att

Passing arguments from C to 64bit linux Assembly

烂漫一生 提交于 2019-12-08 09:39:03
问题 I'm trying to write my first assembly function to be called from a C program. Here is the .c part: #include <stdio.h> extern int sum(int a, int b); int main() { printf("2+3 = %d\n", sum(2,3)); return 0; } And the assembly part: .text .global sum sum: push %rbp mov %rsp, %rbp mov 8(%rbp), %rax add 12(%rbp), %rax pop %rbp ret According to some tutorials, arguments should be stored in %ebp (in 32bit version) register. However, when I compile the code above, I'm getting this result: 2+3 = 4195607

AT&T Assembly Masked Input

别来无恙 提交于 2019-12-08 09:12:59
问题 I'm trying to create a simple password program in AT&T Assembly but i'm having trouble with masking an input. What i want to happen is when the user enters characters, they appear on the screen as asterisk'. In intel syntax it's relatively simple: mov ah, 08h int 21h mov dl,2ah mov ah,02h int 21h This uses the intel command to read an input without echoing it and instead print an asterisk. I'm trying to solve this problem in AT&T syntax and I'm having some trouble. Any input would be greatly

64 bit assembly on Mac OS X runtime errors: “dyld: no writable segment” and “Trace/BPT trap”

混江龙づ霸主 提交于 2019-12-07 06:15:30
问题 When attempting to run the following assembly program: .globl start start: pushq $0x0 movq $0x1, %rax subq $0x8, %rsp int $0x80 I am receiving the following errors: dyld: no writable segment Trace/BPT trap Any idea what could be causing this? The analogous program in 32 bit assembly runs fine. 回答1: OSX now requires your executable to have a writable data segment with content, so it can relocate and link your code dynamically. Dunno why, maybe security reasons, maybe due to the new RIP

What does JL mean in at&t syntax?

不羁岁月 提交于 2019-12-06 21:34:24
I'm working on a project for school and I cannot find anything on the what JL means in at&t syntax. For reference, the question is to find the value of %eax when NOP runs. Here is the code it's used in: MOV $492,%ebx MOV $2494,%eax MOV $28063,%ecx CMP %eax,%ebx JL L1 JMP L2 L1: IMUL %eax,%ebx ADD %eax,%ebx MOV %ebx,%eax SUB %ecx,%eax JMP L3 L2: IMUL %eax,%ebx SUB %eax,%ebx MOV %ebx,%eax ADD %ecx,%eax L3: NOP Also I would appreciate what JMP does as well as how the addition/subtraction/multiplication works ( ADD / SUB / IMUL ). I don't want to cheat, I just want to understand what's happening.

is it certain in which register arguments and variables are stored?

时光怂恿深爱的人放手 提交于 2019-12-06 13:27:15
I'm still uncertain how registers are being used by the assembler say I have a program: int main(int rdi, int rsi, int rdx) { rdx = rdi; return 0; } Would this in assembly be translated into: movq %rdx, %rdi ret rax; I'm new to AT&T and have hard time predicting when a certain register will be used. Looking at this chart from Computer Systems - A programmer's perspective , third edition, R.E. Bryant and D. R. O'Hallaron: charter Is it certain in which register arguments and variables are stored? Only at entry and exit of a function. There is no guarantee as to what registers will be used

Puma Cell Venom Overkill skor sneakers Cycle

不想你离开。 提交于 2019-12-05 19:30:55
Din innersula kan vara 3mm solid plus yttersula kan vara 8mm solid. Därför hittar du 11mm som involverar säkerhet - men inte så mycket dessutom. För mig är denna specifika utformad för att gå på promenader tillsammans med ryggsäck för att utveckla din bas tillsammans med minskade fyrhjulingar med Nike M2K Tekno sneakers som är ett underordentligt skadahot jämfört med jogging. Jag sprang över det omedelbart efter långvariga guider i konkreta såväl som betong som vår fot hamnade ytterligare smärtsamt jämfört med att följa mycket samma långa sträcka i spår. Min make och jag gick med avseende på

