如何在c中进行位集/字节数组转换 [英] how to make a bit-set/byte-array conversion in c

问题描述

给出一个数组，
无符号字符q [32] = 1100111 ... ，

我如何生成一个4字节的位集 unsigned char p [4] ，使得该位集的位等于内部值数组，例如，第一字节p [0] ＝ q [0] ... q [7]；第二个字节p [1] = q [8] ... q [15]，等等。

how can I generate a 4-bytes bit-set, unsigned char p[4], such that, the bit of this bit-set, equals to value inside the array, e.g., the first byte p[0]= "q[0] ... q[7]"; 2nd byte p[1]="q[8] ... q[15]", etc.

以及相反的操作方法，即给定位设置，生成数组？

and also how to do it in opposite, i.e., given bit-set, generate the array?

我自己的第一部分试用版。

my own trial out for the first part.

unsigned char p[4]={0};
for (int j=0; j<N; j++) 
{
    if (q[j] == '1')
    {
        p [j / 8] |= 1 << (7-(j % 8)); 
    }            
}

以上是正确的吗？有什么条件要检查吗？有什么更好的办法吗？

Is the above right? any conditions to check? Is there any better way?

编辑-1

我想知道上述方法是否有效？由于数组的大小可能高达4096甚至更大。

I wonder if above is efficient way? As the array size could be upto 4096 or even more.

推荐答案

我认为这不太可行。您正在将每个位与 1 进行比较，而实际上它应该是‘1’。您还可以通过除去 if 来使其效率更高：

I don't think that will quite work. You are comparing each "bit" to 1 when it should really be '1'. You can also make it a bit more efficient by getting rid of the if:

unsigned char p[4]={0};
for (int j=0; j<32; j++) 
{
    p [j / 8] |= (q[j] == `1`) << (7-(j % 8));           
}

反向操作也很简单。

unsigned char q[32]={0};
for (int j=0; j<32; j++) {
  q[j] = p[j / 8] & ( 1 << (7-(j % 8)) ) + '0';
}

您会注意到（布尔值）+'0'在1/0和'1'/'0'之间转换。

You'll notice the creative use of (boolean) + '0' to convert between 1/0 and '1'/'0'.

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