I've been programming in C# for a couple years now and I'm trying to learn the basics of assembly with masm before I take a course in it next semester. I'm supposed to know the basics before starting, and there is a worksheet I should be able to figure out. I'm having some trouble getting into it without any instruction.
The first half of it is converting numbers to/from different bases, which I remembered well enough from my intro to CS classes a few years ago.
1) Here is one question, that I believe I have correct but was hoping you guy could confirm it for me.
Complete the following table. All numbers are to be represented in the base 16
Quote
ASSEMBLY LANGUAGE CODE EAX
mov eax, 1 1
sub eax, 9ADF01 FF652100
I had to fill in the EAX column. I used visual studio, put the lines in, did a break point, and used the intermediate window to find out what was in eax. Is this correct?
2) Next question:
QuoteAssume in an AL program:
x DWORD 110101111010101000110101101101b
0011 0101 1110 1010 1000 1101 0110 1101
The additive inverse of x is
My understanding is that for binary numbers, to get the negative number, you inverse the 1/0s and then add an extra 1 to the right of the number. However, on this worksheet where the anwser goes its split up into boxes that imply 4 numbers should go per box, which doesn't leave room for the extra 1. This has left me a little confused... Anyone have any idea here? I'll probably end up with more questions as I go down the sheet, but I'm going to give them a go first and see if I can figure those out on my own. These first two I seem to be stuck on.
Quote from: Xycron on December 27, 2012, 11:21:19 AM
The additive inverse of x is
Wiki: In mathematics, the additive inverse, or opposite, of a number a is the number that, when added to a, yields zero.
include \masm32\MasmBasic\MasmBasic.inc ; download (http://masm32.com/board/index.php?topic=94.0)
Init
mov eax, 1
sub eax, 9ADF01h
Print Hex$(eax) ; FF652100 indeed
mov ecx, 110101111010101000110101101101b
mov ebx, ecx ; keep a copy in a safe register
Print Bin$(ecx), CrLf$
neg ecx ; there it is!
add ebx, ecx ; the test for additive inverse
Print Bin$(ecx), CrLf$
Inkey Str$("ebx=%i", ebx)
Exit
end start
Output:
00110101111010101000110101101101
11001010000101010111001010010011
ebx=0q.e.d.
1) Is correct -- try doing it by hand first, then check your answers with the debugger.
2) You invert the bits and then add 1 (the value, not append an extra 1.)
example:
5 = 00000101
invert: 11111010
plus 1: 11111011
11111011 = -5
Quote from: Tedd on December 28, 2012, 03:03:02 AM
2) You invert the bits and then add 1
Any difference from neg eax?
flags :P
also - NEG on a 32-bit machine is good for bytes, words, dwords
if you want to negate other sizes or use SSE to do them in parallel,
then the invert/add1 or sub1/invert method is nice to know
Quote from: jj2007 on December 28, 2012, 05:16:02 AM
Quote from: Tedd on December 28, 2012, 03:03:02 AM
2) You invert the bits and then add 1
Any difference from neg eax?
Neg performs the same function, but knowing that doesn't help with understanding the theory and why it works.
Thank you both for the help!
I have another one... about div. My question comes from a chart below. On the last line you can see pseduo code is diving eax / m. but the assemlby language code is just div m. How does it know to divide divide m by eax? is it just because eax was the last variable? I'm not sure I understand what is happening there.
QuotePSEUDO-CODE AL -CODE EAX
N :=ff ff ff fah n dword 0ff ff ff fah
M:= 10h m dword 10h
EAX:= N mov eax, n
EDX:= 0 mov edx, 0
EAX÷M div m
Thanks again for the help everyone.
for the DIV instruction (as well as MUL), the size of the operand determines the behaviour of the instruction
the assembler knows what the size of the divisor is and selects the opcode accordingly
for a dword size divisor, EDX:EAX is treated as one large 64-bit dividend
http://www.masmforum.com/board/index.php?topic=18841.msg159697#msg159697 (http://www.masmforum.com/board/index.php?topic=18841.msg159697#msg159697)
for byte or word sizes, it uses different register sets
but, the dividend is always twice as many bits as the divisor
The description in the MASM reference is:
Quote
DIV
Unsigned Divide
Divides an implied destination operand by a specified source operand.
For a 32-bit source operand the implied destination operand is EDX:EAX.
