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Genix



Joined: 23 Jan 2006
Posts: 3
Location: Ostrava, Czech Republic
Genix
Can anyone help me with my problem? I need to left rotate 64bit value stored in register pair (ex. EDX:EAX). I am just beginner in asm so this is out of my knowledge. Thanks.
Post 23 Jan 2006, 11:04
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Tomasz Grysztar



Joined: 16 Jun 2003
Posts: 7797
Location: Kraków, Poland
Tomasz Grysztar
If you need to rotate just by 1 bit, you can do it through carry:
Code:
bt eax,31 ; CF := the highest bit of EAX
rcl edx,1
rcl eax,1    

To rotate by larger amount you'd have to use some intermediating register, perhaps with help of SHLD instruction.
Post 23 Jan 2006, 11:18
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Genix



Joined: 23 Jan 2006
Posts: 3
Location: Ostrava, Czech Republic
Genix
So if i need to rotate by N bit, can i use cycle with rotation by 1 bit or can i rotate it at once?
Post 23 Jan 2006, 13:07
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Genix



Joined: 23 Jan 2006
Posts: 3
Location: Ostrava, Czech Republic
Genix
Tomasz Grysztar wrote:
If you need to rotate just by 1 bit, you can do it through carry:
Code:
bt eax,31 ; CF := the highest bit of EAX
rcl edx,1
rcl eax,1    

To rotate by larger amount you'd have to use some intermediating register, perhaps with help of SHLD instruction.



Is it right?

rol64start:
mov ecx, 15 ; 15xrepeat


bt eax, 31
cycle_start:
rcl edx, 1

rcl eax, 1

loop cycle_start
Post 23 Jan 2006, 16:55
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Tomasz Grysztar



Joined: 16 Jun 2003
Posts: 7797
Location: Kraków, Poland
Tomasz Grysztar
No, you are duplicating one bit this way - you would have to put the BT instruction also inside the loop.

Also, assuming you've got the shift count in CL, you can it without loop using SHLD, something like:
Code:
shld ebx,edx,cl
shld edx,eax,cl
ror ebx,cl
shld eax,ebx,cl    

with EBX as a temporary storage.
Post 23 Jan 2006, 17:24
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