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Aug 8th, 2005, 3:19 PM
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Basic examples-what does it means
I started learning Asembler recently,from the book "The art of Assembly Language" and I came to the part with some examples.Well,the problem is that I dont understand some of them.Here are the example proggies:
1. ; Some variables we will access indirectly (using pointers):
dseg segment para public 'data'
J word ?
; Some simple uninitialized array declarations:
ByteAry byte 4 dup (?)
dseg ends
; The following program demonstrates how to access each of the above
; variables.
cseg segment para public 'code'
assume cs:cseg, ds:dseg
Main proc
mov ax, dseg ;These statements are provided by
mov ds, ax ; shell.asm to initialize the
mov es, ax ; segment register.
mov J, 0
; The following code shows how to access elements of the arrays using
; simple 80x86 addressing modes:
mov bx, J ;AL := ByteAry[J]
mov al, ByteAry[bx]2.The values of vairable K and J are: K integer 4 J integer ? lea bx, L ;Point bx at first word in L.
mov ax, [bx] ;Fetch word at L.
add ax, 2[bx] ;Add in word at L+2 (the "1").
add ax, 4[bx] ;Add in word at L+4 (the "2").
add ax, 6[bx] ;Add in word at L+6 (the "3").
mul K ;Compute (0+1+2+3)*123.
mov J, ax ;Save away result in J.
les bx, PtrVar2 ;Loads es:di with address of L.
mov di, K ;Loads 4 into di
mov ax, es:[bx][di] ;Fetch value of L+4.I dont understand what does the "1","2","3" means and this part: mul K ;Compute (0+1+2+3)*123 -isn't the value of K=4 ? 3. string byte "Hello world",0dh,0ah,0 What does odh,oah,0 means? 4. J word 0, 0, 0, 0
K word 1, 2, 3, 4
L word 5, 6, 7, 8
Ptr1 nWrdPtr ?
Ptr2 nWrdPtr K ;Initialize with K's address.
Ptr3 fWrdPtr L ;Initialize with L's segmented adrs
Add the four words in variables J, K, and L together using pointers to
; these variables:
mov bx, Ptr1 ;Get near ptr to J's 1st word.
mov si, Ptr2 ;Get near ptr to K's 1st word.
les di, Ptr3 ;Get far ptr to L's 1st word.
mov ax, ds:[si] ;Get data at K+0.
add ax, es:[di] ;Add in data at L+0.
mov ds:[bx], ax ;Store result to J+0.
add bx, 2 ;Move to J+2.
add si, 2 ;Move to K+2.
add di, 2 ;Move to L+2.
mov ax, ds:[si] ;Get data at K+2.
add ax, es:[di] ;Add in data at L+2.
mov ds:[bx], ax ;Store result to J+2.
add bx, 2 ;Move to J+4.
add si, 2 ;Move to K+4.
add di, 2 ;Move to L+4.
mov ax, ds:[si] ;Get data at K+4.
add ax, es:[di] ;Add in data at L+4.
mov ds:[bx], ax ;Store result to J+4.
add bx, 2 ;Move to J+6.
add si, 2 ;Move to K+6.
add di, 2 ;Move to L+6.
mov ax, ds:[si] ;Get data at K+6.
add ax, es:[di] ;Add in data at L+6.
mov ds:[bx], ax ;Store result to J+6.
Quit: mov ah, 4ch ;Magic number for DOS
int 21h ; to tell this program to quit.
Main endp
cseg ends
sseg segment para stack 'stack'
stk byte 1024 dup ("stack ")
sseg endsI don't understand pretty much everything in this code.I dont see how nor where do we 'Add the four words in variables J, K, and L together using pointers to these variables' If someone could help me with this,I would higly epriciate it. |
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Aug 8th, 2005, 4:24 PM
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#2 |
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Resident Grouch
 
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1. ByteAry [bx] doesn not indicate that an element of ByteAry contains bx. The address of ByteAry is offset by the value in bx. Consider that you may have an integer in C/C++ serve as an index into a byte array. The index does NOT have to be the same type as the element, rather, just an offset. In addition, don't expect assembler syntax to match one for one with any other language's syntax.
2. The comments are very self-explanatory. However, there is some contradiction concerning the value of K. You don't show enough; one doesn't know if it's 4, as you state, or 123 as the comment states. 3. 0d is the ASCII code for a carriage return. 0a is the ASCII code for a line feed (newline). The 0 is a marker saying the string ends there. Sometimes a $ is used. 4. There's no point in addressing this if you don't understand indirect references (pointers). Post back with your level of expertise, including any high level languages.
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Abstraction doesn't make it impossible to write bad code; it makes it possible to write superior code. Contributor's Corner: Grumpy on C++ Exceptions DaWei on Pointers |
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Aug 9th, 2005, 6:28 AM
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#3 |
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Newbie
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I studied C++ earlier from one book.I am not a programer just preparing myself for the university.I think that i understand pointers in C++ and it looked like the pointers are even eaiser in Asembler.
This is what I forgot to wrote: lea ax, J mov Ptr1, ax ;this is needed to Initialize Ptr1 with J's address Can u just explain me this part of code add bx, 2 ;Move to J+2. Before adding bx containes the address of variable J and that address is actually just the address of the first word in J.So when we add 2 does that means that we added 2 bytes and moved to the address of the seond word in variable J? |
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Aug 9th, 2005, 7:09 AM
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#4 |
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Resident Grouch
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When you add to or increment a pointer in C/C++, the compiler sees to it that the addition is in terms of the type pointed to. Assuming a 32-bit integer and byte addressing, if you add one to an integer pointer you will actually add four, to point to the next integer. Not true in assembler. These things are not "typed". BX holds an address. If you add two, it points two bytes (assuming byte addressing) farther into memory. Address 0x1000 would become 0x1002.
Address Data 0x1000 'A' 0x1001 'B' 0x1002 'C' 0x1003 'D' "word" is just a word.  (Never got to say that before.) Your use of it conveys nothing to your auditor unless you both know how you've defined it.
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Abstraction doesn't make it impossible to write bad code; it makes it possible to write superior code. Contributor's Corner: Grumpy on C++ Exceptions DaWei on Pointers |
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