At first I thought I understood it as:
So, 'div x' will do x / eax and the result will be in edx?
but now it sounds like the result is stored in edx and eax? and it will always divide by what was in eax first? I'm still a bit confused.
Could you explain what the : syntax means in EDX:EAX?
Sorry, I really am brand new to assembly.
Don't worry, we are good at teaching Chinese ;-)
edx::eax is a notation used for "take two 32-bit registers together and see them as a 64-bit number"
div memvar divides the 64-bit number in edx::eax by memvar. The quotient will be stored in eax, the remainder in edx
With mov edx, 0 you declare implicitly that you want an unsigned division of eax; because now the highest bit of the edx::eax is zero.
To get signed division, you can use cdq to sign-extend the value of eax into edx, i.e.
.if sdword ptr eax<0
... edx=-1
.else
... edx=0
.endif
Of course, you can also explicitly load a 64-bit number into edx::eax and divide it by something. The result is always 32-bit only.
Here is an example:
include \masm32\MasmBasic\MasmBasic.inc ; download (http://masm32.com/board/index.php?topic=94.0)
.data
MyQW QWORD 1234567890123456789
My1e18 QWORD 1000000000000000000 ; 1e18
Init
PrintLine Str$("A fat number:\t%i", MyQW)
mov eax, dword ptr MyQW ; dword ptr means override qword size
mov edx, dword ptr MyQW[4] ; load the high dword of MyQ
pushad ; save all regs - eax and edx will be trashed by the printing
PrintLine Str$("As edx::eax:\t%i", edx::eax)
popad ; still MyQW
mov ecx, 123
if 0 ; conditional assembly: if 0 means "don't do that"
div ecx ; this will cause an exception, because the result of the division does not fit into 32 bits
PrintLine Str$("Divided by 123:\t%i", eax)
else
cdq ; sign-extend eax into edx; since eax is positive, edx will be zero
div ecx ; divide the lower dword of MyQW by ecx
PrintLine Str$("eax div 123:\t%i", eax)
endif
mov eax, dword ptr My1e18
mov edx, dword ptr My1e18[4]
mov ecx, 1000000000 ; 1e9
div ecx
PrintLine Str$("My1e18 div 1e9:\t%i", eax)
Inkey
Exit
end start
Output:
A fat number: 1234567890123456789
As edx::eax: 1234567890123456789
eax div 123: 17174430
My1e18 div 1e9: 1000000000
I think the convention (at least in the US, but what do we know) is to use a single colon to represent a 64-bit value in two 32-bit registers (EDX:EAX). The double colon notation (EDX::EAX) is for invoke.
Quote from: MichaelW on December 29, 2012, 07:20:50 PM
I think the convention (at least in the US, but what do we know) is to use a single colon to represent a 64-bit value in two 32-bit registers (EDX:EAX). The double colon notation (EDX::EAX) is for invoke.
Michael,
Thanks for correcting me on that. You are right, of course, but I'll keep the edx::eax syntax for MasmBasic Str$() as shown above because it's closer to an "invoke situation". Is there any case where you can use edx:eax in assembly, I mean in real code that gets assembled, not in comments?
Thanks. I think I have a better understanding now. I'll probably be posting soon on 'jumps'.. but I'm going to see if I can figure this one out first as well.
Quote from: jj2007 on December 29, 2012, 07:35:31 PM
Is there any case where you can use edx:eax in assembly, I mean in real code that gets assembled, not in comments?
None that I know of.
Would someone mind double checking my anwesers for there two? I'm nearly positive the first one is correct, not as sure about the second.
Does a jump occur?
Quote
mov al,255
cmp al,2;
jnle xyz
xyz: ; a label
I put yes. jnle jumps if al is not less than or equal to (greater than) 2, which is is. correct, right?
Next one is:
Quote
mov al,255
cmp al,2
begin1: jnl end1
mov ah, 10
end1:jl end2
mov ah 20
end2
The above stores ___ in ah.
I believe it enters the first part, and not the second, which means it stores 10 in ah?
Quote from: Xycron on December 29, 2012, 09:28:00 PM
Does a jump occur?
