Given the following variable definitions:
.data
var2 WORD 1000h, 2000h, 3000h, 4000h
var3 SWORD -16, -42
The value, according to the book, the following instructions calculate to:
mov ax,[var2 + 4] ;3000h Why is this not 1000h, (since the last element in the var2 word list is 4000h at index [3]) wouldn't the [4] have no where to go but 1000h at index 0?
mov ax,[var3 - 2] ;4000h ;Where is 4000h coming from? It's not evened signed, unlike -16 and -42
Quote from: RedSkeleton007 on September 07, 2015, 02:47:27 PM
Given the following variable definitions:
.data
var2 WORD 1000h, 2000h, 3000h, 4000h
var3 SWORD -16, -42
The value, according to the book, the following instructions calculate to:
mov ax,[var2 + 4] ;3000h Why is this not 1000h, (since the last element in the var2 word list is 4000h at index [3]) wouldn't the [4] have no where to go but 1000h at index 0?
mov ax,[var3 - 2] ;4000h ;Where is 4000h coming from? It's not evened signed, unlike -16 and -42
Okay imagine those variables as a list:
1000h ; offset 0
2000h ; offset 2
3000h ; offset 4
4000h ; 0ffset 6
-16 ; offset 8
-42 ; offset 10
It has to do with 'indexing'
I don't have a good technical explanation, but that should give you a good idea
You might want to look up the "lea" instruction as well
The way you have the code, it is referencing the address where var2 is at for the first one,
and the ADDRESS of var3 for the second.
word values are 2 bytes long.
so mov ax, [var2 + 0] == var2 == 1000h
mov ax, [var2 + 2] == 2000h
mov ax, [var2 + 4] == 3000h
mov ax, [var2 +6] == 4000h
and continuing on the same path,
mov ax, [var2 + 8] == -16
mov ax, [var2 + 10] == -42
mov ax, [var2 + 12] == whatever word size value follows
----------------------------------
which means that
mov ax, [var3 - 2] == 4000h is correct
and also means that
mov ax, [var3 + 2] would equal -42
mov ax, word ptr [var2 + 4]
is NOT the same as
mov ax, var2
add ax, 4
----------------------------
It is the same as
-------------------------------
mov eax, offset var2
add eax, 4
mov ax, word ptr [eax]
--------- or ---------------
lea eax, var2
add eax, 4
mov ax, word ptr [eax]
--------------------------------
I think what you want or expect is:
---------------------------
add word ptr [var2], 4 ; changes the value of var2
mov ax, var2
- - - - - - - - - - - - -
or even more simply
- - - - - - - - - - - - - -
add var2, 4 ; changes the value of var2
mov ax, var2
- - - - - - -- - - - - - -
or by adding in ax
- - - - - - - - - - - - - -
mov ax, var2 ; does not change the value of var2
add ax, 4
---------------------------
if I am not mistaken
Do some more research, you'll get the hang of it
Hope this helps
edit = clarification, and more clarification.
by the way Red, which book are you referring to?
Quote from: zedd151 on September 07, 2015, 03:27:35 PM
Okay imagine those variables as a list:
1000h ; offset 0
2000h ; offset 2
3000h ; offset 4
4000h ; 0ffset 6
-16 ; offset 8
-42 ; offset 10
It has to do with 'indexing'
Red,
The secret is that asm != C 8)
There is no automatic indexing in assembler, so var[2] is "what you find at byte offset 2",
not "element 2, i.e. the third element".
Re signed: The value -42 can be interpreted as signed or unsigned. A mov ax, -42 does not impress the register ax, it just sees 0000FFF0h. Your print routine may then decide whether fff0h is "-42" or "+65520". You can force an interpretation with the movzx and movsx instructions.
