問題描述
我使用 64 位整數中的位存儲標志.
我想知道在 64 位整數中的位置是否設置了一個位(即我不關心任何特定位的位置).
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 Bit Twiddling Hacks 計算位數(肖恩·埃隆·安德森(Sean Eron Anderson))，但我想知道僅檢測是否設置了一個位的最有效方法是什么...
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...
我發現了一些其他相關的問題:
I found some other related questions:
- (8051) 檢查是否設置了單個位
- 檢測整數內的單個一位流
還有一些維基百科頁面:
and also some Wikipedia pages:
- 查找第一個
- 位操作
- 漢明權重
注意:我的應用程序是用 java 編寫的，但我對使用其他語言的優化感到好奇...
NB: my application is in java, but I am curious about optimizations using other languages...
編輯:L?u V?nh Phúc 指出我的問題中的第一個鏈接已經得到了答案:請參閱確定整數是否為 2 的冪部分em>Bit Twiddling Hacks(作者 Sean Eron Anderson).我沒有意識到一位與二的冪是一樣的.
EDIT: L?u V?nh 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.
推薦答案
如果您只是想檢查是否設置了一個位，那么您實際上是在檢查該數字是否是 2 的冪.要做到這一點，您可以做:
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) ...
這也將 0 視為 2 的冪，因此如果這很重要，您應該檢查不是 0 的數字.那么:
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) ...
這篇關于檢查整數中是否只設置了一個位(無論其位置如何)的文章就介紹到這了，希望我們推薦的答案對大家有所幫助，也希望大家多多支持html5模板網！