esp 3250

data that is larger than 4 bytes occupies a 4-byte integer multiple in the stack.3) The two registers associated with the operation of the stack are the EBP register and the ESP register, and in this article you only need to interpret EBP and ESP as 2 pointers. The ESP register always points to the top of the stack, and when the push command presses the data int

results are as follows:　　One of the first two cases overflowed, and only the third was normal. And then we'll look at their assembly code, which is the assembler code I disassembled with Objdump:　　1 intMainintArgc,char *argv[])2 {3 8048394: - Push%EBP4 8048395: theE5mov%ESP,%EBP5 8048397: theE4 F8 and$0xfffffff8,%esp6804839a: theEc - Sub$0x30,%esp7Long Muln =203879;8804839d: C7 - - 0c the1cGenevamovl $0x31c6

caches.Iv. Types of registersThe early x86 CPU had only 8 registers, and each had a different purpose. Now there are more than 100 registers, all become general-purpose registers, not specifically used, but the names of the early registers are preserved. EAX EBX Ecx EDX Edi Esi Ebp Esp Of the above 8 registers, the first seven are universal. ESP registers have a sp

, click OK, and come here.0045E317/$8B4424 08 mov eax, dword ptr ss: [ESP + 8]0045E31B |. 56 PUSH ESI0045E31C |. 85C0 test eax, EAX0045E31E |. 8BF1 mov esi, ECX0045E320 |. 75 08 jnz short 1_ra_m.0045e32a0045E322 |. E8 AEED0000 CALL history ra_m.0046d0d50045E327 |. 8B40 10 mov eax, dword ptr ds: [EAX + 10]0045E32A |> 85F6 test esi, ESI0045E32C |. 75 04 jnz short 1_ra_m.0045e3320045E32E |. 33C9 xor ecx, ECX0045E330 |. EB 03 jmp short 1_ra_m.0045e3350045

Gas multi-file Engineering From: programming from groundup Chapter 6 The main purpose is to understand how to organize multi-file projects (Forms) File record-def.s: . Equ record_firstname, 0 . Equ record_lastname, 40 . Equ record_address, 80 . Equ record_age, 320 . Equ record_size, 324 File Linux. S: # Common Linux Definitions # System call numbers . Equ sys_exit, 1 . Equ sys_read, 3 . Equ sys_write, 4 . Equ sys_open, 5 . Equ sys_close, 6 . Equ sys_brk, 45 # System Call interrupt number . Eq

consists of Security Consortium (Security Association, SA) [1] and security protocols (including authentication header (AH) [2]. Security load ESP (Encapsulating Security Payload, ESP) [3]. Key Management (Internet Key Management, Ike) [4] and authentication and encryption algorithms constitute a complete security system. Structure 1.2.1 Security Alliance (SA)Security Alliance is the basis of IPSec. It det

I hope you can come up with some ideas ~ Check whether there is any shell... Microsoft Visual C ++ 6.0, indicating no shell ~ Let's enter an account and password for trial run ~ The error message "incorrect registration name or registration code!" is displayed !" Okay. Let's load it with OD ~~ Search for asii, find the error message, and double-click it to go to the code ~ 00401d00/0f85 df1_00 jnz ultradic.00401de500401d06. | 68 01100000 push 100100401d0b. | 68 f4704300 push ultradic.004370f4; A

plays an important role in the program operation. Most importantly, the stack holds the maintenance information needed for a function call, known as a stackframe, and a stack frame of a function (the called function) generally includes the following elements: (1) Function parameters, the default invocation convention, in the Order of the right and left to press the parameters into the stack in turn. Executed by the function caller. (2) The return address of the function, that is, the address of

