Finding N contiguous zero bits in an integer to the left of the MSB position of another integer

Posted by James Morris on Stack Overflow See other posts from Stack Overflow or by James Morris
Published on 2010-05-07T21:52:31Z Indexed on 2010/05/08 0:58 UTC
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The problem is: given an integer val1 find the position of the highest bit set (Most Significant Bit) then, given a second integer val2 find a contiguous region of unset bits, with the minimum number of zero bits given by width to the left of the position (ie, in the higher bits).

Here is the C code for my solution:

typedef unsigned int t;
unsigned const t_bits = sizeof(t) * CHAR_BIT;

_Bool test_fit_within_left_of_msb(  unsigned width,
                                    t val1,
                                    t val2,
                                    unsigned* offset_result)
{
    unsigned offbit = 0;
    unsigned msb = 0;
    t mask;
    t b;

    while(val1 >>= 1)
        ++msb;

    while(offbit + width < t_bits - msb)
    {
        mask = (((t)1 << width) - 1) << (t_bits - width - offbit);
        b = val2 & mask;

        if (!b)
        {
            *offset_result = offbit;
            return true;
        }

        if (offbit++) /* this conditional bothers me! */
            b <<= offbit - 1;

        while(b <<= 1)
            offbit++;
    }
    return false;
}

Aside from faster ways of finding the MSB of the first integer, the commented test for a zero offbit seems a bit extraneous, but necessary to skip the highest bit of type t if it is set.

I have also implemented similar algorithms but working to the right of the MSB of the first number, so they don't require this seemingly extra condition.

How can I get rid of this extra condition, or even, are there far more optimal solutions?

Edit: Some background not strictly required. The offset result is a count of bits from the high bit, not from the low bit as maybe expected. This will be part of a wider algorithm which scans a 2D array for a 2D area of zero bits. Here, for testing, the algorithm has been simplified. val1 represents the first integer which does not have all bits set found in a row of the 2D array. From this the 2D version would scan down which is what val2 represents.

Here's some output showing success and failure:

t_bits:32
     t_high: 10000000000000000000000000000000 ( 2147483648 )
---------

-----------------------------------
*** fit within left of msb test ***
-----------------------------------
      val1:  00000000000000000000000010000000 ( 128 )
      val2:  01000001000100000000100100001001 ( 1091569929 )
msb:   7
offbit:0 + width: 8 = 8
      mask:  11111111000000000000000000000000 ( 4278190080 )
      b:     01000001000000000000000000000000 ( 1090519040 )
offbit:8 + width: 8 = 16
      mask:  00000000111111110000000000000000 ( 16711680 )
      b:     00000000000100000000000000000000 ( 1048576 )
offbit:12 + width: 8 = 20
      mask:  00000000000011111111000000000000 ( 1044480 )
      b:     00000000000000000000000000000000 ( 0 )
offbit:12
iters:10


***** found room for width:8 at offset: 12 *****

-----------------------------------
*** fit within left of msb test ***
-----------------------------------
      val1:  00000000000000000000000001000000 ( 64 )
      val2:  00010000000000001000010001000001 ( 268469313 )
msb:   6
offbit:0 + width: 13 = 13
      mask:  11111111111110000000000000000000 ( 4294443008 )
      b:     00010000000000000000000000000000 ( 268435456 )
offbit:4 + width: 13 = 17
      mask:  00001111111111111000000000000000 ( 268402688 )
      b:     00000000000000001000000000000000 ( 32768 )
 ***** mask: 00001111111111111000000000000000 ( 268402688 )
offbit:17
iters:15


***** no room found for width:13 *****

(iters is the count of iterations of the inner while loop)

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