x86-64

一. 乱糟糟的概念 指令集体系结构（ISA ）：一个处理器支持的指令和指令的字节级编码（每条指令被编码成由一个字节序列或多个字节序列组成的二进制格式） 不同的处理器家族有不同的ISA（例如Intel IA32和x86-64、IBM/Freescale Power和ARM处理器家族）有不同的ISA 同一处理器家族的不同处理器有相同的ISA ISA为编译器编写者和处理器设计人员之间提供了一个概念抽象层：编译器编写者只需知道允许哪些指令，指令的字节级编码是什么；而处理器设计者只需建造出执行这些指令对应的二进制编码的处理器。 计算机科学的重要思想：用巧妙的方法在提高性能的同时又保持一个更简单、更抽象模型的功能。 数字硬件设计：基本构件块，如何连接以及操作，硬件控制语言（硬件系统控制部分的简单语言，用它来描述处理器设计） 二. 创建Y86-64指令集 较x86-64处理器支持的指令集而言，Y86-64指令集的 数据类型 、 指令 、 寻址方式 较少。 好处在于： 字节级编码简单，CPU译码逻辑简单 。 可以实现一些处理整数的程序。 在这一章中，我们会学习到如何用这个新定义出来的Y86-64指令集来设计处理器，并自己设计出来一个处理器， CMU设计了一些研究和测试处理器设计的工具，包括：Y86-64的汇编器，以及运行Y86-64程序的模拟器

Lets say, you have an application, which is consuming up all the computational power. Now you want to do some other necessary work. Is there any way on Linux, to interrupt that application and checkpoint its state, so that later on it could be resumed from the state it was interrupted? Especially I am interested in a way, where the application could be stopped and restarted on another machine. Is that possible too? In general terms, checkpointing a process is not entirely possible (because a process is not only an address space, but also has other resources likes file descriptors, and TCP/IP

How can I see what values the registers hold? I have the following line of assembly: mov 0x8(%rax), %rax cpm %ebx, (%rax) Using the command: (gdb) p/x $ebx (gdb) p/x $rbx $3 = 0xb I get the value that is stored in this register. However, when I try to see what it is stored int the memory location (%rax) I have the following problem: (gdb) display *(int *)$rax Disabling display 10 to avoid infinite recursion. 10: *(int *)$rax = Cannot access memory at address 0x17 I cannot not understand why this happens and moreover how to find out what (%rax) has in. To see the values of the registers just

I have been looking for an practical tool that would print the opcodes of any Intel 64-bit or 32-bit instruction in Linux, eg. something like Hiew's assembler in DOS. A web-based service would be one option too. As I wasn't able to find any, I made my own bash script, that creates an assembly source file from command line parameters (instruction[s] and <32/64>), compiles, links and disassembles it and shows the correct rows of disassembly. But is there already some program that would show all the possible encodings for any given instruction, eg. for mov eax,ebx ? My approach using nasm , ld

2019年10月14日 1.指令集体系结构（Instruction-Set Architecture, ISA）:一个处理器支持的指令和指令的字节级编码. 2.硬件控制语言（Hardware Control Language, HCL）:一种描述硬件系统控制部分的简单语言. 4.1 Y86-64指令集体系结构 状态单元、指令集和他们的编码、一组编程规范、异常事件处理 4.1.1 程序员可见的状态 4.1.2 Y86-64指令 1. Y86-64 vs x86-64 ： Y86-64指令集是x86-64的子集，只包含8字节(q)操作 前者简单，后者紧凑(常数值编码灵活、字段不固定) 2. Y86-64指令 (1)movq指令:irmovq,rrmovq,mrmovq,rmmovq（格式为: 源+目的+movq,源和目的包括i(instant number),r(register),m(memory);注意,不允许immovq或mmmovq） (2)4个整数操作:addq,subq,andq,xorq（注意，只对寄存器操作，同时设置条件码ZF、SF、OF） (3)7个跳转指令:jmp,jle,jl,je,jne,jge,jg (4)6个条件传送指令:cmovle,cmovl,cmove,cmovne,cmovge,cmovg（注意：只能寄存器->寄存器） (5)call,ret (6

I know that OS X is 16 byte stack align, but I don't really understand why it is causing an error here. All I am doing here is to pass an object size (which is 24) to %rdi, and call malloc. Does this error mean I have to ask for 32 bytes ? And the error message is: libdyld.dylib`stack_not_16_byte_aligned_error: -> 0x7fffc12da2fa <+0>: movdqa %xmm0, (%rsp) 0x7fffc12da2ff <+5>: int3 libdyld.dylib`_dyld_func_lookup: 0x7fffc12da300 <+0>: pushq %rbp 0x7fffc12da301 <+1>: movq %rsp, %rbp Here is the code: Object_copy: pushq %rbp movq %rbp, %rsp subq $8, %rsp movq %rdi, 8(%rsp) # save self address