Who knows ::)
include \masm32\include\masm32rt.inc
.code
start:
; Does a jump occur?
mov al,255
cmp al,2;
jnle xyz
inkey "NO"
exit
xyz: ; a label
inkey "YES"
exit
end start
Seriously: At a certain point you must decide how you want to learn. I suggest that you install the Masm32 package and learn how to use \Masm32\qeditor.exe ... and read the tips & tricks in my signature :icon14:
if you
mov al,255
the high bit is set (sign bit in context where signed numbers are used)
JL, JNL, JNLE are used for signed comparison
so, 255 is treated as -1
now see if you can figure it out without executing it :P
Thanks for the help from both of you. I was trying to get both of them to compile to check myself but they wouldn't compile in VS and I wasn't sure why, perhaps THATS what I should have asked about.
i tried adding the include line, and that made the first one compile(although i lost the ability to use the debugger for some reason) but the second one still does not compile.
I'm trying the link in your signature but am having trouble installing the MASM32 SDK... Getting errors when installing about ordinal 200 AcGenral.DLL. and 202 AcLayers.dll. not being able to be located. I think it may be a windows 8 issue, but I will google further...
And thanks for the tip on signed numbers and how 255 is really -1. I would have been confused as to why it did not jump.
The AcLayers thing is a known issue with Win8, see here (http://masm32.com/board/index.php?topic=1110.0).
Using VS for simple Masm examples is an overkill; it may work, of course, but you also may run into setup problems. And it is so incredibly slow.... better try \Masm32\qeditor.exe with the Console assemble & link.
My own editor comes with MasmBasic (http://www.masm32.com/board/index.php?topic=94.0) and sits, after extraction, at \Masm32\RichMasm\RichMasm.exe - it looks very different 8), but among the useful features is that you can open practically all examples in \Masm32\examples, hit F6 and see the result immediately.
I seem to finally have the SDK installed and running. I am able to compile the same things i could compile in visualstudio, but the second example posted below (and a few posts above) still does not compile:
mov al,255
cmp al,2
begin1: jnl end1
mov ah, 10
end1:jl end2
mov ah 20
end2
syntax error : in instruction
syntax error: end2
undefined symbol : end2
I don't really understand what the end2's are doing. Can anyone point me in the right direction for this?
If there is a jump to end2, then logically end2 must be a label:
end2:
And don't forget at the end of a program there must be
end start
or
end Main
or whatever the entry point was called.
Quote from: Xycron on December 30, 2012, 08:58:38 AM
I seem to finally have the SDK installed and running. I am able to compile the same things i could compile in visualstudio, but the second example posted below (and a few posts above) still does not compile:
mov al,255
cmp al,2
begin1: jnl end1
mov ah, 10
end1:jl end2
mov ah 20
end2
syntax error : in instruction
syntax error: end2
undefined symbol : end2
I don't really understand what the end2's are doing. Can anyone point me in the right direction for this?
When you have a label, it's a good idea to start your code on the next line.
Where is end1 at ?
Do you mean end2: ?
Another idea is to comment your code in some places.
Quote from: Magnum on December 30, 2012, 09:05:03 AM
Quote from: Xycron on December 30, 2012, 08:58:38 AM
I seem to finally have the SDK installed and running. I am able to compile the same things i could compile in visualstudio, but the second example posted below (and a few posts above) still does not compile:
mov al,255
cmp al,2
begin1: jnl end1
mov ah, 10
end1:jl end2
mov ah 20
end2
syntax error : in instruction
syntax error: end2
undefined symbol : end2
I don't really understand what the end2's are doing. Can anyone point me in the right direction for this?
When you have a label, it's a good idea to start your code on the next line.
Where is end1 at ?
Do you mean end2: ?
Another idea is to comment your code in some places.
This is code the teacher gave us. I have no idea what he meant, perhaps he gave us code with an error...
Here is my entire code for how I'm trying to compile it...
Quote.486
.MODEL FLAT
.data
.code
main PROC
mov al,255
cmp al,2
begin1: jnl end1
mov ah, 10
end1:jl end2
mov ah 20
end2
main ENDP
END main
Quote from: jj2007 on December 30, 2012, 09:02:56 AM
If there is a jump to end2, then logically end2 must be a label:
end2:
.486
.MODEL FLAT
.data
.code
main PROC
mov al,255
cmp al,2
begin1: jnl end1
mov ah, 10
end1:jl end2
mov ah, 20 ; <<<<<<<<<<<
end2: ; <<<<<<<<<<<
ret ; <<<<<<<<<<<
main ENDP
END main
Hi Xycron,
so, you've the answer from Jochen. But please have a look at the forum rules: http://masm32.com/board/index.php?topic=4.0 (http://masm32.com/board/index.php?topic=4.0), especially the The NO HOMEWORK Rule.
Gunther