Java 8类型推断错误

时间:2017-11-06 15:22:56

标签: java java-8 java-stream

我尝试使用Java 8流来推广我的地图转置方法。这是代码

public static <K, V> Map<V, Collection<K>> trans(Map<K, Collection<V>> map,
                                                     Function<? super K, ? extends V> f,
                                                     Function<? super V, ? extends K> g) {
        return map.entrySet()
                .stream()
                .flatMap(e -> e.getValue()
                        .stream()
                        .map(l -> {
                            V iK = f.apply(e.getKey());
                            K iV = g.apply(l);
                            return Tuple2.of(iK, iV);
                        }))
                .collect(groupingBy(Tuple2::getT2, mapping(Tuple2::getT1, toCollection(LinkedList::new))));
    }

public class Tuple2<T1, T2> {

    private final T1 t1;
    private final T2 t2;

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

    // constructor and getters omitted
}

但是我收到了此错误消息

Error:(66, 25) java: incompatible types: inference variable K has incompatible bounds
    equality constraints: V
    lower bounds: K

我需要改变什么才能发挥作用?

1 个答案:

答案 0 :(得分:3)

实际上,您实际上将值转换为原始输入的键和副词,但由于您应用的函数保留了与原始映射中相同的键值类型,因此最终得到{{1}在flatmap操作之后,collect会再次返回Stream<Tuple2<V, K>>

所以方法标题应为:

Map<K, Collection<V>>