include \masm32\MasmBasic\MasmBasic.inc
.data
var2 WORD 1000h, 2000h, 3000h, 4000h
var3 SWORD -16, -42
Init
mov ax, var2[2]
mov dx,[var3 - 2]
movsx ecx, [var3 + 2]
movzx esi, var3
movsx edi, var3
deb 1, "Test", x:ax, x:dx, ecx, x:esi, x:edi, esi, edi ; x: means "use hex"
Exit
end startOutput:
x:ax 2000
x:dx 4000
ecx -42
x:esi 0000FFF0
x:edi FFFFFFF0
esi 65520
edi -16
Quote from: RedSkeleton007 on September 07, 2015, 02:47:27 PM
Given the following variable definitions:
.data
var2 WORD 1000h, 2000h, 3000h, 4000h
var3 SWORD -16, -42
The value, according to the book, the following instructions calculate to:
mov ax,[var2 + 4] ;3000h Why is this not 1000h, (since the last element in the var2 word list is 4000h at index [3]) wouldn't the [4] have no where to go but 1000h at index 0?
mov ax,[var3 - 2] ;4000h ;Where is 4000h coming from? It's not evened signed, unlike -16 and -42
If you look at the declaration below.. it's an array of WORD (16 bits) sized values
var2
WORD 1000h, 2000h, 3000h, 4000h
The instruction says ...
mov ax,[var2 + 4] --> [var2 + 4] means Address of Var2 plus 4 bytes offset (offsets are always in Bytes)
Move a 16 bit value (you using the AX register which is 16 bits wide (EAX is 32 bits)) from the
[address] (Address of Var2 with 4 bytes added to it) to AX.
As a byte is 8 bits wide and each array value is (WORD) 16 bits wide.. so 8x4 = 32 bits, but 16x2 = 32 bits.. so the
[address] points to 2 variables further on in the array (=3000h)
Var2 + 0 = 1000h
Var2 + 2 = 2000h
Var2 + 4 = 3000h
Now the same idea with
mov ax,[var3 - 2]
..which is nothing more than 2 bytes backwards = 1x 16 bit array value before Var3... === 4000h, which is the last value of the Var2 array.
;)
Damm.. wrong button.. how does one delete.. :biggrin:
Edt: = Wrong button pressed :greensml:
Quotedelete contents of the post
She.. no worka !
Quote from: K_F on September 07, 2015, 05:36:00 PM
Damm.. wrong button.. how does one delete.. :biggrin:
Click on Modify
delete contents of the post :biggrin:
I also always try to add a footnote
edit = <reason>
:biggrin:
Quote from: zedd151 on September 07, 2015, 04:22:47 PM
by the way Red, which book are you referring to?
This one:
http://kipirvine.com/asm/index6th.htm (http://kipirvine.com/asm/index6th.htm)
Try Art of Asm (https://courses.engr.illinois.edu/ece390/books/artofasm/artofasm.html) for comparison, especially chapter 4 onwards. And attention, both books have a 16-bit history and are likely to focus on obsolete 16-bit stuff. See also http://www.webalice.it/jj2006/Masm32_Tips_Tricks_and_Traps.htm, Iczelion tutorials.
Quote from: jj2007 on September 09, 2015, 11:25:34 AM
Try Art of Asm (https://courses.engr.illinois.edu/ece390/books/artofasm/artofasm.html) for comparison, especially chapter 4 onwards. And attention, both books have a 16-bit history and are likely to focus on obsolete 16-bit stuff.
No offense dude, but my book and the art of assembly have a 14 year difference, so I don't even understand why you would bother comparing them. Also, I want to learn MS-DOS and 16-bit stuff eventually, because video games for the Super Nintendo run on 16-bit assembly, and are thus very fun to hack. In fact, I've been a member of SMW Central far longer than I've been a member of this site ;)
Quote from: RedSkeleton007 on September 09, 2015, 12:09:56 PMvideo games for the Super Nintendo run on 16-bit assembly, and are thus very fun to hack
That makes sense :t
i don't think super nintendo uses intel 8086 processors
so, you are in for a lesson about segmented memory that you may not use elsewhere
it uses a processor that's an off-shoot from rockwell 6500 series
Quote from: dedndave on September 09, 2015, 09:12:22 PM
i don't think super nintendo uses intel 8086 processors
so, you are in for a lesson about segmented memory that you may not use elsewhere
it uses a processor that's an off-shoot from rockwell 6500 series
That's true. The CPU of the SNES has a 65c816 core, so SNES programming is done with 65c816 assembly. But asm is still asm. My hope is that someday I will be able to port 16-bit MS-DOS games onto an SNES cartridge. Imagine Jazz Jackrabbit on the SNES 8)