java - 为什么Stream.reduce采用BinaryOperator 而不是BiFunction ？

Question

This question already has answers here:

What are good reasons for choosing invariance in an API like Stream.reduce()?

(2个答案)


去年关闭。




对于我的特定情况，我想简化使用功能组合；例如:

BiFunction<ImmutableSet<Integer>, ImmutableSet<Integer>, Sets.SetView<Integer>> f = Sets::intersection;
Function<Sets.SetView<Integer>, ImmutableSet<Integer>> g = Sets.SetView::immutableCopy;
BiFunction<ImmutableSet<Integer>, ImmutableSet<Integer>, ImmutableSet<Integer>> biFunction = f.andThen(g);
ImmutableSet<Integer> intersection = Stream.of(ImmutableSet.of(1, 2, 3), ImmutableSet.of(1, 2), ImmutableSet.of(4))
    .reduce(biFunction)
    .orElse(ImmutableSet.of());

这有一个编译错误:

argument mismatch BiFunction cannot be converted to BinaryOperator

相反，我需要这样做:

ImmutableSet<Integer> intersection = Stream.of(ImmutableSet.of(1, 2, 3), ImmutableSet.of(1, 2), ImmutableSet.of(4))
    .reduce((a, b) -> Sets.intersection(a, b).immutableCopy())
    .orElse(ImmutableSet.of());

但是，这会丢失组合提供的 point-free style。

为什么Stream API是这样设计的？ BinaryOperator是 BiFunction，所以用父类(super class)型声明 reduce方法的参数更有意义吗？

最佳答案

reduce操作必须采用相同类型的参数并返回相同类型。如果不是，则类型不匹配。这就是BinaryOperator的确切含义:BinaryOperator<T> extends BiFunction<T,T,T>
您可以创建BiFunction，而不必使用lambda。然后创建一个BinaryOperator:

import java.util.function.BinaryOperator;
import java.util.function.BiFunction;
import java.util.function.Function;

import java.util.stream.Stream;

import com.google.common.collect.ImmutableSet;
import com.google.common.collect.Sets;

public class StackOverflowTest {
  public static void main(String[] args) {

    BiFunction<ImmutableSet<Integer>, ImmutableSet<Integer>, Sets.SetView<Integer>> f = Sets::intersection;
    Function<Sets.SetView<Integer>, ImmutableSet<Integer>> g = Sets.SetView::immutableCopy;

    BiFunction<ImmutableSet<Integer>, ImmutableSet<Integer>, ImmutableSet<Integer>> biFunction = f.andThen(g);

    BinaryOperator<ImmutableSet<Integer>> biOperator = biFunction::apply;

    ImmutableSet<Integer> intersection =
       Stream.of(ImmutableSet.of(1, 2, 3),
                 ImmutableSet.of(1, 2),
                 ImmutableSet.of(1, 4)) // added a 1
             .reduce(biOperator)
             .orElse(ImmutableSet.of());

    System.out.println(intersection);

/*
prints:
[1]
*/
  }
}

关于java - 为什么Stream.reduce采用BinaryOperator <T>而不是BiFunction <T，T，T>？，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/60271793/

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Answer 1

reduce操作必须采用相同类型的参数并返回相同类型。如果不是，则类型不匹配。这就是BinaryOperator的确切含义:BinaryOperator<T> extends BiFunction<T,T,T>
您可以创建BiFunction，而不必使用lambda。然后创建一个BinaryOperator:

import java.util.function.BinaryOperator;
import java.util.function.BiFunction;
import java.util.function.Function;

import java.util.stream.Stream;

import com.google.common.collect.ImmutableSet;
import com.google.common.collect.Sets;

public class StackOverflowTest {
  public static void main(String[] args) {

    BiFunction<ImmutableSet<Integer>, ImmutableSet<Integer>, Sets.SetView<Integer>> f = Sets::intersection;
    Function<Sets.SetView<Integer>, ImmutableSet<Integer>> g = Sets.SetView::immutableCopy;

    BiFunction<ImmutableSet<Integer>, ImmutableSet<Integer>, ImmutableSet<Integer>> biFunction = f.andThen(g);

    BinaryOperator<ImmutableSet<Integer>> biOperator = biFunction::apply;

    ImmutableSet<Integer> intersection =
       Stream.of(ImmutableSet.of(1, 2, 3),
                 ImmutableSet.of(1, 2),
                 ImmutableSet.of(1, 4)) // added a 1
             .reduce(biOperator)
             .orElse(ImmutableSet.of());

    System.out.println(intersection);

/*
prints:
[1]
*/
  }
}