从reduce的文档中:
身份值必须是累加器功能的身份。这意味着对于所有t,accumulator.apply(identity,t)等于t。
在您的情况下,这是不正确的-“”和“ a”创建了“ / a”。
我已经提取了累加器功能并添加了打印输出以显示发生了什么:
BinaryOperator<String> accumulator = (s1, s2) -> { System.out.println("joining "" + s1 + "" and "" + s2 + """); return s1 + "/" + s2;};System.out.println(Stream .of("a", "b", "c", "d", "e", "f") .parallel() .reduce("", accumulator));
这是示例输出(运行之间有所不同):
joining "" and "d"joining "" and "f"joining "" and "b"joining "" and "a"joining "" and "c"joining "" and "e"joining "/b" and "/c"joining "/e" and "/f"joining "/a" and "/b//c"joining "/d" and "/e//f"joining "/a//b//c" and "/d//e//f"/a//b//c//d//e//f
您可以在函数中添加if语句以分别处理空字符串:
System.out.println(Stream .of("a", "b", "c", "d", "e", "f") .parallel() .reduce((s1, s2) -> s1.isEmpty()? s2 : s1 + "/" + s2));
正如Marko Topolnik所注意到的,
s2由于累加器不必是可交换的函数,因此不需要进行检查。
欢迎分享,转载请注明来源:内存溢出
评论列表(0条)