11
Website and Forum / Re: nasm.us certificate expired this morning
« Last post by netjunki on February 24, 2021, 12:59:00 AM »Seems to be fixed now! Thanks to whoever (or the bot) which cycled the cert. :-)
Recent Posts
12
Website and Forum / nasm.us certificate expired this morning
« Last post by netjunki on February 23, 2021, 07:16:17 PM »Hey all,
It looks like the nasm.us https certificate expired this morning. Looking at the details maybe the script which is run to regenerate it through Let's Encrypt failed to run.
Cheers,
Ben
It looks like the nasm.us https certificate expired this morning. Looking at the details maybe the script which is run to regenerate it through Let's Encrypt failed to run.
Cheers,
Ben
13
Programming with NASM / Re: How are you multiplying by 3
« Last post by debs3759 on February 22, 2021, 07:04:20 PM »I can't figure out what you are not seeing? (EDX * 2) + EDX is obviously 3 * EDX, where are you seeing the "1" that you don't understand?
0 can be multiplied by any number...
0 can be multiplied by any number...
14
Programming with NASM / Re: How are you multiplying by 3
« Last post by gygzkunw on February 22, 2021, 04:34:33 PM »I didn't say it wasn't elementary. To get 3 you have to add 1. The question is where does the 1 come from? Up till this point in the author does not explain it.edx+2 * edx = 3edx-> This is my question, how is this possible?
EDX + 2*EDX = EDX * (1 + 2) = EDX * 3
Elementary arithmetic.QuoteBut earlier mov edx,ecx -> 0 was moved into edxYep... this code is obviously WRONG. Or, at least, is incomplete (and ECX is updated in a loop).
edx(0)+ 2*edx(0) =should equal to 0, right?
Do you understand what I am saying here?
15
Programming with NASM / Re: How are you multiplying by 3
« Last post by fredericopissarra on February 22, 2021, 03:40:15 PM »edx+2 * edx = 3edx-> This is my question, how is this possible?
EDX + 2*EDX = EDX * (1 + 2) = EDX * 3
Elementary arithmetic.
Quote
But earlier mov edx,ecx -> 0 was moved into edxYep... this code is obviously WRONG. Or, at least, is incomplete (and ECX is updated in a loop).
edx(0)+ 2*edx(0) =should equal to 0, right?
Do you understand what I am saying here?
16
Programming with NASM / Re: How are you multiplying by 3
« Last post by gygzkunw on February 22, 2021, 02:11:02 PM »edx+2 * edx = 3edx-> This is my question, how is this possible?
But earlier mov edx,ecx -> 0 was moved into edx
edx(0)+ 2*edx(0) =should equal to 0, right?
Do you understand what I am saying here?
But earlier mov edx,ecx -> 0 was moved into edx
edx(0)+ 2*edx(0) =should equal to 0, right?
Do you understand what I am saying here?
17
Programming with NASM / Re: How are you multiplying by 3
« Last post by Frank Kotler on February 22, 2021, 05:21:03 AM »edx plus two times edx equals three times edx. I don't think I understand the question.
I have not read Jeff Duntemann's book. I did not mean to give the impression I didn't like it! It has an excellent reputation. I don't have another suggestion. Dr. Paul Carter's Tutorial, perhaps?
Best,
Frank
I have not read Jeff Duntemann's book. I did not mean to give the impression I didn't like it! It has an excellent reputation. I don't have another suggestion. Dr. Paul Carter's Tutorial, perhaps?
Best,
Frank
18
Programming with NASM / Re: How are you multiplying by 3
« Last post by gygzkunw on February 22, 2021, 04:27:45 AM »Well... it's a "trick" using lea (load effective address) to "load" edx with edx plus edx times two... As you observe, zero times three is zero. You don't show it, but we then increment edx.. and ,,,I get its a trick but i don't see where the 3 is coming from...
I've never read "Step by Step"... I ASSume Jeff explains this...?
Best,
Frank
"Not only is this virtually always faster than shifts combined with adds, it’s also clearer from your source code what sort of calculation is actually being done. The fact that what ends up in EDX may not in fact be the legal address of anything is unimportant.LEA does not try to reference the address it calculates.It does the math on the stuff inside the brackets and drops it into the destination operand. Job over. Memory is not touched, and the flags are not affected.Of course, you’re limited to what calculations can be done on effective addresses; but right off the top, you can multiply any GP register by 2, 3, 4, 5, 8,and 9, while tossing in a constant too! It’s not arbitrary math, but multiplying by 2, 3, 4, 5, 8, and 9 comes up regularly in assembly work, and you can combine LEA with shifts and adds to do more complex math and ‘‘fill in the holes.’’ You can also use multiple LEA instructions in a row."
Is there a book you like to recommend? You dont seem to like the book...
19
Programming with NASM / Re: How are you multiplying by 3
« Last post by Frank Kotler on February 22, 2021, 03:07:34 AM »Well... it's a "trick" using lea (load effective address) to "load" edx with edx plus edx times two... As you observe, zero times three is zero. You don't show it, but we then increment edx.. and ,,,
I've never read "Step by Step"... I ASSume Jeff explains this...?
Best,
Frank
I've never read "Step by Step"... I ASSume Jeff explains this...?
Best,
Frank
20
Programming with NASM / How are you multiplying by 3
« Last post by gygzkunw on February 22, 2021, 12:21:15 AM »Code: [Select]
SECTION .bss
BUFFLEN equ 16 ; ----> Bufflen = 16
Buff: resb BUFFLEN ; ----> Buff will read 16 bytes from the buffer
SECTION .data ; Section containing initialised data
HexStr: db " 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00",10
HEXLEN equ $-HexStr ; ---->length of Hexstr
Digits: db "0123456789ABCDEF"
SECTION .text ; Section containing code
global _start ; Linker needs this to find the entry point!
_start:
nop ; This no-op keeps gdb happy...
; Read a buffer full of text from stdin:
Read:
mov eax,3 ; ----> syscall read
mov ebx,0 ; ----> read from standard input (keyboard)
mov ecx,Buff ; ----> where the string read is stored
mov edx,BUFFLEN ; ----> how many bytes is supposed to read
int 80h ; ----> sys call
mov ebp,eax ; ----> copy the number that where actually read into ebp
cmp eax,0 ; ----> sub 0 from eax and set the appropriate flag? (if user enters 0 shouldn't the program end?)
je Done ; ----> if the zero flag is set move tone the label done
; Set up the registers for the process buffer step:
mov esi,Buff ; ----> copy the content of Buff into esi
mov edi,HexStr ; ----> copy hexstr into edi
xor ecx,ecx ; ----> Clears ecx
; Go through the buffer and convert binary values to hex digits:
Scan:
s xor eax,eax ; ----> clears eax
; Here we calculate the offset into the line string, which is ecx X 3
mov edx,ecx ; ----> copy ecx which is 0 into edx
lea edx,[edx*2+edx] ; ----> This is where i am stuck. How are you able to multiply by 3 over here?
LEA = [BASE +( INDEX * SCALE) + DISP)]
if edx is 0, how are we able to multiply by 3I am not understand how the author is using the lea instruction
The code is from Assembly Language Step-By-Step by Jeff Duntemann
Thank you