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Linux x86-64 Hello World and register usage for parameters

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问题 I found this page which has a Hello World example for x86-64 on Linux: http://blog.markloiseau.com/2012/05/64-bit-hello-world-in-linux-assembly-nasm/ ; 64-bit "Hello World!" in Linux NASM global _start ; global entry point export for ld section .text _start: ; sys_write(stdout, message, length) mov rax, 1 ; sys_write mov rdi, 1 ; stdout mov rsi, message ; message address mov rdx, length ; message string length syscall ; sys_exit(return_code) mov rax, 60 ; sys_exit mov rdi, 0 ; return 0

Linux x86-64 Hello World and register usage for parameters

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自制操作系统Antz(1)——Boot Sector

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0. 引子　　最近在看操作系统底层方面的东西，最开始的为什么是07c00h这个问题就让我对操作系统有了很大的兴趣。所以准备在看书之余顺便写一个操作系统(Anz)。至于为什么这个系统会被叫做Antz,可以参考Antz Uhl Kone(某个日漫男主的名字), 日语为アインズ·ウール·ゴウン，与之对应的还有接下来准备写的自制脚本语言AntzScript，因为准备是用Java实现解释器，所以如何把AntzScript运行在Antz上是一个很大问题（其实问题就是引入Java）。　　Antz系统更新地址： https://www.cnblogs.com/LexMoon/category/1262287.html 　　Linux内核源码分析地址： https://www.cnblogs.com/LexMoon/category/1267413.html 1.关于Boot Sector 　　引导扇区(Boot Sector) 通常指设备的第一个扇区，用于加载并转让处理器控制权给操作系统。　　 1.1 主引导扇区　　　　硬盘的0柱面、0磁头、1扇区称为主引导扇区，也叫主引导记录 MBR ，该记录占用512个字节，它用于硬盘启动时将系统控制权转给用户指定的、在分区表中登记了某个操作系统分区。MBR的内容是在硬盘分区时由分区软件写入该扇区的，MBR不属于任何一个操作系统

Intel assembly syntax OFFSET

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问题 Now that i know u can use gcc for Intel syntax instead of default at&t with gcc -S -masm=intel test.c There is this line mov DWORD PTR [ebp-16], OFFSET FLAT:base Is it the same as mov dword[ebp-16], base ? Otherwise what must i do? 回答1: Yes, mov dword [ebp - 16], base should be fine. I haven't seen offset flat: for a while - I think it's obsolete, but it's what AT&T's idea of .intel_syntax used to demand (I had to look at Gas's source code to find that out). Means the same as offset to Masm,

x86汇编入门

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Intel® 64 and IA-32 Architectures Software Developer’s Manual 文档官方地址： https://software.intel.com/sites/default/files/managed/39/c5/325462-sdm-vol-1-2abcd-3abcd.pdf 很好的入门文档：https://www.ibm.com/developerworks/cn/linux/l-assembly/ 汇编语言由三部分组成：汇编指令：跟机器指令一一对应，实际就算一个个的助记符。有 AT&T 和 INTEL 两种语法伪指令：给编译器看，用于指示编译器该如何做。不同编译器的语法不同，有 NASM、GNU as 和 MASM 三种常用编译器运算符：±*/ 之类的符号汇编指令具体可以参考：https://blog.csdn.net/kennyrose/article/details/7575952 AT&T 和 INTEL 两种语法的主要差异有： - AT&T INTEL 操作数方向从左向右从右向左立即数表示方式 $0x01 30h 寄存器表示方式 %eax eax 助记符指定操作数长度 b8位，w16位，l32位， movl $lb, %eax mov eax, dw ptr lb 长跳转和调用 ljmp $sect,

Loading a register from a “db 0” doesn't load a 0 into EAX?

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问题 I've been bashing my head against the wall for over an hour and I can't understand why the below doesn't work. If I change b: db 1 to b: db 0 then it should print 10, otherwise it should print 0. Instead, the program always prints 10. I've been writing a project that writes assembly and this is one of the unit test that fails and I just don't get it. It has to be something simple. extern printf, exit section .bss section .data b: db 1 x: dd 5 y: dd 5 z: dd 0 int_pattern: db "%i", 10, 0 global

Loading a register from a “db 0” doesn't load a 0 into EAX?

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Function that takes a char array and 2 indices; swapping the chars in those indices

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问题 This is my function prototype: char* swap(char* array, int index1, int index2); This is my assembly code: segment .text global swap swap: mov r14,[rdi+rsi] mov r15,[rdi+rdx] mov [rdi+rsi],r15 ;this line segfaults mov [rdi+rdx],r14 mov rax,rdi ret The lines mov [rdi+rsi],r15 and mov [rdi+rdx],r14 give me a segfault; I'm not sure where I'm going wrong The calling function: #include <stdio.h> #include <stdlib.h> extern char* swapLetters(char* str, int indexA, int indexB); int main() { char* st=