Factor out the hexdumper functionality from iov for all to use. Useful for
creating verbose debug printfery that dumps packet data.
Signed-off-by: Peter Crosthwaite <peter.crosthwaite@xilinx.com>
Message-id: faaac219c55ea586d3f748befaf5a2788fd271b8.1361853677.git.peter.crosthwaite@xilinx.com
Signed-off-by: Peter Maydell <peter.maydell@linaro.org>
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: Laszlo Ersek <lersek@redhat.com>
Reviewed-by: Eric Blake <eblake@redhat.com>
Signed-off-by: Paolo Bonzini <pbonzini@redhat.com>
Signed-off-by: Kevin Wolf <kwolf@redhat.com>
