adressing problem

Question

i have two numbers , and want add it .

mov eax,3[90fffffh]

mov ebx,4

add eax,ebx

i have extern C structure and result in Asm said ...

The result is Asm-1862270970

I use .model flat

i ought have result Asm7

Where is my mistake ? Can We help me ?

Answer 1

On 8/21/2012 2:11 PM, rafal_bator wrote:

i have two numbers , and want add it .

mov eax,3[90fffffh]

If you want to load 3 into eax, just write

mov eax, 3

What you have is something like "load into eax the four bytes that are in memory at offset 3 from address 0x90fffff". What is that supposed to achieve?

---

Igor Tandetnik

Answer 2

Yes i want load in eax 3 from 0x90fffff . And add it  to 4 .

Answer 3

On 8/21/2012 3:13 PM, rafal_bator wrote:

Yes i want load in eax 3 from 0x90fffff . And add it  to 4 .

What exactly do you mean by "from 0x90fffff"? What is the significance of the number 0x90fffff ?

If you want to load 3 into eax, you just write "mov eax, 3". If you want to load some value X other than 3 into eax, then why do you expect that X+4 should equal 7? I'm confused.

---

Igor Tandetnik

Answer 4

how can i load "3" to eax under 0x90fffff and add  "4" to ebx under 0x91ffffff ? 

Answer 5

On 8/21/2012 4:31 PM, rafal_bator wrote:

how can i load "3" to eax under 0x90fffff and add  "4" to ebx under 0x91ffffff ?

What do you mean by "under 0x90fffff"? How is it different from simply loading 3 into eax?

---

Igor Tandetnik

Answer 6

i want load to eax cell capacity from segment number 3 and move about 0x90fffff in relation to segment begin , and load 4 to ebx , and get result .

Answer 7

On 8/21/2012 5:48 PM, rafal_bator wrote:

i want load to eax cell capacity from segment number 3 and move about 0x90fffff in relation to segment begin , and load 4 to ebx , and get result .

What do you mean by a "cell"? What do you mean by a "segment"?

---

Igor Tandetnik

Answer 8

cell -> cellule

segment -> bit slice

Answer 9

On 8/21/2012 7:22 PM, rafal_bator wrote:

cell -> cellule

segment -> bit slice

This doesn't explain much. I'm not familiar with the terms "cellule" or "bit slice" either.

---

Igor Tandetnik

Answer 10

i think about segment:offset

Answer 11

On 8/21/2012 7:35 PM, rafal_bator wrote:

i think about segment:offset

If you are talking about segmented memory model as used in, say, MS DOS, then Windows hasn't used it since 1995. Win32 uses a flat address space - every byte is uniquely identified by one 32-bit address.

---

Igor Tandetnik

Answer 12

Ok . Thank You .

How write number 3 into 0x90fffff ? next add it to 4 in 0x91fffff ?

Answer 13

On 8/21/2012 8:09 PM, rafal_bator wrote:

How write number 3 into 0x90fffff ?

mov eax, 3
mov [90fffffh], eax

next add it to 4 in 0x91fffff ?

I don't understand this part. Do you want to read the contents of memory at address 0x91fffff? How do you know this address contains the value 4 - and if you do somehow already know that, why do you want to read from memory?

Or are you saying you want to add 4 to 3, and then write 7 to the memory at address 0x91fffff?

Where do the numbers 0x91fffff and 0x90fffff come from? What's so special about them?

---

Igor Tandetnik

Answer 14

ok , must i write 4 as 3 ?eg.

mov ebx,4

mov[91fffffh],ebx

?

and add as :

add [91fffffh],[90fffffh]

?

Answer 15

On 8/21/2012 8:30 PM, rafal_bator wrote:

ok , must i write 4 as 3 ?eg.

mov ebx,4
mov[91fffffh],ebx

Yes, if you want to save the value 4 into memory at address 91fffffh

and add as :

add [91fffffh],[90fffffh]

You can't add two values from memory directly - you have to load them into registers first. But since at that point you already have them in registers, just do

add eax, ebx

---

Igor Tandetnik

Answer 16

mov[91fffffh],eax and ebx is not allowed . I have that communicate .

Answer 17

I give up.  What does any of this have to do with Visual C?

Answer 18

I do not understand ? Can You answer on my problem ?

Answer 19

rafal_bator wrote:

mov[91fffffh],eax and ebx is not allowed . I have that communicate .

I don't understand what you are trying to say.

---

Igor Tandetnik

Answer 20

this is communicate after build .

Answer 21

rafal_bator wrote:

this is communicate after build .

