在位流中设置位

Question

I have encountered the following C function while working on a legacy code and I am compeletely baffled, the way the code is organized. I can see that the function is trying to set bits at given position in bit stream but I can't get my head around with individual statements and expressions. Can somebody please explain why the developer used divison by 8 (/8) and modulus 8 (%8) expressions here and there. Is there an easy way to read these kinds of bit manipulation functions in c?

static void setBits(U8 *input, U16 *bPos, U8 len, U8 val)
{
  U16 pos;
  if (bPos==0)
  {
    pos=0;
  }
  else
  {
    pos = *bPos;
    *bPos += len;
  }
  input[pos/8] = (input[pos/8]&(0xFF-((0xFF>>(pos%8))&(0xFF<<(pos%8+len>=8?0:8-(pos+len)%8)))))
                |((((0xFF>>(8-len)) & val)<<(8-len))>>(pos%8));
  if ((pos/8 == (pos+len)/8)|(!((pos+len)%8)))
    return;
  input[(pos+len)/8] = (input[(pos+len)/8]
                        &(0xFF-(0xFF<<(8-(pos+len)%8))))
                       |((0xFF>>(8-len)) & val)<<(8-(pos+len)%8);
}

Answer 1

please explain why the developer used divison by 8 (/8) and modulus 8 (%8) expressions here and there

First of all, note that the individual bits of a byte are numbered 0 to 7, where bit 0 is the least significant one. There are 8 bits in a byte, hence the "magic number" 8.

Generally speaking: if you have any raw data, it consists of n bytes and can therefore always be treated as an array of bytes uint8_t data[n] . To access bit x in that byte array, you can for example do like this:

Given x = 17 , bit x is then found in byte number 17/8 = 2 . Note that integer division "floors" the value, instead of 2.125 you get 2.

The remainder of the integer division gives you the bit position in that byte, 17%8 = 1 .

So bit number 17 is located in byte 2, bit 1. data[2] gives the byte.

To mask out a bit from a byte in C, the bitwise AND operator & is used. And in order to use that, a bit mask is needed. Such bit masks are best obtained by shifting the value 1 by the desired amount of bits. Bit masks are perhaps most clearly expressed in hex and the possible bit masks for a byte will be (1<<0) == 0x01 , (1<<1) == 0x02 , (1<<3) == 0x04 , (1<<4) == 0x08 and so on.

In this case (1<<1) == 0x02 .

C code:

uint8_t data[n];
...
size_t  byte_index = x / 8;
size_t  bit_index  = x % 8;
bool    is_bit_set;

is_bit_set = ( data[byte_index] & (1<<bit_index) ) != 0;