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  • Paolo Bonzini's avatar
    add hierarchical bitmap data type and test cases · e7c033c3
    Paolo Bonzini authored
    
    
    HBitmaps provides an array of bits.  The bits are stored as usual in an
    array of unsigned longs, but HBitmap is also optimized to provide fast
    iteration over set bits; going from one bit to the next is O(logB n)
    worst case, with B = sizeof(long) * CHAR_BIT: the result is low enough
    that the number of levels is in fact fixed.
    
    In order to do this, it stacks multiple bitmaps with progressively coarser
    granularity; in all levels except the last, bit N is set iff the N-th
    unsigned long is nonzero in the immediately next level.  When iteration
    completes on the last level it can examine the 2nd-last level to quickly
    skip entire words, and even do so recursively to skip blocks of 64 words or
    powers thereof (32 on 32-bit machines).
    
    Given an index in the bitmap, it can be split in group of bits like
    this (for the 64-bit case):
    
         bits 0-57 => word in the last bitmap     | bits 58-63 => bit in the word
         bits 0-51 => word in the 2nd-last bitmap | bits 52-57 => bit in the word
         bits 0-45 => word in the 3rd-last bitmap | bits 46-51 => bit in the word
    
    So it is easy to move up simply by shifting the index right by
    log2(BITS_PER_LONG) bits.  To move down, you shift the index left
    similarly, and add the word index within the group.  Iteration uses
    ffs (find first set bit) to find the next word to examine; this
    operation can be done in constant time in most current architectures.
    
    Setting or clearing a range of m bits on all levels, the work to perform
    is O(m + m/W + m/W^2 + ...), which is O(m) like on a regular bitmap.
    
    When iterating on a bitmap, each bit (on any level) is only visited
    once.  Hence, The total cost of visiting a bitmap with m bits in it is
    the number of bits that are set in all bitmaps.  Unless the bitmap is
    extremely sparse, this is also O(m + m/W + m/W^2 + ...), so the amortized
    cost of advancing from one bit to the next is usually constant.
    
    Reviewed-by: default avatarLaszlo Ersek <lersek@redhat.com>
    Reviewed-by: default avatarEric Blake <eblake@redhat.com>
    Signed-off-by: default avatarPaolo Bonzini <pbonzini@redhat.com>
    Signed-off-by: default avatarKevin Wolf <kwolf@redhat.com>
    e7c033c3