunctionalInterface
public interface ChurchNumeral<T> {                // This line changed.
  public static ChurchNumeral<?> valueOf(int n) {  // This line changed.
    if (n < 0) {
      throw new IllegalArgumentException("Argument n must be non-negative.");
    }
    if (n == 0) {
      return (f, arg) -> arg;
    }
    return (f, arg) -> f(valueOf(n-1).apply(f, arg));
  }

  T apply(UnaryOperator<T> f, T arg);              // This line changed.
}

第一个lambda表达式现在编译，但第二个表达式失败并出现此错误：

该方法适用于（UnaryOperator，捕获＃1-of？）类型ChurchNumeral不适用于参数（UnaryOperator，Object）

此外，我不希望每种可能的函数/参数类型都有不同版本的ChurchNumeral.ZERO。

有什么建议吗？

Answer 1

static interface Church<T> extends UnaryOperator<UnaryOperator<T>> {

    static <T> Church<T> of(int n) {
        if (n < 0) {
            throw new IllegalArgumentException();
        } else if (n == 0) {
            return f -> (t -> t);
        } else {
            return sum(f -> f, Church.of(n-1));
        }
    }

    static <T> Church<T> sum(Church<T> a, Church<T> b) {
        return f -> b.apply(f).andThen(a.apply(f))::apply;
    }
}

public static void main(String[] args) {
    Church<Integer> five = Church.of(5);
    Church<Integer> three = Church.of(3);
    Church<Integer> eight = Church.sum(five, three);

    assert 3 == three.apply(x -> x + 1).apply(0);
    assert 5 == five.apply(x -> x + 1).apply(0);
    assert 8 == eight.apply(x -> x + 1).apply(0);
}

修改：如果您希望教会界面中有apply(UnaryOperator<T> f, T arg)，那么您就可以拨打.apply(x->x+1,0)而不是.apply(x->x+1).apply(0)，可以添加默认值像这样的方法到上面的教会界面：

default T apply(UnaryOperator<T> f, T t) {
    return this.apply(f).apply(t);
}

编辑2：以下是可以在不同类型之间进行转换的更新类。我还添加了一个mul方法来乘法，看看它是如何工作的：

static interface Church<T> extends UnaryOperator<UnaryOperator<T>> {

    static <T> Church<T> of(int n) {
        if (n < 0) {
            throw new IllegalArgumentException();
        } else if (n == 0) {
            return zero();
        } else {
            return sum(one(), Church.of(n - 1));
        }
    }

    static <T> Church<T> zero() {
        return f -> (t -> t);
    }

    static <T> Church<T> one() {
        return f -> f;
    }

    static <T> Church<T> sum(Church<T> a, Church<T> b) {
        return f -> b.apply(f).andThen(a.apply(f))::apply;
    }

    static <T> Church<T> mul(Church<T> a, Church<T> b) {
        return f -> a.apply(b.apply(f))::apply;
    }

    default <U> Church<U> convert() {
        return (Church<U>) this;
    }
}

public static void main(String[] args) {
    Church<Integer> zero = Church.zero();
    Church<Integer> five = Church.of(5);
    Church<Integer> three = Church.of(3);
    Church<Integer> eight = Church.sum(five, three);
    Church<Integer> fifteen = Church.mul(three, five);

    assert 0 == zero.apply(x -> x + 1).apply(0);
    assert 3 == three.apply(x -> x + 1).apply(0);
    assert 5 == five.apply(x -> x + 1).apply(0);
    assert 8 == eight.apply(x -> x + 1).apply(0);
    assert 15 == fifteen.apply(x -> x + 1).apply(0);

    Church<String> strOne = Church.one();     
    Church<String> strThree = three.convert();  // make Church<String>
                                                // from a Church<Integer>

    assert "foo:bar".equals(strOne.apply("foo:"::concat).apply("bar"));
    assert "foo:foo:foo:bar".equals(strThree.apply("foo:"::concat).apply("bar"));

}

Answer 2

有没有办法做到这一点所以我不需要创建一个ChurchNumeral 每种可能的类型？我想能够申请ZERO（例如）任何UnaryOperator和类型为T的参数

我假设你的意思是你想做这样的事情：

ChurchNumeral five = ChurchNumeral.valueOf(5);
five.apply(s -> s + s, "s");
five.apply(Math::sqrt, Double.MAX_VALUE);

表示第一个示例中的方法签名：

<T> T apply(UnaryOperator<T> f, T arg);

是需要的。

然而，you can't use a lambda expression for a functional interface, if the method in the functional interface has type parameters.

解决方法是创建一个与lambdas兼容的子接口，并将对apply的调用委托给其方法，如下所示。

public static void main(String[]a){
    ChurchNumeral five = ChurchNumeral.valueOf(5);
    System.out.println(five.apply(s -> s + s, "s"));
    System.out.println(five.apply(Math::sqrt, Double.MAX_VALUE));
}
@FunctionalInterface
private interface ChurchNumeralT<T> extends ChurchNumeral {
    @SuppressWarnings({ "rawtypes", "unchecked" })
    @Override
    default<U> U apply(UnaryOperator<U> f, U arg){
        return (U)((ChurchNumeralT)this).tapply(f, arg);
    }
    T tapply(UnaryOperator<T> f, T arg);
}
public interface ChurchNumeral {

    <T> T apply(UnaryOperator<T> f, T arg);

    static ChurchNumeral valueOf(int n) {
        if (n < 0) {
            throw new IllegalArgumentException("Argument n must be non-negative.");
        }
        if (n == 0) {
            return (ChurchNumeralT<?>)(f, arg) -> arg;
        }
        return (ChurchNumeralT<?>)(f, arg) -> f.apply(valueOf(n - 1).apply(f, arg));
    }
}

如何使用Java 1.8实现Church Numerals

2 个答案: