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Index > Windows > I/O in assmbly

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ishkabible



Joined: 13 Sep 2010
Posts: 54
ishkabible
how do i get the port of file path? i see that I/O with out any clib functions requires the in and out instructions witch reqiure a port. how do i obtain this port?
Post 14 Nov 2010, 04:07
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Tyler



Joined: 19 Nov 2009
Posts: 1216
Location: NC, USA
Tyler
You don't. in and out is for other uses. You won't use them unless you get into OS/driver dev. You must use the interfaces provided, such as WinAPI and libc. You would use those APIs just as you would in C, but you have to add the functions you wish to use to an import list. vid has a good example of calling libc from asm, here.
Post 14 Nov 2010, 06:18
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ishkabible



Joined: 13 Sep 2010
Posts: 54
ishkabible
ok ya OS/driver dev is beyond me at this point. so i could access my gpu this way? not that it is at my skill level to do any thing with it just wondering
Post 15 Nov 2010, 01:38
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ishkabible



Joined: 13 Sep 2010
Posts: 54
ishkabible
i have a question that dose not pertain to this. i was using idiv and i devided -45 by 15 but the answer that came out was 286331150. what is causing this? any time i divide a negative number it acts like its unsigned.
Post 15 Nov 2010, 02:33
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Tyler



Joined: 19 Nov 2009
Posts: 1216
Location: NC, USA
Tyler
Quote:
ok ya OS/driver dev is beyond me at this point. so i could access my gpu this way? not that it is at my skill level to do any thing with it just wondering
Not necessarily, you'd have to get access to proprietary(ie closed) information to find out exactly how to manipulate your GPU.

Quote:
i have a question that dose not pertain to this. i was using idiv and i devided -45 by 15 but the answer that came out was 286331150. what is causing this? any time i divide a negative number it acts like its unsigned.
Try dividing 45 by -15. Just a guess.
Post 15 Nov 2010, 03:05
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ishkabible



Joined: 13 Sep 2010
Posts: 54
ishkabible
Quote:
Quote:
i have a question that dose not pertain to this. i was using idiv and i devided -45 by 15 but the answer that came out was 286331150. what is causing this? any time i divide a negative number it acts like its unsigned.
Try dividing 45 by -15. Just a guess.


that works but not the other way around whats the deal?

edit: about the other thing. how come some one hasn't figured out this information. i mean it's our computers right? maybe there is no use for this sort of thing. i guess gpu's are meant to be for graphics and aren't much good much else. who knows the infmation? hardware manufacturers like ATI and Nvidia or software developers like OpenGL and DirectX?
Post 15 Nov 2010, 03:34
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Tyler



Joined: 19 Nov 2009
Posts: 1216
Location: NC, USA
Tyler
Quote:

that works but not the other way around whats the deal?
My hypothesis was that the first operand is unsigned and the second is signed, apparently I was right.

Quote:

edit: about the other thing. how come some one hasn't figured out this information. i mean it's our computers right? maybe there is no use for this sort of thing. i guess gpu's are meant to be for graphics and aren't much good much else. who knows the infmation? hardware manufacturers like ATI and Nvidia or software developers like OpenGL and DirectX?
The people who code the libraries for accessing GPUs either have access to the information from ATI/Nvidia, or have reverse engineered one of their drivers. I think there's a project for a reversed Nvidia driver for Linux. ATI recently released a small amount of documentation, but it's only a small step in the right direction. http://www.x.org/docs/AMD/
Post 15 Nov 2010, 03:56
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ishkabible



Joined: 13 Sep 2010
Posts: 54
ishkabible
so i how do i avoid this unsigned vs signed deal, i thought those kind of things weren't specified in assembly and that it was the instruction that dealt with this. i used idiv and idiv is sigend no?

here is my sample program that shows my problem.

Code:
format PE console
entry main

include 'INCLUDE\MACRO\import32.inc'
include 'INCLUDE\MACRO\proc32.inc'

section '.data' data readable writeable

pausemsg db "pause>nul",0
saynum db "%i is the number",10,0
mynumber1 dd ?
mynumber2 dd -45
mynumber3 dd 15

section '.code' code readable executable

main:
        mov edx,0
        mov eax,[mynumber2]
        mov ebx,[mynumber3]
        div ebx
        mov [mynumber1],eax
        push eax
        push saynum
        call [printf]
        add esp,8
        push pausemsg
        call [system]
        add esp,4
        ret


section '.idata' import data readable
library msvcrt,'msvcrt.dll'

import msvcrt,\
printf,'printf',\
scanf,'scanf',\
system,'system',\
getchar,'getchar',\
realloc,'realloc',\
malloc,'malloc',\
free,'free',\
calloc,'calloc',\
exit,'exit'
    
Post 15 Nov 2010, 04:10
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Tyler



Joined: 19 Nov 2009
Posts: 1216
Location: NC, USA
Tyler
I mean that it expects the first operand to be unsigned(ie the highest bit makes it bigger), instead of signed(ie the highest bit makes it negative). Although they aren't specified, they still make a big difference when interpreted differently. For example, take 45, and print it as "%u" to printf, then negate it, using neg, and print that as unsigned.
Post 15 Nov 2010, 05:37
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Alphonso



Joined: 16 Jan 2007
Posts: 294
Alphonso
Code:
        mov eax,[mynumber2]
        cdq                   ;sign extend eax -> edx:eax
        mov ebx,[mynumber3]
        idiv ebx    
Post 15 Nov 2010, 09:37
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ishkabible



Joined: 13 Sep 2010
Posts: 54
ishkabible
ok so so what dose cdq do exactly?
Post 15 Nov 2010, 17:11
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baldr



Joined: 19 Mar 2008
Posts: 1651
baldr
ishkabible,

Alphonso already wrote that: cdq sign-extends eax into edx, i.e. edx==-1 if eax<0, edx==0 otherwise. After this edx:eax contains 64-bit signed representation of the same number as 32-bit signed in eax.
Post 15 Nov 2010, 18:51
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ishkabible



Joined: 13 Sep 2010
Posts: 54
ishkabible
ok i get it uses eax joined with edx so that if edx is zero it would be like eax was unsigned because the highest order bit of edx:eax is 0 not 1. thanks guys Smile
Post 15 Nov 2010, 21:35
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