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> Main > Clearing leftmost set bit? |
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revolution 14 Jun 2010, 14:45
By "leftmost" do you mean most-significant?
Don't forget about BSR and BSF opcodes. |
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14 Jun 2010, 14:45 |
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bitshifter 14 Jun 2010, 14:48
like input = 11111111b, ouput = 01111111b
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14 Jun 2010, 14:48 |
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ass0 14 Jun 2010, 14:51
Code: and al, 0x7f;for 8 bits and ax, 0x7fff ; for 16-bits xor ecx, ecx dec ecx shr ecx, 1 and eax, ecx ;for 32-bits and eax, 0x7fffffff; also 32-bits but not sure if legal too easy hahaha many times edited =p _________________ 
Nombre: Aquiles Castro. Location2: about:robots Last edited by ass0 on 14 Jun 2010, 15:06; edited 5 times in total |
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14 Jun 2010, 14:51 |
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bitshifter 14 Jun 2010, 14:52
Could you provide BSR example please
I am new to this instruction... (For 32bit unsigned integer) |
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14 Jun 2010, 14:52 |
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ass0 14 Jun 2010, 14:57
I though you were an intel manual's reader bitshifter
_________________
Nombre: Aquiles Castro. Location2: about:robots |
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14 Jun 2010, 14:57 |
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Tyler 14 Jun 2010, 15:02
You don't even need Intel's manuals. Fasm's has a good example. http://flatassembler.net/docs.php?article=manual#2.1.5
"btr eax,31" clears the 31st bit and sets the flags depending if the bit was already clear or not. It doesn't work on bytes though. |
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14 Jun 2010, 15:02 |
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bitshifter 14 Jun 2010, 15:03
I am, but never had a reason to use those instructions...
*just made print_binary routine to try them out* |
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14 Jun 2010, 15:03 |
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revolution 14 Jun 2010, 15:27
Code: movzx eax,al bsr ecx,eax jz .eax_is_zero btr eax,ecx |
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14 Jun 2010, 15:27 |
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bitshifter 14 Jun 2010, 15:32
Kick ass!
Thanx revo! So for 32 bit unsigned integer i can just omit movzx? This is what i came up with... (It uses some BIOS functions and some 32 bit registers) Code: org 0x0100 use16 mov ax,0x0003 int 0x10 mov eax,7 call print_binary call print_newline call clear_leftmost_bit call print_binary xor ah,ah int 0x16 ret print_binary: ; EAX = input pusha mov ebx,eax mov ecx,32 @@: xor al,al shl ebx,1 adc al,'0' call print_character sub ecx,1 jnz @b mov al,'b' call print_character popa ret print_newline: push ax mov al,13 call print_character mov al,10 call print_character pop ax ret print_character: ; AL = character pusha mov ah,0x0e mov bh,0x00 int 0x10 popa ret clear_leftmost_bit: ; Input: EAX = 32bit unsigned integer push ecx bsr ecx,eax jz .is_zero btr eax,ecx .is_zero: pop ecx ; Output: EAX = 32bit unsigned integer with leftmost bit cleared ret Last edited by bitshifter on 14 Jun 2010, 15:48; edited 1 time in total |
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14 Jun 2010, 15:32 |
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revolution 14 Jun 2010, 15:41
bitshifter wrote: So for 32 bit unsigned integer i can just omit movzx? |
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14 Jun 2010, 15:41 |
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Tyler 24 Aug 2010, 05:01
Hmm... should I revive it? Yep.
Bored, nothing to do. I was looking for an old post, but I found this instead. Made me think of this: Code: scl ax,1 shr ax,1 It clears the left most bit, and leaves it's set-ness in the carry flag. Ironic how you could've done this with bitshifting  . |
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24 Aug 2010, 05:01 |
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Madis731 24 Aug 2010, 09:43
http://www.jjj.de/fxt/
1.6: Operations on high bits or blocks of a word 1.6.1 Isolating the highest one and fi nding its index will suggest the same BSR instruction. @Tyler: scl + shr combo will yield the same result as shl + shr and they both clear *the leftmost bit* NOT the *leftmost SET bit* bitshifter needs to have the correct instruction sequence to convert 00001111b to 00000111b. This is what BSR instruction is for. |
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24 Aug 2010, 09:43 |
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LocoDelAssembly 24 Aug 2010, 13:09
What is scl?
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24 Aug 2010, 13:09 |
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Tyler 24 Aug 2010, 15:21
> scl + shr combo will yield the same result as shl + shr and they both clear *the leftmost bit* NOT the *leftmost SET bit*
I just now got what he meant. I was thinking: clear leftmost bit, and tell whether it was set or not. 
> What is scl? I meant rcl, I think. |
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24 Aug 2010, 15:21 |
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LocoDelAssembly 24 Aug 2010, 15:38
I suspected so
But note that what RCL does is fill the vacant bit with CF. ROL, RCL and SHL always set CF to the value of the leftmost bit. In your code, once scl is replaced with either SHL or ROL, SHR will always ensure that CF is zero and if RCL was used first then CF will preserve the value prior to executing both instructions. |
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24 Aug 2010, 15:38 |
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edfed 24 Aug 2010, 15:54
Code: ;eax=number mov cl,0 @@: inc cl add eax,eax ;shl eax,1 jc @f jne @b ret @@: shr eax,cl ret something like that. |
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24 Aug 2010, 15:54 |
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Tyler 24 Aug 2010, 18:19
edfed wrote:
bitshifter wrote: Also, no looping can be performed... |
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24 Aug 2010, 18:19 |
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edfed 24 Aug 2010, 18:45
if i read the title, i see:
clear the leftmost set bit (not the most significant bit), without a loop, i don't see how it can be done... i see that for a dword = 5, the leftmost set bit is bit 2. and is 101b result will give 1. for 01001101b, result will be 1101b, and for 111101011b, it will output 11101011b. to clear the LSB, it is very easy. shl eax,1 shr eax,1 and eax,80000000h btc eax,31 |
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24 Aug 2010, 18:45 |
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bitRAKE 25 Aug 2010, 00:42
I play along...
Single register method: Code: stc @@: rcl rax,1 jnc @B clc @@: rcr rax,1 jnbe @B |
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25 Aug 2010, 00:42 |
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