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In some cases it may be usefull to reverse the order of the bits. Consider 01011001 and reversed 10011010. One solution is to use a lookuptable. Andi Smithers wrote slower but smaller and more elegant solution at the cost of no table, this version uses 32 bit registers but these can be replaced so that the gem may run on a 8086:
xor eax,eax
mov ebx,data2invert ; your input data
mov cl,revbits ; number of bits to reverse
revloop:
ror ebx,1
rcl eax,1
dec cl
jnz revloopThis routine is small and general but slow. Vesa Karvonen presents a faster routine for inverting 32 bit numbers:bswap eax ; Data in EAX mov edx,eax ; Trashes EDX and eax,00F0F0F0Fh ; "AND eax,0F0F0F0Fh" < "AND edx,0F0F0F0Fh" xor edx,eax ; Smaller than "AND edx,0F0F0F0F0h"! rol eax,8 ; Rotate instead of 2 shifts! or eax,edx mov edx,eax and eax,033333333h xor edx,eax rol eax,4 or eax,edx ; Shift version could use LEAs, mov edx,eax ; but it may not be faster. and eax,055555555h xor edx,eax rol eax,2 or eax,edx ror eax,7 ; Rotate back...According to my "in memory" 486 info (it has been a while since I last used one) it should take about 21 clock cycles. It is not the fastest possible, but it also doesn't use a LUT. The rotates are NP on Pentium so you might want use shifts on Pentium instead and schedule the instructions appropriately.
return byterev[a >> 24] + (byterev[(a >> 16) & 255] << 8) +
(byterev[(a >> 8) & 255] << 16) + (byterev[a & 255] << 24);The actual bit reverse is done with the following snippet:;
; input: eax
; output: eax
; destroys: ebx, ecx, edx
xor ebx,ebx
mov bl,al
xor ecx,ecx
mov cl,ah
bswap eax
xor edx,edx
mov dl,ah
and eax,255
mov bl,byterev[ebx]
mov bh,byterev[ecx]
bswap ebx
mov al,byterev[eax]
mov ah,byterev[edx]
or eax,ebx