【发布时间】:2017-06-28 14:13:29
【问题描述】:
我正在学习 Stream,在一次这样的练习中,我试图通过以下代码加入 2 个不同类型的字符串和整数流:
Stream<String> fruitStream = Stream.of("Apple", "Mango", "Muskmalon", "Guvava");
Stream<Integer> vegetablesStream = Stream.of(1, 2, 3, 4);
Stream<String> mixStream = Stream.concat(fruitStream, vegetablesStream.map(i -> i.toString()));
mixStream.forEach(System.out :: println);
它工作正常并给了我想要的结果。
然后我尝试使用方法参考:
Stream<String> mixStream = Stream.concat(fruitStream, vegetablesStream.map(Integer :: toString));
它开始抛出错误:
Wrong 1st argument type. Found: 'java.util.stream.Stream<java.lang.String>', required: 'java.util.stream.Stream<? extends T>'
concat (java.util.stream.Stream<? extends T>, java.util.stream.Stream<? extends T>) in Stream cannot be applied to (java.util.stream.Stream<java.lang.String>, java.util.stream.Stream<R>)
reason: No compile-time declaration for the method reference is found
当我用 String :: ValueOf 替换我的方法参考时,它工作正常:
Stream<String> mixStream = Stream.concat(fruitStream, vegetablesStream.map(String :: valueOf));
我无法理解为什么 Integer :: toString 失败但 String :: valueOf 已通过?
【问题讨论】:
-
我知道这一定是重复的......
标签: java-8 stream java-stream method-reference