给定一个uint8_t数组，提取位的任何子序列作为uint32_t的好方法是什么？

Question

I have run into an interesting problem lately:

Lets say I have an array of bytes (uint8_t to be exact) of length at least one. Now i need a function that will get a subsequence of bits from this array, starting with bit X (zero based index, inclusive) and having length L and will return this as an uint32_t. If L is smaller than 32 the remaining high bits should be zero.

Although this is not very hard to solve, my current thoughts on how to do this seem a bit cumbersome to me. I'm thinking of a table of all the possible masks for a given byte (start with bit 0-7, take 1-8 bits) and then construct the number one byte at a time using this table.

Can somebody come up with a nicer solution? Note that i cannot use Boost or STL for this - and no, it is not a homework, its a problem i run into at work and we do not use Boost or STL in the code where this thing goes. You can assume that: 0 < L <= 32 and that the byte array is large enough to hold the subsequence.

One example of correct input/output:

array: 00110011 1010 1010 11110011 01 101100
subsequence: X = 12 (zero based index), L = 14
resulting uint32_t = 00000000 00000000 00 101011 11001101

Answer 1

Only the first and last bytes in the subsequence will involve some bit slicing to get the required bits out, while the intermediate bytes can be shifted in whole into the result. Here's some sample code, absolutely untested -- it does what I described, but some of the bit indices could be off by one:

uint8_t bytes[];
int X, L;

uint32_t result;

int startByte  = X / 8,  /* starting byte number */
    startBit   = 7 - X % 8,  /* bit index within starting byte, from LSB */
    endByte    = (X + L) / 8, /* ending byte number */
    endBit     = 7 - (X + L) % 8; /* bit index within ending byte, from LSB */

/* Special case where start and end are within same byte:
   just get bits from startBit to endBit */
if (startByte == endByte) {
  uint8_t byte = bytes[startByte];
  result = (byte >> endBit) & ((1 << (startBit - endBit)) - 1);
}
/* All other cases: get ending bits of starting byte,
                    all other bytes in between,
                    starting bits of ending byte */
else {
  uint8_t byte = bytes[startByte];
  result = byte & ((1 << startBit) - 1);

  for (int i = startByte + 1; i < endByte; i++)
    result = (result << 8) | bytes[i];

  byte = bytes[endByte];
  result = (result << (8 - endBit)) | (byte >> endBit);
}

Answer 2

看一下std :: bitset和boost :: dynamic_bitset。

Answer 3

I would be thinking something like loading a uint64_t with a cast and then shifting left and right to lose the uninteresting bits.

uint32_t extract_bits(uint8_t* bytes, int start, int count)
{
    int shiftleft =  32+start;
    int shiftright = 64-count;
    uint64_t *ptr = (uint64_t*)(bytes);
    uint64_t hold = *ptr;
    hold <<= shiftleft;
    hold >>= shiftright;
    return (uint32_t)hold;
}

Answer 4

For the sake of completness, i'am adding my solution inspired by the comments and answers here. Thanks to all who bothered to think about the problem.

static const uint8_t firstByteMasks[8] = { 0xFF, 0x7F, 0x3F, 0x1F, 0x0F, 0x07, 0x03, 0x01 };

uint32_t getBits( const uint8_t *buf, const uint32_t bitoff, const uint32_t len, const uint32_t bitcount )
{
    uint64_t result = 0;

    int32_t startByte = bitoff / 8; // starting byte number
    int32_t endByte = ((bitoff + bitcount) - 1) / 8; // ending byte number
    int32_t rightShift = 16 - ((bitoff + bitcount) % 8 );

    if ( endByte >= len ) return -1;

    if ( rightShift == 16 ) rightShift = 8; 

    result = buf[startByte] & firstByteMasks[bitoff % 8];
    result = result << 8;

    for ( int32_t i = startByte + 1; i <= endByte; i++ )
    {
        result |= buf[i];
        result = result << 8;
    }
    result = result >> rightShift;
    return (uint32_t)result;
}

Few notes: i tested the code and it seems to work just fine, however, there may be bugs. If i find any, i will update the code here. Also, there are probably better solutions!