pubg esp

// Big map. Text: 6f3a2060 sub_6f3a2060 proc near; Code xref: sub_6f38d120 + 67 P. Text: 6f3a2060; sub_6f39bca0 + 67 P.... Text: 6f3a2060. Text: 6f3a2060 var_8 = dword ptr-8. Text: 6f3a2060 var_4 = dword ptr-4. Text: 6f3a2060 arg_0 = dword ptr 4. Text: 6f3a2060 arg_4 = dword ptr 8. Text: 6f3a2060 arg_8 = dword ptr 0ch. Text: 6f3a2060 arg_c = dword ptr 10 h. Text: 6f3a2060. Text: 6f3a2060 sub ESP, 8. Text: 6f3a2063 mov eax, [

---restore content starts---Content one: Experimental report related instructions.Real name Chering HelpOriginal works reproduced please indicate the sourceLessons learned: Linux kernel Analysis MOOC courseLinks:http://mooc.study.163.com/course/USTC-1000029000Virtual Lab ExperimentContent two: Analysis of stack changes during the working process of assembler codeThe analysis is divided into two parts: (The label of the stack address is not aligned, please understand)The first part is the stack t

Explains stack changes for MAIN.C assembler codeThe experiment looks like this:The following analysis begins with the main function (three variable values from top to bottom are 4,7,4):Suppose you start with an empty stack, the initial position is 0,EBP=ESP, and the following is marked with a line number18:esp point to position (1), the value of position (1) is the address of the ESP19: Make EBP point to th

Transfer from http://blog.csdn.net/dongtingzhizi/article/details/6680050In-depth analysis of C + + function call processBrotherWeibo:-bing son of the dongting0. Introduction　　The process of a function call is actually a process of interruption, so how does a function call be implemented in C + +? How to implement parameters in the stack, function jump, protection site, reply to the scene? In this paper, the procedure of function call is analyzed and explained in detail, and it is demonstrated in

the contents of RAM inside pull. You can go to "see the snow" to learn a simple assembler command. 004f3b9c/$ PUSH EBX 004f3b9d |. 83C4 F8 ADD esp,-8 004f3ba0 |. 8BDA MOV Ebx,edx; Data Destination address after decryption 004f3ba2 |. 8bd4 MOV Edx,esp; Data Delivery Destination Address 004f3ba4 |. B9 04000000 MOV ecx,4; The number of passes is 4 004f3ba9 |. E8 12eef8ff call client.004829c0; Pass the 4 valu

the number of records replaced 0X2E, that is, ".", the completion of the processing. The function that handles the incoming domain name in Symdns.sys is at the Symdns.sys base address +0xa76, which allocates enough space in the stack (in fact, the last Shellcode execution is not performed on the stack, but in the non-paged pool). The incoming domain name has a maximum length limit and cannot exceed 0x40 bytes, so i shellcode length is 0x3f (63) bytes per paragraph. After overwriting 532 bytes,

Reverse basic OS-specpacific (1) Chapter 64 methods for passing Parameters 64.1 cdcel This method of passing parameters is popular in C/C ++. As shown in the following code snippet, the caller places parameters in the stack in reverse order: the last parameter, the second to the last parameter, and the first parameter. The caller must also restore the stack pointer (ESP) to the initial state after the function returns. Listing64.1: cdecl push arg3pus

// N // t/"/N ");Fprintf (Out, "/T/" movl/T % ESP, % EBP // N // t/"/N ");Fprintf (Out, "/T/" mov/T $ % d, % eax // N // t/"/N", sys_call_idx );Fprintf (Out, "/T/" Lea/T8 (% EBP), % edX // N // t/"/N ");Fprintf (Out, "/T/" int/T $ 0x2e // N // t/"/N ");Fprintf (Out, "/T/" popl/T % EBP // N // t/"/N ");Fprintf (Out, "/T/" RET/T $ % S // N // t/");/n", nr_args );......} '/T' in the Code indicates the Tab character. It should be fine for readers to read