What is communicated after build?

---

Igor Tandetnik

Answer 22

Maybe eax and ebx are allowed, but you do not use them correctly. For example if you write ‘_asm mov [91fffffh], eax’ in Visual Studio, you will receive an error, but when you write ‘_asm mov ds:[91fffffh], eax’, nu such compiler error.

So show the error messages that are displayed in console window, and the text of the program that you are trying to execute. And maybe rephrase the question giving more details.

Answer 23

mov [90fffffh],eax

mov [91fffffh],ebx

build it

1>asm.asm(7): error A2001: immediate operand not allowed

1>asm.asm(9): error A2001: immediate operand not allowed

ok .

Viorel .

How add this ?

must i define load from adress ?

Answer 24

how add this ?

main.cpp :

#include<iostream>

#include<conio.h>

using namespace std;

extern "C" int Asm();

int main()

{

cout<<"Asm"<<Asm()<<endl;

_getch();

return 0;

}

asm.asm:

.486

.model flat

.code

Asm proc C

mov eax,3

mov ds:[80fffffh],eax

mov eax,ds:[80fffffh]

mov ebx,4

mov es:[90fffffh],ebx

mov ebx,es:[90fffffh]

add ebx,eax

ret

Asm endp

end

Build is ok .

Debuging

i have :

Unhandled exception at 0x01232045 in project1.exe: 0xC0000005: Access violation writing location 0x080fffff.

where is my adress space , is it my mistake ?

Answer 25

The error means that cells situated at addresses 80fffffh and 90fffffh do not belong to your program, therefore you do not have rights to access them.

Instead of numeric addresses, try variables. For example define data segment before ‘end’ statement:

.data

    X dd ?

    Y dd ?

Then try ‘mov X, eax’ instead of ‘mov ds:[80fffffh],eax’ and ‘mov Y, ebx’ instead of ‘mov es:[90fffffh],ebx’ and so on.

Answer 26

i want to place datas in  definite cells .
how can i reserve memory  cells  about definite adress for simply datas ?

what do
You think about dynamic allocation ?

mov ah,49h

mov es,[ds:2ch]

int 21h

mov ax,3

mov ds,ax

mov ah,48h

mov bx,10

int 21h

is this slow down and allocated in 2ch number 3 ? how can i make next slow down and allocation for other number in this same thread ?

if i definite from  2ch , and slow down by int 21h , can i put this sime as previous ?:

Answer 27

rafal_bator wrote:

i want to place datas in definite cells .

Why? What's the purpose of this exercise?

how can i reserve memory cells about definite adress for simply datas  ?

You can't. Again, why do you want to?

what do
You think about dynamic allocation ?

mov ah,49h
mov es,[ds:2ch]
int 21h

You seem to be reading some very old manual about programming for MS  DOS, where "int 21h" was a system call. This manual is obsolete by 17  years - Windows doesn't work this way since 1995. Throw it away, forget  everything you read in it.

---

Igor Tandetnik

Answer 28

The address 90FFFFFh does not belong to your program, but you can try something like this:

.data

    db 90FFFFFh+4 dup(?)

Now you should have enough space for your experiments. However a declaration like this usually slows down the compilation (http://social.msdn.microsoft.com/Forums/en-US/Vsexpressvc/thread/dea7357d-0650-4182-a55e-8cf951aba97e). Maybe you first try some smaller addresses.

Answer 29

what belong to my free adress space , how count it ? 

Answer 30

This is a Visual C++ forum. If you need help with asm, please ask in appropriate forum.

-- pa

Answer 31

I took with You , Sir Pavel A . Part of Visual C++ is Masm . VC++ have Masm32 and Masm64 , it is depends of version . Masm is as _asm or Asm in You , Sir code . Before if You want to use it , Sir have to definite it .

Answer 32

what are different between diferent MASMs ? Masm work in cmd too .

Answer 33

what is instead int 21h ?

Answer 34

Unhandled exception at 0x00e72044 in project1.exe: 0xC0000005: Access violation writing location 0x00000300.

What is it ?

Answer 35

On 8/23/2012 5:22 PM, rafal_bator wrote:

Unhandled exception at 0x00e72044 in project1.exe: 0xC0000005: Access violation writing location 0x00000300.

What is it ?

Which part of the error message is unclear? You attempt to write to a memory location at address 0x00000300, but your process doesn't have access to write to that location.

---

Igor Tandetnik

Answer 36

On 8/23/2012 4:59 PM, rafal_bator wrote:

what is instead int 21h ?

