（可能很奇怪）长增量行为？
我问这个问题有点尴尬，但以下代码片段的结果让我难住了：
System.out.println("incrementResultResponses() has been invoked!");
final long oldValue = resultResponses;
final long newValue = resultResponses++;
System.out.println("Old value = " + oldValue);
System.out.println("New value = " + newValue);
输出如下：
incrementResultResponses() has been invoked!
Old value = 0
New value = 0
为什么？并发会对这里的结果产生任何影响吗？顺便说一句，resultResponses 是一个long。
I'm a bit embarrassed in asking this question, but the result of the following code snippet has me stumped:
System.out.println("incrementResultResponses() has been invoked!");
final long oldValue = resultResponses;
final long newValue = resultResponses++;
System.out.println("Old value = " + oldValue);
System.out.println("New value = " + newValue);
That outputs the following:
incrementResultResponses() has been invoked!
Old value = 0
New value = 0
Why? Would concurrency have any influence upon the result here? By the way, resultResponses is a long.
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后缀
++运算符返回旧值（递增之前）。您想使用前缀++：The postfix
++operator returns the old value (before incrementing). You want to use prefix++:因为增量会在分配后增加值（后增量）。这正是
resultResponses++应该做的事情。如果希望resultResponses为1，则需要使用Pre-Increment，即
++resultResponsesBecause the increment increases the value after it was assigned (Post-Increment). That's exactly what
resultResponses++is supposed to do.If you want resultResponses to be 1, you need to use Pre-Increment, which is
++resultResponses如果您想在赋值之前递增 oldValue，则必须在变量之前放置
++：这意味着递增发生在语句执行之前而不是之后。
If you want to increment oldValue before the assignment you will have to place
++before the variable:This means that the increment takes place before the statement is executed instead of after.
请参阅 此，了解如何
postfix和prefix有效。正如上面的答案中提到的，您可以使用这个：
或者为了让它更漂亮，您也可以使用：
这也会产生相同的输出。
Refer this, to know how
postfixandprefixwork.As mentioned in the above answers you can use this:
Or to make it fancier you can also use:
which will also result in the same output.