. Global _ % S // N // t/"/N ", name );Fprintf (Out, "/". Global _ % S // N // t/"/N", name2 );Fprintf (Out, "/" _ % s: // N // t/"/N", name );Fprintf (Out, "/" _ % s: // N // t/"/N", name2 );# EndifFprintf (Out, "/T/" pushl/T % EBP // N // t/"/N ");Fprintf (Out, "/T/" movl/T % ESP, % EBP // N // t/"/N ");Fprintf (Out, "/T/" mov/T $ % d, % eax // N // t/"/N", sys_call_idx );Fprintf (Out, "/T/" Lea/T8 (% EBP), % edX // N // t/"/N ");Fprintf (Out, "/T/"

course, the disadvantage of simple instruction is inefficiency.The virtual machine protection technique we're talking about here is to change the register-based CPU code to the pseudo-code of the stack-based CPU. The pseudo-code is then interpreted by the stack-based virtual machine (CPU).Instruction SystemThe key is to design a virtual stack-based virtual machine (CPU) instruction system. The more concise the command system, the better the reusability.Or take the add command as an example. The

First of all, the children who have assembled the computer should know that the computer is composed of CPU, memory, hard disk, motherboard, power supply, of course, playing games of children's shoes will also be obsessed with video cards (such as GTX980 Tactical nuclear graphics)As long as these things are available, the computer can run.But how do these things work together and let the program run?First, we need to know something called the operating system, essentially, it is also a program,

#define SWITCH_TO (Prev,next,last) do { unsigned long esi,edi; ASM volatile ("pushfl\n\t" "PUSHL%%ebp\n\t" "MOVL%%esp,%0\n\t"/* Save ESP */ "MOVL%5,%%esp\n\t"/* Restore ESP */" MOVL $1f,%1\n\t"/* Save EIP */"

Student Bérénice Angremy Course Content Three magic weapon of computer• Stored program computer working model, the most basic logical structure of computer system;• Function call stack, the basis for high-level language operation, only machine language and assembly language when the stack mechanism for the computer is not so important, but with high-level language and functions, the stack becomes the basic function of the computer;• Interrupts, the base point of the multi-cha

, where the compaction stack is not really a stack, but the data is stored in a stacked way.Then there are two loops to place p corresponding byte values, note that this is p++, because the previous copy of the time is--p.Returns the SP when the data population is complete.Next, assign the start_stack of the current process to the page where P is located.Finally, it is worth noting that:Replace the code pointer on the stack with the program that originally called the system break with the entry

dtor_idx.7023 080483C0 T Frame_dummy 080483e4 T Main//Address of main function U [email protected] @GLIBC_2.0 Call Gcc-s xx.c to get the assembly code:[CPP]View Plaincopy . File "name_of_array.c" . Globl A . Data . Align 4 . Type A, @object . Size A, 8 //From here we know that sizeof (a) equals 8 A: . long 1 //From here you can see that the compiler directly converts the int in the. c file to a long type . Long 2 . section. Rodata . LC0: . String "a =

Author: Sandy Original works reproduced please indicate the source"Linux kernel Analysis" MOOC course http://mooc.study.163.com/course/USTC-1000029000 "Experimental environment: C+linux64 bit (32-bit system may result in different)In accordance with the academic integrity of the terms, I guarantee that the answer for my original, all the references to the external materials have been marked by provenance. ———————————-Everyone, I'm a split line ———————————————— C source code is as follows /******

. Now we WinDbg stopped on mysqld!mysql_parse . The one million question Is:everytime that any MYSQL Query executes something, it'll freeze my app until press F5 app? The answer is no, if we use a more intelligent breakpoint. We know the function mysql_parse , but are which memory address it is stored? A Call stack theory: Let's explain this if the process is starting a function and it pushes the function parameters to be used. Then what happens with

currently executing statement will be displayed each time the next assembly statement is executed. The following shows the changes in%esp,%EBP, and stack contents at each step: Procedure (the instruction in the command is actually a pending command) Initial Push $0x7 Call 0x80483e8 calls F (0x80483e8) Push%EBP executes the F function, F initializes the frame pointer, moves the base address of the

Label:1. Memory view command x.x/x with hexadecimal outputX/D Output in decimalX/C output with single characterX/i Disassembly – Typically, we will usex/10i $ip-20 来查看当前的汇编（$ip是指令寄存器）X/S output as a string(GDB) x/10i main 0X80483BB Push%ebp 0X80483BC 1:mov%esp,%ebp 0x80483be 3: and$0xfffffff0,%esp 0x80483c1 6:Sub$0x20,%esp 0x80483c4 9: Movl $0x0,0x1c (%

", and then press into the function return address (RET), and then jump to the function address followed by execution (here to add, The article describing the principle of buffer overflow under Unix mentions that after the RET is pressed in, it continues to press into the current EBP and replaces the EBP with the current ESP. However, there is an article that describes a function call under Windows that also has this step, but according to my actual d