Windows API functions:

http://msdn.microsoft.com/en-us/library/windows/desktop/hh447209.aspx

---

Igor Tandetnik

Answer 37

Thank You

i wrote :

.486

.model flat

.code

Asm proc C

mov eax,4

mov ecx,[30]

mov ecx,eax

mov ebx,5

mov edx,[40]

mov edx,ebx

add ecx,edx

ret

Asm endp

end

my asm see number 4 , why do not add 4 to 5 ?

Answer 38

On 8/23/2012 6:24 PM, rafal_bator wrote:

my asm see number 4 , why do not add 4 to 5 ?

A function returns a result by placing it into EAX register. In other words, whatever is in EAX after the function returns, becomes the function's return value.

Your code loads 4 into EAX in the very first line, and never touches that register afterwards. So when the function returns, EAX still holds 4, which becomes the function's return value. All the other lines don't serve any useful purpose, and can be removed without changing the outcome.

---

Igor Tandetnik

Answer 39

to ecx , is not load adress [30] ? And eax with 4 to change into ecx as adressed [30]?

Answer 40

On 8/23/2012 6:52 PM, rafal_bator wrote:

to ecx , is not load adress [30] ? And eax with 4 to change into ecx as adressed [30]?

In a mov instruction, the first argument is the destination and the second argument is the source. Thus, "mov ecx, eax" means "take value from EAX register, copy it into ECX register". EAX register remains unchanged.

---

Igor Tandetnik

Answer 41

Thank You .

how adressed into memory eg. 4 ?

Answer 42

On 8/23/2012 7:08 PM, rafal_bator wrote:

how adressed into memory eg. 4 ?

I don't understand the question.

---

Igor Tandetnik

Answer 43

how can i put into definite memory address eg.4 ?

Answer 44

On 8/23/2012 7:25 PM, rafal_bator wrote:

how can i put into definite memory address eg.4 ?

Why do you want to? Where is this "definite" memory address coming from? There is never a good reason to access a hard-coded address. But if you insist, you can just write

mov [10000000h], 4

Replace 10000000h with the actual address you want to access.

---

Igor Tandetnik

Answer 45

1>asm.asm(5): error A2001: immediate operand not allowed

i have it .

Answer 46

On 8/23/2012 7:41 PM, rafal_bator wrote:

1>asm.asm(5): error A2001: immediate operand not allowed

In this case, try

mov eax, 4
mov [10000000h], eax

---

Igor Tandetnik

Answer 47

this same result

Answer 48

On 8/23/2012 7:57 PM, rafal_bator wrote:

this same result

How about

mov eax, 10000000h
mov [eax], 4

---

Igor Tandetnik

Answer 49

then take value in hexadecimal 10000000h to eax , and 4 to eax ?

how is purpose it ?

when i meke mov [eax],4 i have it :

1>asm.asm(6): error A2070: invalid instruction operands

Answer 50

On 8/23/2012 8:20 PM, rafal_bator wrote:

then take value in hexadecimal 10000000h to eax , and 4 to eax ?

No. [eax] (note square brackets) means "the memory location whose address is stored in EAX".

when i meke mov [eax],4 i have it :

1>asm.asm(6): error A2070: invalid instruction operands

Last attempt, and I give up. It's been 19 years since I last coded in assembler.

mov eax, 10000000h
mov dword ptr [eax], 4

---

Igor Tandetnik

Answer 51

i have it :Unhandled exception at 0x01052045 in project1.exe: 0xC0000005: Access violation writing location 0x10000000.

Answer 52

On 8/23/2012 8:43 PM, rafal_bator wrote:

i have it :Unhandled exception at 0x01052045 in project1.exe: 0xC0000005: Access violation writing location 0x10000000.

Of course. I told you it never makes sense to access a hard-coded address.

---

Igor Tandetnik

Answer 53

if i use it in e.g. 7h , i have this same . this is free space , windows contain self in up memory .
what is your time ?

Answer 54

On 8/23/2012 8:55 PM, rafal_bator wrote:

if i use it in e.g. 7h , i have this same.

Of course. Why would you expect any different?

this is free space , windows contain self in up memory.

A 32-bit process can address 2GB of address space. But that doesn't mean that every process has 2GB of physical RAM available to it. You simply don't have that much RAM in your machine. Thus, most of that address space is not in fact backed by physical memory. If you try to access a virtual memory address that is not backed by physical RAM, you will get an access violation. Which is exactly what you observe in practice.

You cannot just go poking at random addresses. Before you can access memory, you must allocate it.

See if this helps clarify the picture:

http://en.wikipedia.org/wiki/Virtual_address_space
http://en.wikipedia.org/wiki/Virtual_memory

what is your time ?

