算法生成位掩码 [英] Algorithm to generate bit mask

问题描述

我面临基于所述输入参数的位掩码的这种独特的问题。例如，

I was facing this unique problem of generating a bit-mask based on the input parameter. For example,

如果参数= 2，则遮罩将是0x3中（11B） 如果param = 5，则遮罩将是为0x1F（1 1111B）

if param = 2, then the mask will be 0x3 (11b) if param = 5, then the mask will be 0x1F (1 1111b)

这个我实现了用一个for循环中的C，像

This I implemented using a for-loop in C, something like

int nMask = 0;
for (int i = 0; i < param; i ++) {

    nMask |= (1 << i);
}

我想知道是否有更好的算法~~~

I would like to know if there is a better algorithm ~~~

推荐答案

有一点要注意的一样，位掩码是，他们总是比2的幂小。

One thing to notice about bitmasks like that is that they are always one less than a power of two.

这位前pression1＆LT; n种，是最简单的方式来获得2的n次方

The expression "1 << n" is the easy way to get the n-th power of two.

您不想零提供00000001的位掩码，你想让它提供零。所以，你需要减一。

You don't want Zero to provide a bitmask of "00000001", you want it to provide zero. So you need to subtract one.

mask = (1 << param) - 1;

编辑：的

如果你想为param>的32个特例：

If you want a special case for param > 32:

int sizeInBits = sizeof(mask) * BITS_PER_BYTE; // BITS_PER_BYTE = 8;
mask = (param >= sizeInBits ? -1 : (1 <<  param) - 1);

此方法应该为16，32，或64位整数，但您可能必须明确地键入1。

This method should work for 16, 32, or 64 bit integers, but you may have to explicitly type the '1'.

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