为什么 Stream::flatMap 的这种用法是错误的?

2022-01-22 00:00:00 java-8 java java-stream flatmap

我希望能够像这样使用 Stream::flatMap

I expected to be able to use Stream::flatMap like this

public static List<String> duplicate(String s) {

    List<String> l = new ArrayList<String>();
    l.add(s);
    l.add(s);

    return l;
}


listOfStrings.stream().flatMap(str -> duplicate(str)).collect(Collectors.toList());

但我得到以下编译器错误

But I get the following compiler error

Test.java:25:错误:不兼容的类型:无法推断类型变量R listOfStrings.stream().flatMap(str -> duplicate(str)).collect(Collectors.toList());

Test.java:25: error: incompatible types: cannot infer type-variable(s) R listOfStrings.stream().flatMap(str -> duplicate(str)).collect(Collectors.toList());

(参数不匹配；lambda 表达式中的返回类型错误列表无法转换为流)
其中 R,T 是类型变量:R 扩展了在方法 flatMap(Function>) 中声明的对象T 扩展接口 Stream 中声明的 Object

(argument mismatch; bad return type in lambda expression List cannot be converted to Stream)
where R,T are type-variables: R extends Object declared in method flatMap(Function>) T extends Object declared in interface Stream

在 scala 中，我可以做我认为等效的事情

In scala I can do what I believe to be equivalent

scala> List(1,2,3).flatMap(duplicate(_))
res0: List[Int] = List(1, 1, 2, 2, 3, 3)

为什么在 java 中这不是 flatMap 的有效用法?

Why is this not a valid usage of flatMap in java?

推荐答案

flatMap 需要返回一个Stream，可以看出通过 flatMap 类型的参数 Function<?超级T，?扩展流.

The lambda expression in flatMap needs to return a Stream, as can be seen by the argument of flatMap which is of type Function<? super T, ? extends Stream<? extends R>>.

以下代码将编译并运行良好:

The following code will compile and run fine:

listOfStrings.stream()
             .flatMap(str -> duplicate(str).stream()) // note the .stream() here
             .collect(Collectors.toList());

因为 lambda 表达式 str ->;duplicate(str).stream() 是 Function> 类型.

because the lambda expression str -> duplicate(str).stream() is of type Function<String, Stream<String>>.

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