blocks|key|2924023|text|来自reduce的文档：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2924024|2924025|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。|blockquote|2924026|2924027|2924028|这在你的例子中不是真的-+"“和"a”创建了"/a“。|2924029|我已经提取了累加器函数，并添加了一个打印输出来显示发生了什么：|2924030|BinaryOperator<String>+accumulator+=+(s1,+s2)+->+{
++++System.out.println("joining+\""+%2B+s1+%2B+"\"+and+\""+%2B+s2+%2B+"\"");
++++return+s1+%2B+"/"+%2B+s2;
};
System.out.println(Stream
++++++++++++++++.of("a",+"b",+"c",+"d",+"e",+"f")
++++++++++++++++.parallel()
++++++++++++++++.reduce("",+accumulator)
);|code-block|syntax|javascript|2924031|这是示例输出(每次运行都有所不同)：|2924032|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|2924033|您可以将if语句添加到函数中，以单独处理空字符串：|2924034|System.out.println(Stream
++++++++.of("a",+"b",+"c",+"d",+"e",+"f")
++++++++.parallel()
++++++++.reduce((s1,+s2)+->+s1.isEmpty()?+s2+:+s1+%2B+"/"+%2B+s2)
);|2924035|正如Marko+Topolnik注意到的，检查s2不是必需的，因为累加器不一定是交换函数。|offset|length|style|CODE|2924036|entityMap^0|0|0|0|0|0|0|0|0|0|0|0|0|N|2|0^^$0|@$1|2|3|4|5|6|7|16|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|17|8|@]|9|@]|A|$]]|$1|C|3|D|5|E|7|18|8|@]|9|@]|A|$]]|$1|F|3|-4|5|6|7|19|8|@]|9|@]|A|$]]|$1|G|3|-4|5|6|7|1A|8|@]|9|@]|A|$]]|$1|H|3|I|5|6|7|1B|8|@]|9|@]|A|$]]|$1|J|3|K|5|6|7|1C|8|@]|9|@]|A|$]]|$1|L|3|M|5|N|7|1D|8|@]|9|@]|A|$O|P]]|$1|Q|3|R|5|6|7|1E|8|@]|9|@]|A|$]]|$1|S|3|T|5|N|7|1F|8|@]|9|@]|A|$O|P]]|$1|U|3|V|5|6|7|1G|8|@]|9|@]|A|$]]|$1|W|3|X|5|N|7|1H|8|@]|9|@]|A|$O|P]]|$1|Y|3|Z|5|6|7|1I|8|@$10|1J|11|1K|12|13]]|9|@]|A|$]]|$1|14|3|-4|5|6|7|1L|8|@]|9|@]|A|$]]]|15|$]]

From reduce's documentation:

<blockquote>
 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. 
</blockquote>

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

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

<pre><code>BinaryOperator&lt;String&gt; accumulator = (s1, s2) -&gt; {
 System.out.println("joining \"" + s1 + "\" and \"" + s2 + "\"");
 return s1 + "/" + s2;
};
System.out.println(Stream
 .of("a", "b", "c", "d", "e", "f")
 .parallel()
 .reduce("", accumulator)
);
</code></pre>

This is example output (it differs between runs):

<pre><code>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
</code></pre>

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

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

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

blocks|key|138214|text|为了补充其他答案，|type|unstyled|depth|inlineStyleRanges|entityRanges|data|138215|您可能想要使用可变还原，文档中指定这样做|offset|length|style|BOLD|138216|String+concatenated+=+strings.reduce("",+String::concat)|code-block|syntax|javascript|138217|会给出不好的性能结果。|138218|138219|我们会得到想要的结果，甚至可以并行工作。然而，我们可能对性能不满意！这样的实现将执行大量的字符串复制，并且运行时的字符数将为O(n%5E2)。一种性能更好的方法是将结果累积到一个StringBuilder，中，这是一个用于累积字符串的可变容器。我们可以使用与普通归约相同的技术来并行化可变归约。|blockquote|138220|138221|138222|因此，您应该使用StringBuilder。|138223|entityMap^0|0|7|4|0|0|0|0|2F|D|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|Z|8|@$D|10|E|11|F|G]]|9|@]|A|$]]|$1|H|3|I|5|J|7|12|8|@]|9|@]|A|$K|L]]|$1|M|3|N|5|6|7|13|8|@]|9|@]|A|$]]|$1|O|3|-4|5|6|7|14|8|@]|9|@]|A|$]]|$1|P|3|Q|5|R|7|15|8|@$D|16|E|17|F|G]]|9|@]|A|$]]|$1|S|3|-4|5|6|7|18|8|@]|9|@]|A|$]]|$1|T|3|-4|5|6|7|19|8|@]|9|@]|A|$]]|$1|U|3|V|5|6|7|1A|8|@]|9|@]|A|$]]|$1|W|3|-4|5|6|7|1B|8|@]|9|@]|A|$]]]|X|$]]

