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Using the "MUL/IMUL" and "DIV/IDIV" opco

 
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deama1234
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PostPosted: Thu Jan 22, 2015 12:57 pm    Post subject: Using the "MUL/IMUL" and "DIV/IDIV" opco Reply with quote
I read on some website how to do it, but it confuses me and so I thought you guys might help me...

ok so, I want this "[ebx+00000480]" to be multiplied by 2, how do I do this?
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Dark Byte
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PostPosted: Thu Jan 22, 2015 5:25 pm    Post subject: Reply with quote
this is easier: shl [ebx+00000480], 1

mul multiplies the value in eax with the value specified in eax and stores the result in edx:eax

(edx is used when the result is too big to fit in 32 bit)
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deama1234
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PostPosted: Thu Jan 22, 2015 6:55 pm    Post subject: Reply with quote
Dark Byte wrote:
this is easier: shl [ebx+00000480], 1

mul multiplies the value in eax with the value specified in eax and stores the result in edx:eax

(edx is used when the result is too big to fit in 32 bit)


Thanks; though, should I use "shl [ebx+00000480],2" if I wanted to multiply it by 3?

EDIT: No, it multiplies it by 4... guess I'll go try that mul thing again then.

EDIT AGAIN: Oh and what did you mean by "mul multiplies the value in eax with the value specified in eax"? So if I had eax = 5 then mul eax will make edx = 25?
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Dark Byte
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PostPosted: Thu Jan 22, 2015 7:14 pm    Post subject: Reply with quote
if eax was 5 amd you did mul eax
then eax would become 25 and edx 0
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panraven
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PostPosted: Thu Jan 22, 2015 7:18 pm    Post subject: Reply with quote
LEA (load effective address) can multiply certain constant too.

Code:
lea  eax,[eax*2] ; x2
lea  eax,[eax+eax*2] ; x3
lea  eax,[ax*4] ; x4
lea  eax,[eax+eax*4] ; x5
lea  eax,[eax*8] ; x8
lea  eax,[eax+eax*8] ; x9


no flags changed, nor overflow error.

combining shl or other multiplication with some more register, other constant multiple is possible without using imul.
Code:
lea ebx,[eax*2]
lea eax,[eax+eax*8]
sub eax,ebx ; x7
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deama1234
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PostPosted: Thu Jan 22, 2015 7:28 pm    Post subject: Reply with quote
Dark Byte wrote:
if eax was 5 amd you did mul eax
then eax would become 25 and edx 0

Ah, then it would be easier to use "shl" XD.


panraven wrote:
LEA (load effective address) can multiply certain constant too.

Code:
lea  eax,[eax*2] ; x2
lea  eax,[eax+eax*2] ; x3
lea  eax,[ax*4] ; x4
lea  eax,[eax+eax*4] ; x5
lea  eax,[eax*8] ; x8
lea  eax,[eax+eax*8] ; x9


no flags changed, nor overflow error.

Oh, nice. Now, how would I divide? Is it the same way except I'd use a "/"?
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panraven
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PostPosted: Thu Jan 22, 2015 7:44 pm    Post subject: Reply with quote
For divide by 2,4,8,...
It is simpler to use shr or sar (shift)

Code:
shr eax,4 ; /16 = / 2^4


Other may use use IDIV (signed div) or DIV (unsigned div)

If dividend (EAX) is 32bit (ie, higher 32bit of dividend EDX = 0x0), then to divided by ECX

Code:
xor edx,edx
idiv ecx  ; eax implied     

(sorry, typo 'imul' -> 'idiv')

quotient is in EAX, remainder in edx

btw, there is way to divide by using multiplication if remainder can be ignore, can do approximate division on something like floating number.

By using MUL(unsigned), theoretically multiple eax (edx=0) with 0x100000000 is shifting edx:eax to eax:0.

If we multiple something smaller than 0x100000000, say 0x49249249
It will be approximate equivalent to multiply 0x49249249 / 0x100000000
or divided by 0x100000000 / 0x49249249 = 3.5

After such multiplication, the division result will be in edx.

Code:

mov eax,0x2160ec0  ; 35000000 in decimal
xor edx,edx
mov ecx,0x49249249
mul ecx ; now edx = 0x98967f = 9999999 ~= 35000000 / 3.5 = 10000000


Last edited by panraven on Thu Jan 22, 2015 8:13 pm; edited 2 times in total
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deama1234
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PostPosted: Thu Jan 22, 2015 7:58 pm    Post subject: Reply with quote
panraven wrote:
You can use IDIV (signed div) or DIV (unsigned div)

If dividend (EAX) is 32bit (ie, higher 32bit of dividend EDX = 0x0), then to divided by ECX

Code:
xor edx,edx
imul ecx  ; eax implied


quotient is in EAX, remainder in edx

Well, I got this:
Code:
push eax
push edx
push ecx

mov eax,02
mov edx,06
mov ecx,03

xor edx,edx
imul ecx

add [ebx+00000480],eax

pop ecx
pop edx
pop eax


but all it does is ecx*eax = eax

panraven wrote:
For divide by 2,4,8,...
It is simpler to use shr or sar (shift)

Code:
shr eax,4 ; /16 = / 2^4


Other may use use IDIV (signed div) or DIV (unsigned div)

If dividend (EAX) is 32bit (ie, higher 32bit of dividend EDX = 0x0), then to divided by ECX

Code:
xor edx,edx
idiv ecx  ; eax implied     

(sorry, typo 'imul' -> 'idiv')

quotient is in EAX, remainder in edx

btw, there is way to divide by using multiplication if remainder can be ignore, can do approximate division on something like floating number.

By using IMUL, theoretically multiple eax (edx=0) with 0x100000000 is shifting edx:eax to eax:0.

If we multiple something smaller than 0x100000000, say 0x49249249
It will be approximate equivalent to multiply 0x49249249 / 0x100000000
or divided by 0x100000000 / 0x49249249 = 3.5

After such multiplication, the division result will be in edx.

Code:

mov eax,0x2160ec0  ; 35000000 in decimal
xor edx,edx
mov ecx,0x49249249
mul ecx ; now edx = 0x98967f = 9999999 ~= 35000000 / 3.5 = 10000000


Oh, now it works... thanks! Oh, and it works with multiplication fine too!
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