I beg your pardon?

---

Igor Tandetnik

Answer 55

good night , i will be for 7 hours . thank You .

Answer 56

Igor, my compliments for your patience.

Answer 57

1.how can i check where i have free memory , and all  memory structure ?

2.have i use to space of application app.exe only in Visual C++ or can i definite this space in free space on outside this aplication app.exe ?

3.how can i allocate my application app.exe ?

4. is my asm.asm inside main.cpp , or outside ?

5. is my _asm inside main.cpp ?

6. who is who ?

Answer 58

rafal_bator wrote:

1.how can i check where i have free memory , and all memory structure  ?

There is no free memory just lying around waiting to be picked up. Free  memory is made available to you upon request. When you need a certain  amount of memory, you allocate it, then you can use it.

2.have i use to space of application app.exe only in Visual C++ or can  i definite this space in free space on outside this
aplication app.exe ?

I don't understand this question. But I suspect you have deep  misconceptions about how virtual memory works.

3.how can i allocate my application app.exe ?

If you are asking how you can allocate memory for use by your  application, then calling HeapAlloc API function would be one  possibility.

4. is my asm.asm inside main.cpp , or outside ?

I don't understand this question.

5. is my _asm inside main.cpp ?

What do you mean by "your _asm"? What do you mean by being "inside"?

6. who is who ?

I am me. I can't speak for others.

---

Igor Tandetnik

Answer 59

mov eax,1h // load to eax 1

mov ebx,eax // load eax to edx  ,free pointer of datas

mov esi,201h//201h , load to source pointer esi

mov edx,esi//load esi to data register edx

i build and debug

my result is : 1

but i have it :

Run-Time Check Failure #0 - The value of ESP was not properly saved across a function call.  This is usually a result of calling a function declared with one calling convention with a function pointer declared with a different calling convention.

what is it ?

who is who // this is mistake , sorry// i ought : what is what ?
inside -> when You have aplication start - to - finish , when You use this _asm into and Asm outside .

Answer 60

rafal_bator wrote:

but i have it :

Run-Time Check Failure #0 - The value of ESP was not properly saved  across a function call. This is usually a result of calling a
function declared with one calling convention with a function pointer  declared with a different calling convention.

what is it ?

Calling convention mismatch. Show how you declare this function in your  C++ code, and show the complete assembly code, from "proc" to "endp".

---

Igor Tandetnik

Answer 61

main.cpp:

#include <iostream>

#include <conio.h>

using namespace std;

extern "C" int Asm();

int main()

{

cout<<"Asm"<<Asm()<<endl;

_getch();

return 0;

}

asm.asm:

.486

.model flat

.code

Asm proc C

mov eax,1h

mov ebx,eax

mov esi,201h

mov edx,esi

ret

Asm endp

end

Answer 62

rafal_bator wrote:

Asm proc C
mov esi,201h

With __cdecl calling convention (which is the one you are using), the  function must preserve all registers except EAX, ECX, and EDX. This  doesn't mean the function cannot use these registers for its own needs -  but if it does, it must save the original values at the beginning, and  restore them at the end, before returning.

Your code fails to preserve the value of ESI register.

---

Igor Tandetnik

Answer 63

how can i use ds segment register in 32 mode ?

how can i replace this segment register in 32 bit mode ?

why cannot i use 16 bit registers , x86 architecture have this ?

Answer 64

rafal_bator wrote:

how can i use ds segment register in 32 mode ?

You can't.

how can i replace this segment register in 32 bit mode ?

You can't.

why cannot i use 16 bit registers , x86 architecture have this ?

You can happily use AX, BX, CX and DX 16-bit registers. Probably also SI  and DI, but I'm not sure.

---

Igor Tandetnik

Answer 65

Thank You .

I know , that i cannot use DS in 64 bit mode . Can i replace this by descriptor ?

Answer 66

rafal_bator wrote:

I know , that i cannot use DS in 64 bit mode . Can i replace this by  descriptor ?

What do you mean by "descriptor"? In any case, I know very little about  64-bit assembly, and won't be able to help you with it, I'm afraid.

---

Igor Tandetnik

Answer 67

Descriptor it is data structure who is use by segment motion :

Segment descriptor describe segment , and contain place in virtual space , size , safety and more ...

Answer 68

Hi rafal,

Thanks for your active participation.

Apparently, you have asked a lot of question in this thread. I temporarily marked two of the replies, which in my view have answered the questions in your original post. If you disagree with them, you can unmarked them.

Thanks,

---

Damon Zheng [MSFT]
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