Java 8中谓词如何维护状态

Question

public static <T> Predicate<T> distinctByKey(Function<? super T,Object> keyExtractor) {
    Map<Object,Boolean> seen = new ConcurrentHashMap<>();
    return t -> seen.putIfAbsent(keyExtractor.apply(t), Boolean.TRUE) == null;
}

BigDecimal totalShare = orders.stream()
    .filter(distinctByKey(o -> o.getCompany().getId()))
    .map(Order::getShare)
    .reduce(BigDecimal.ZERO, BigDecimal::add);

我的问题是每次调用distinctByKey并产生新的ConcurrentHashMap。如何使用新的ConcurrentHashMap<>()维护状态；？

Answer 1

由于这是一个捕获λ,因此在每次调用DistinctByKey时，实际上都会返回一个新的谓词实例；但这将发生在整个流，而不是每个单独的元素。

如果您愿意使用以下内容运行示例：

djdk.internal.lambda.dumpProxyClasses=/your/path/here

对DistinctByKey的每次调用都将生成一个不同的实例，但该实例将被重用用于流的每个元素。如果运行这个示例，情况可能会更加明显：

 public static <T> Predicate<T> distinctByKey(Function<? super T,Object> keyExtractor) {
    Map<Object,Boolean> seen = new ConcurrentHashMap<>();
    Predicate<T> predicate =  t -> {
        Object obj = keyExtractor.apply(t);
        System.out.println("stream element : " + obj);
        return seen.putIfAbsent(obj, Boolean.TRUE) == null;
    };

    System.out.println("Predicate with hash :" + predicate.hashCode());
    return predicate;
}

@Getter
@AllArgsConstructor
static class User {
    private final String name;
}

public static void main(String[] args) {
    Stream.of(new User("a"), new User("b"))
          .filter(distinctByKey(User::getName))
          .collect(Collectors.toList());

}

这将输出：

Predicate with hash :1259475182
stream element : a
stream element : b

流的两个元素都有一个谓词。

Stream.of(new User("a"), new User("b"))
      .filter(distinctByKey(User::getName))
      .filter(distinctByKey(User::getName))
      .collect(Collectors.toList());

Predicate with hash :1259475182
Predicate with hash :1072591677
stream element : a
stream element : a
stream element : b
stream element : b