Java8流顺序和并行执行产生不同的结果？ [英] Java8 streams sequential and parallel execution produce different results?

问题描述

在Java8中运行以下流示例：

Running the following stream example in Java8:

    System.out.println(Stream
        .of("a", "b", "c", "d", "e", "f")
        .reduce("", (s1, s2) -> s1 + "/" + s2)
    );

收益率：

/a/b/c/d/e/f

这是 - 的当然 - 毫不奇怪。
由于 http：// docs .oracle.com / javase / 8 / docs / api / index.html？overview-summary.html 流是按顺序还是并行执行无关紧要：

Which is - of course - no surprise. Due to http://docs.oracle.com/javase/8/docs/api/index.html?overview-summary.html it shouldn't matter whether the stream is executed sequentially or parallel:

除了标识为显式非确定性的操作（例如findAny（））之外，流是顺序执行还是并行执行不应更改计算结果。

Except for operations identified as explicitly nondeterministic, such as findAny(), whether a stream executes sequentially or in parallel should not change the result of the computation.

AFAIK reduce（）是确定性的，（ s1，s2） - > s1 +/+ s2 是关联的，因此添加 parallel（）应该产生相同的结果：

AFAIK reduce() is deterministic and (s1, s2) -> s1 + "/" + s2 is associative, so that adding parallel() should yield the same result:

    System.out.println(Stream
            .of("a", "b", "c", "d", "e", "f")
            .parallel()
            .reduce("", (s1, s2) -> s1 + "/" + s2)
    );

然而，我机器上的结果是：

However the result on my machine is:

/a//b//c//d//e//f

这里有什么问题？

BTW：使用（首选） .collect（Collectors.joining（/） ）而不是 reduce（...）产生相同的结果 a / b / c / d / e / f 用于顺序执行和并行执行。

BTW: using (the preferred) .collect(Collectors.joining("/")) instead of reduce(...) yields the same result a/b/c/d/e/f for sequential and parallel execution.

JVM详细信息：

java.specification.version: 1.8
java.version: 1.8.0_31
java.vm.version: 25.31-b07
java.runtime.version: 1.8.0_31-b13

推荐答案

来自reduce的文档：

From reduce's documentation:

标识值必须是累加器函数的标识。这意味着对于所有t，accumulator.apply（identity，t）等于t。

The identity value must be an identity for the accumulator function. This means that for all t, accumulator.apply(identity, t) is equal to t.

在你的情况下不是这样 - 和a创建/ a。

Which is not true in your case - "" and "a" creates "/a".

我已经提取累加器函数并添加了一个打印输出以显示发生的情况：

I have extracted the accumulator function and added a printout to show what happens:

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（运行之间不同）：

This is example output (it differs between runs):

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

你可以在你的函数中添加一个if语句来分别处理空字符串：

You can add an if statement to your function to handle empty string separately:

System.out.println(Stream
        .of("a", "b", "c", "d", "e", "f")
        .parallel()
        .reduce((s1, s2) -> s1.isEmpty()? s2 : s1 + "/" + s2)
);

正如Marko Topolnik注意到的那样，检查 s2 是不需要因为累加器不必是交换函数。

As Marko Topolnik noticed, checking s2 is not required as accumulator doesn't have to be commutative function.

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