Defining Bytes in GCC Inline Assembly in Dev-C++(.ascii in AT&T syntax on Windows)

三世轮回 提交于 2019-12-05 16:42:30
问题 The code below is just showing a Message Box on the screen. The addresses are hardcoded to facilitate: int main () { asm("xorl %eax, %eax \n" "xorl %ebx, %ebx \n" "xorl %ecx, %ecx \n" "xorl %edx, %edx \n" "pushl %ecx \n" //$0x0 "pushl $0x20206c6c \n" //" ll" "pushl $0x642e3233 \n" //"d.23" "pushl $0x72657375 \n" //"resu" "movl %esp, %ecx \n" //store "user32.dll" address in %ecx "movl $0x7c801d7b, %ebx \n" //store address of LoadLibraryA in %ebx "pushl %ecx \n" "call *%ebx \n" "movl

Adidas Tubular Shadow skor vara mest involverad

我的未来我决定 提交于 2019-12-05 12:01:57
Skulle du Köp Adidas Tubular Shadow skor vara mest involverad att sikla i de mest uppdaterade Nike Glide Turbo bara två såväl som Adidas Ultraboost tjugo löparskor, jag är bara Nike-sneakers på antalet rabatterade försäljningar, du kommer att glömma. Du kommer att hitta otaliga mycket bra jogging. Puma-företaget Mobil eller bärbar Venom Overkill sneakers sneakers gör ingen person verkligen referenser utanför husets joggingarenor, tillsammans med en kan också vara Hoka En speciell En speciell. Noll, det är inte en ny typ . Exakt varför du behöver förstå vad gäller ett företag som liknar Hoka

64 bit assembly on Mac OS X runtime errors: “dyld: no writable segment” and “Trace/BPT trap”

拥有回忆 提交于 2019-12-05 09:16:36
When attempting to run the following assembly program: .globl start start: pushq $0x0 movq $0x1, %rax subq $0x8, %rsp int $0x80 I am receiving the following errors: dyld: no writable segment Trace/BPT trap Any idea what could be causing this? The analogous program in 32 bit assembly runs fine. OSX now requires your executable to have a writable data segment with content, so it can relocate and link your code dynamically. Dunno why, maybe security reasons, maybe due to the new RIP register. If you put a .data segment in there (with some bogus content), you'll avoid the "no writable segment"

CSS3常用新特性

本小妞迷上赌 提交于 2019-12-04 21:37:11
CSS3的新特性 新增CSS3特性有兼容性问题,ie9+才支持 移动端支持优于PC端 新增选择器和盒子模型以及其他特性 CSS新增选择器 属性选择器 属性选择器可以根据元素特定属性来选择元素,这样就可以不用借助于类或者id选择器 选择符 简介 E[att] 选择具有att属性的E元素 E[att="val"] 选择具有att属性切属性值等于val的E元素 E[att^="val"] 匹配具有att属性且值以val开头的E元素 E[att$="val"] 匹配具有att属性且值以val结尾的E元素 E[att*="val"] 匹配具有att属性且值中含有val的E元素 【注意】类选择器、属性选择器、伪类选择器,权重为10 结构伪类选择器 选择符 简介 E:first-child 匹配父元素中元素的第一个子元素E E:last-child 匹配父元素中最后一个E元素 E:nth-child(n) 匹配父元素中的第n个子元素E E:first-of-type 指定类型E的第一个 E:last-of-type 指定类型E的最后一个 E:nth-of-type(n) 指定类型E的第n个 nth-child(n)选择某个父元素的一个或多个特定的子元素 n可以是数字,关键字和公式 n如果是数字,就是选择第n个子元素,里面数字从1开始 n可以是关键字:even偶数,odd奇数 n可以是公式