如何在Java中實現函數參數的元組解包？

Question

我創建了一些tuple類，並將它們放入Java集合中。 但是我不想在迭代集合時直接使用tuple作為函數參數。所以我實現了元組解包，如下面的代碼。

基本上它可以工作，但問題是需要進行類型轉換：

map((Func2<Long, Long, Long>) (a, b) -> a + b)

有沒有辦法在這里刪除類型轉換？

---

編輯：

也許我沒有說清楚，不僅是tuple2 ，還有tuple3 ， tuple4 ......應該得到支持。 Tuple2的答案對Tuple2很大Tuple2 ，但在此期間對tuple2 ， tuple3 ， tuple4

package test;


import com.google.common.collect.Iterables;

import java.util.Arrays;
import java.util.function.Function;

import static test.TupleIterable.Tuple.tuple;

public interface TupleIterable<T> {

    Iterable<T> apply();

    static <E> TupleIterable<E> from(Iterable<E> iterable) {
        return () -> iterable;
    }

    default <E> TupleIterable<E> map(Function<? super T, ? extends E> op) {
        return () -> Iterables.transform(TupleIterable.this.apply(), op::apply);
    }

    interface Func2<T1, T2, R> extends Function<Tuple.Tuple2<T1, T2>, R> {

        R apply(T1 t1, T2 t2);

        @Override
        default R apply(Tuple.Tuple2<T1, T2> t) {
            return apply(t.t1, t.t2);
        }
    }

    interface Func3<T1, T2, T3, R> extends Function<Tuple.Tuple3<T1, T2, T3>, R> {

        R apply(T1 t1, T2 t2, T3 t3);

        @Override
        default R apply(Tuple.Tuple3<T1, T2, T3> t) {
            return apply(t.t1, t.t2, t.t3);
        }
    }

    interface Tuple {

        static <T1, T2> Tuple2<T1, T2> tuple(T1 t1, T2 t2) {
            return new Tuple2<>(t1, t2);
        }

        static <T1, T2, T3> Tuple3<T1, T2, T3> tuple(T1 t1, T2 t2, T3 t3) {
            return new Tuple3<>(t1, t2, t3);
        }

        class Tuple2<T1, T2> implements Tuple {

            public T1 t1;

            public T2 t2;

            public Tuple2(T1 t1, T2 t2) {
                this.t1 = t1;
                this.t2 = t2;
            }
        }

        class Tuple3<T1, T2, T3> implements Tuple {

            public T1 t1;

            public T2 t2;

            public T3 t3;

            public Tuple3(T1 t1, T2 t2, T3 t3) {
                this.t1 = t1;
                this.t2 = t2;
                this.t3 = t3;
            }
        }
    }

    public static void main(String[] args) {
        TupleIterable.from(Arrays.asList(1L, 2L))
                .map(x -> tuple(x, x)) // map long to tuple2
                .map((Func2<Long, Long, Tuple.Tuple3<Long, Long, Long>>) (a, b) -> tuple(a, b, a + b)) // map tuple2 to tuple3
                .map((Func3<Long, Long, Long, Long>) (a, b, c) -> a + b + c) // map tuple3 to Long
                .apply()
                .forEach(System.out::println);
    }
}

Answer 1

你可以像JDK開發人員那樣為原始類型的Stream化做同樣的技巧。 介紹mapToTuple並在不同的接口中unwrap 。

我重寫了一些代碼並使用了一些現有的FunctionalInterfaces 。

import java.util.Arrays;
import java.util.Iterator;
import java.util.function.BiFunction;
import java.util.function.Function;
import java.util.function.Supplier;

public class Test {

  public static void main(String... args) {
    MyIterable.from(Arrays.asList(1L, 2L)).mapToTuple(l -> Tuple2.tuple(l, l + 1L)).unwrap((a, b) -> a + b).get()
        .forEach(System.out::println);
  }
}

final class Tuple2<T1, T2> {

  public static <T1, T2> Tuple2<T1, T2> tuple(T1 t1, T2 t2) {
    return new Tuple2<>(t1, t2);
  }

  public final T1 t1;
  public final T2 t2;

  private Tuple2(T1 t1, T2 t2) {
    this.t1 = t1;
    this.t2 = t2;
  }
}

@FunctionalInterface
interface TupleIterable<T1, T2> extends Supplier<Iterable<Tuple2<T1, T2>>> {

  default <E> MyIterable<E> unwrap(BiFunction<T1, T2, E> func) {
    return () -> Iterables.transform(get(), t -> func.apply(t.t1, t.t2));
  }
}

