检查是否只有一个位整数中设置（无论其位置） [英] Check if only one single bit is set within an integer (whatever its position)
问题描述
我店使用的是64位整数位之内的标志。结果
我想知道，如果有一个位设置到任何64位整数中的位置（E.I.我不关心任何特定的位的位置）。
I store flags using bits within a 64-bits integer.
I want to know if there is a single bit set whatever the position within the 64-bits integer (e.i. I do not care about the position of any specific bit).
boolean isOneSingleBitSet (long integer64)
{
return ....;
}
我可以使用数位数量的的位操作黑客的（肖恩·安德森玉龙），但我想知道什么是只检测一个单个位是否设置最有效的方式...
I could count number of bits using the Bit Twiddling Hacks (by Sean Eron Anderson), but I am wondering what is the most efficient way to just detect whether one single bit is set...
我发现了其他一些相关的问题：
I found some other related questions:
- (8051) Check if a single bit is set
- Detecting single one-bit streams within an integer
，也有一些维基百科页面：
and also some Wikipedia pages:
- Find first one
- Bit manipulation
- Hamming weight
注：我的应用程序是在Java中，但我使用其他语言很好奇优化...
NB: my application is in java, but I am curious about optimizations using other languages...
修改：琉永福指出我的问题在我的第一个环节已经得到了答案：参见确定是否一个整数是2 中的
EDIT: Lưu Vĩnh Phúc pointed out that my first link within my question already got the answer: see section Determining if an integer is a power of 2 in the Bit Twiddling Hacks (by Sean Eron Anderson). I did not realized that one single bit was the same as power of two.
推荐答案
如果你只是从字面上要检查单个位设置，那么你基本上是检查，如果数量为2的幂。要做到这一点，你可以这样做：
If you just literally want to check if one single bit is set, then you are essentially checking if the number is a power of 2. To do this you can do:
if ((number & (number-1)) == 0) ...
这也将算为0 2的幂，所以你应该检查数量不为0，如果这是非常重要的。于是：
This will also count 0 as a power of 2, so you should check for the number not being 0 if that is important. So then:
if (number != 0 && (number & (number-1)) == 0) ...
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