To add to other answer,

You might want to use Mutable reduction, the doc specify that doing something like 

<pre><code>String concatenated = strings.reduce("", String::concat)
</code></pre>

Will give bad performance result.

<blockquote>
 We would get the desired result, and it would even work in parallel.
 However, we might not be happy about the performance! Such an
 implementation would do a great deal of string copying, and the run
 time would be O(n^2) in the number of characters. A more performant
 approach would be to accumulate the results into a StringBuilder,
 which is a mutable container for accumulating strings. We can use the
 same technique to parallelize mutable reduction as we do with ordinary
 reduction.
</blockquote>

So you should use a StringBuilder instead.

blocks|key|138293|text|对于一个刚开始使用lambdas和stream的人来说，花了相当长的时间才达到"AHA“的时刻，直到我真正理解这里发生了什么。对于像我这样的流媒体新手来说，为了让它更简单(至少我希望它得到真正的回答)，我将对此稍作修改。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|138294|这一切都是在reduce文档中声明的：|138295|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。|offset|length|style|BOLD|138296|我们可以很容易地证明代码的方式是，关联性被打破：|138297|static+private+void+isAssociative()+{
+++++BinaryOperator<String>+operator+=+(s1,+s2)+->+s1+%2B+"/"+%2B+s2;
+++++String+result+=+operator.apply("",+"a");
+++++System.out.println(result);+
+++++System.out.println(result.equals("a"));+
}|code-block|syntax|javascript|138298|一个空字符串与另一个字符串连接，应该真正产生第二个字符串；这不会发生，因此累加器(BinaryOperator)是不相关的，因此reduce方法不能保证在并行调用的情况下得到相同的结果。|138299|entityMap^0|0|0|0|1M|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|V|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|W|8|@$F|X|G|Y|H|I]]|9|@]|A|$]]|$1|J|3|K|5|6|7|Z|8|@]|9|@]|A|$]]|$1|L|3|M|5|N|7|10|8|@]|9|@]|A|$O|P]]|$1|Q|3|R|5|6|7|11|8|@]|9|@]|A|$]]|$1|S|3|-4|5|6|7|12|8|@]|9|@]|A|$]]]|T|$]]

For someone who just started with lambdas and streams, it took quite some time to get to the "AHA" moment, until I really understood what is going on here. I'll rephrase this a bit to make a bit easier (at least how I wish it was really answered) for a stream newbie like me.

It's all under the reduce documentation that states:

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.

We can easily prove that the way code is, the associativity is broken:

<pre><code>static private void isAssociative() {
 BinaryOperator&lt;String&gt; operator = (s1, s2) -&gt; s1 + "/" + s2;
 String result = operator.apply("", "a");
 System.out.println(result); 
 System.out.println(result.equals("a")); 
}
</code></pre>

An empty String concatenated with another String, should really produce the second String; which does not happen, thus accumulator (BinaryOperator) is NOT associative and thus the reduce method can not guarantee the same result in case of parallel invocation.

Running the following stream example in Java8:

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

yields:

<pre><code>/a/b/c/d/e/f
</code></pre>

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

<blockquote>
 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. 
</blockquote>

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

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

However the result on my machine is:

<pre><code>/a//b//c//d//e//f
</code></pre>

What's wrong here?

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

JVM details:

<pre><code>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
</code></pre>

Java8 streams sequential and parallel execution produce different results?

容器

在Java8中运行以下流示例：    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当然，这并不令...

问Java8 streams顺序和并行执行会产生不同的结果吗？
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