@FunctionalInterface
interface MyIterable<T> extends Supplier<Iterable<T>> {

  static <E> MyIterable<E> from(Iterable<E> iterable) {
    return () -> iterable;
  }

  default <E> MyIterable<E> map(Function<? super T, ? extends E> mapper) {
    return () -> Iterables.transform(get(), mapper::apply);
  }

  default <T1, T2> TupleIterable<T1, T2> mapToTuple(Function<? super T, ? extends Tuple2<T1, T2>> tupleMapper) {
    return () -> Iterables.transform(get(), tupleMapper::apply);
  }
}

final class Iterables {
  public static <T, E> Iterable<E> transform(Iterable<T> iterable, Function<? super T, ? extends E> mapper) {
    return () -> new Iterator<E>() {

      private final Iterator<T> iter = iterable.iterator();

      @Override
      public boolean hasNext() {
        return iter.hasNext();
      }

      @Override
      public E next() {
        return mapper.apply(iter.next());
      }
    };
  }
}

Answer 2

我沒有進行類型轉換，而是創建了一個unpack函數來解包元組。 它的工作方式類似於map(unpack((a, b) -> tuple(a, b, a + b))) ，但仍然不夠直觀。

package test;


import com.google.common.collect.Iterables;

import java.util.Arrays;
import java.util.function.Function;

import static test.TupleIterable.Func.unpack;
import static test.TupleIterable.Tuple.tuple;

public interface TupleIterable<T> {

    Iterable<T> apply();

    static <E> TupleIterable<E> from(Iterable<E> iterable) {
        return () -> iterable;
    }

    default <E> TupleIterable<E> map(Function<? super T, E> op) {
        return () -> Iterables.transform(TupleIterable.this.apply(), op::apply);
    }

    interface Func {

        static <T1, T2, R> Function<Tuple.Tuple2<T1, T2>, R> unpack(Func2<T1, T2, R> f) {
            return t -> f.apply(t.t1, t.t2);
        }

        static <T1, T2, T3, R> Function<Tuple.Tuple3<T1, T2, T3>, R> unpack(Func3<T1, T2, T3, R> f) {
            return t -> f.apply(t.t1, t.t2, t.t3);
        }

        @FunctionalInterface
        interface Func2<T1, T2, R> {

            R apply(T1 t1, T2 t2);
        }

        @FunctionalInterface
        interface Func3<T1, T2, T3, R> {

            R apply(T1 t1, T2 t2, T3 t3);
        }
    }

    interface Tuple {

        static <T1, T2> Tuple2<T1, T2> tuple(T1 t1, T2 t2) {
            return new Tuple2<>(t1, t2);
        }

        static <T1, T2, T3> Tuple3<T1, T2, T3> tuple(T1 t1, T2 t2, T3 t3) {
            return new Tuple3<>(t1, t2, t3);
        }

        class Tuple2<T1, T2> implements Tuple {

            public T1 t1;

            public T2 t2;

            public Tuple2(T1 t1, T2 t2) {
                this.t1 = t1;
                this.t2 = t2;
            }
        }

        class Tuple3<T1, T2, T3> implements Tuple {

            public T1 t1;

            public T2 t2;

            public T3 t3;

            public Tuple3(T1 t1, T2 t2, T3 t3) {
                this.t1 = t1;
                this.t2 = t2;
                this.t3 = t3;
            }
        }
    }

    public static void main(String[] args) {
        TupleIterable.from(Arrays.asList(1L, 2L))
                .map(x -> tuple(x, x)) // map long to tuple2
                .map(unpack((a, b) -> tuple(a, b, a + b))) // map tuple2 to tuple3
                .map(unpack((a, b, c) -> a + b + c)) // map tuple3 to Long
                .apply()
                .forEach(System.out::println);
    }
}

Answer 3

我無法確定這是否太“臟”，但為什么不在Tuple2為T1和T2提供getter並執行：

.map(x -> tuple(x, x + 1)) // map long to tuple2
.map(a -> a.getT1() + a.getT2()) // map tuple2 to long