Bitwise operators are useful for looping arrays which length is power of 2. As many people mentioned, bitwise operators are extremely useful and are used in Flags, Graphics, Networking, Encryption. Not only that, but they are extremely fast. My personal favorite use is to loop an array without conditionals. Suppose you have a zero-index based array(e.g. first element's index is 0) and you need to loop it indefinitely. By indefinitely I mean going from first element to last and returning to first. One way to implement this is:

int[] arr = new int[8];
int i = 0;
while (true) {
    print(arr[i]);
    i = i + 1;
    if (i >= arr.length) 
        i = 0;
}

这是最简单的方法，如果你想避免if语句，你可以像这样使用模方法:

int[] arr = new int[8];
int i = 0;
while (true) {
    print(arr[i]);
    i = i + 1;
    i = i % arr.length;
}

这两种方法的缺点是，模运算符是昂贵的，因为它在整数除法后寻找余数。第一个方法在每次迭代中运行if语句。然而，如果你的数组长度是2的幂，你可以很容易地生成一个像0 ..长度- 1，使用&(位和)操作符，如I & Length。知道了这些，上面的代码就变成了

int[] arr = new int[8];
int i = 0;
while (true){
    print(arr[i]);
    i = i + 1;
    i = i & (arr.length - 1);
}

下面是它的工作原理。在二进制格式中，所有2的幂减去1的数都只用1表示。例如，二进制的3是11,7是111,15是1111，等等，你懂的。现在，如果你用任意一个数对一个只由1组成的二进制数，会发生什么?假设我们这样做:

num & 7;

如果num小于或等于7，那么结果将是num，因为每个加1的&-ed就是它自己。如果num大于7，在&操作期间，计算机将考虑7的前导零，当然，在&操作后，这些前导零将保持为零，只有后面的部分将保留。比如二进制的9和7

1001 & 0111

结果将是0001，它是十进制的1，并定位数组中的第二个元素。

Bit fields (flags) They're the most efficient way of representing something whose state is defined by several "yes or no" properties. ACLs are a good example; if you have let's say 4 discrete permissions (read, write, execute, change policy), it's better to store this in 1 byte rather than waste 4. These can be mapped to enumeration types in many languages for added convenience. Communication over ports/sockets Always involves checksums, parity, stop bits, flow control algorithms, and so on, which usually depend on the logic values of individual bytes as opposed to numeric values, since the medium may only be capable of transmitting one bit at a time. Compression, Encryption Both of these are heavily dependent on bitwise algorithms. Look at the deflate algorithm for an example - everything is in bits, not bytes. Finite State Machines I'm speaking primarily of the kind embedded in some piece of hardware, although they can be found in software too. These are combinatorial in nature - they might literally be getting "compiled" down to a bunch of logic gates, so they have to be expressed as AND, OR, NOT, etc. Graphics There's hardly enough space here to get into every area where these operators are used in graphics programming. XOR (or ^) is particularly interesting here because applying the same input a second time will undo the first. Older GUIs used to rely on this for selection highlighting and other overlays, in order to eliminate the need for costly redraws. They're still useful in slow graphics protocols (i.e. remote desktop).

这些只是我最先想到的几个例子——这不是一个详尽的清单。

按位&用于屏蔽/提取字节的某一部分。

1字节变量

 01110010
&00001111 Bitmask of 0x0F to find out the lower nibble
 --------
 00000010

特别是移位运算符(<< >>)经常用于计算。

我一直假设按位操作是相当简单的操作，所以当运行时间至关重要时，通过bitset实现的解决方案可以通过恒定的数量提高运行时间，这取决于算法。

一个常见的用法是对齐，例如我需要我的数据在4字节或16字节的边界上对齐。这在RISC处理器中非常常见，其中未对齐的加载/存储要么代价高昂(因为它触发了一个异常处理程序，然后需要修复未对齐的加载)，要么根本不允许。

对于任何以2为幂的对齐，下一个对齐的pos可以计算如下:

aligned_offset = alignment + ((current_offset - 1) & ~(alignment - 1))

所以在4字节对齐和当前偏移量为9的情况下:

aligned_offset = 4 + ((9-1) & ~(4-1)) = 4 + (8 & 0xFFFFFFFC) = 4+ 8  = 12  

所以下一个4字节的对齐偏移量是12

如果你想计算你的数字mod(%) 2的某次方，你可以使用yourNumber & 2^N-1，在这种情况下，它与yourNumber % 2^N相同。

number % 16 = number & 15;
number % 128 = number & 127;

这可能只是作为模数运算的一种替代品有用，它的红利很大，是2^N。但即便如此，在我在。net 2.0上的测试中，它相对于模运算的速度提升也可以忽略不计。我怀疑现代编译器已经执行了这样的优化。有人知道更多吗?