Author: uuk[Software name]: Total Commander[Software Version]: 7.56a[Shelling method]: new version without shelling[Programming language]: Borland Delphi 2.0 [Overlay][Tools]: OD PEID IDA[Operating platform]: Windows XP[Software introduction]: A pretty good dual-column File Management Software[Author's statement]: it is only for research purposes. Please purchase a genuine version for use.We know that Total Commander has self-verification. by tracking the CreateFile and ReadFile functions, it is

carried out in assembly, let's talk about some personal opinions. Next, we will conduct some small tests and explain them in assembly language. You can do it together. (1) Char name [] and char * Name [CPP] View plaincopy 1: 2:VoidProcess () 3 :{ 00401020 push EBP 00401021 mov EBP, ESP 00401023 sub ESP, 4ch 00401026 push EBX 00401027 push ESI 00401028 push EDI 00401029 Le

and receiver are not the hostTunnel is used to provide original group protection for intrusions from the sender and receiver. It seems that the entire group is transmitted through a hypothetical tunnel.1. 2. Two security protocolsIPSec defines two security protocols: the Authentication Header (AH) protocol and the encapsulation security load (ESP) protocol, which provide authentication and/or encryption at the IP layer.1.2.1. header Authentication Pr

Grasping Writer: Li Pengju No.: 20132201(* Original works reproduced please specify the source *)(Study course: "Linux kernel Analysis" MOOC course http://mooc.study.163.com/course/USTC-1000029000)Experimental section (the following command is available for lab building 64-bit Linux virtual machine environment, 32-bit Linux environment may be slightly different)Compile the assembly code using the gcc–s–o main.s main.c-m32 command.The code is as follows:int g (int x) { return8;} int f (int x)

talk about some personal opinions. Next, we will conduct some small tests and explain them in assembly language. You can do it together. (1) char name [] and char * name 1: 2: void process () 3 :{ 00401020 push ebp 00401021 mov ebp, esp 00401023 sub esp, 4Ch 00401026 push ebx 00401027 push esi 00401028 push edi 00401029 lea edi, [ebp-4Ch] 0040102C mov ecx, 13 h 00401031 mov eax, 0 CCCCCCCCh 00401036 re

Wang Yi Original Works reproduced please specify the source "Linux kernel Analysis" MOOC course http://mooc.study.163.com/course/USTC-1000029000The core of this lesson is to complete the scheduling of the process through the interrupt mechanism, in this course, __init My_start_kernel as an entry function, define the T-PCB structure of process No. 0, copy to make other process TPCB data structure, The interrupt Time function is My_timer_handler periodic call to modify the value of the my_need_sch

(void){int pid = 0;int i;/* Initialize Process 0*/Task[pid].pid = pid;Task[pid].state = 0;/*-1 unrunnable, 0runnable, >0 stopped * *Task[pid].task_entry = task[pid].thread.ip= (unsigned long) my_process;TASK[PID].THREAD.SP = (unsignedlong) task[pid].stack[KERNEL_STACK_SIZE-1];Task[pid].next = task[pid];/*fork More Process */for (i=1;i{memcpy (task[i],task[0],sizeof (TPCB));Task[i].pid = i;Task[i].state =-1;TASK[I].THREAD.SP = (unsignedlong) task[i].stack[KERNEL_STACK_SIZE-1];Task[i].next = Task

WusiDeep understanding of the function call stackA stack is a space that the C language program must run with a record call path and parametersThe role of the stack Function call Framework Passing parameters Save return address Provides local variable space Stack-related registers ESP, stack pointer, pointing to the top of the stack EBP, base point pointer, to the bottom of the stack, used in C to record the curr

In the Win32 assembly, we often have to deal with the API, but also will often use their own API-like with the parameters of the subroutine, this article is about the subroutine call in the process of the concept and analysis of parameter transfer. Generally in a program, the passing of parameters is carried out through the stack, that is, the caller presses the parameters to be passed to the subroutine (or the callee), and the subroutine takes the corresponding value out of the stack and uses i

", and the above is just the key value of the registry, the bottom line "you can still use this program % d ". In fact, this is the number of times you can use it. There should be a calculator here, but you don't have to worry about it. first go to the code and find a suitable place for breakpoint.004017DA 68 00014300 multi-open push program. 00430100; SOFTWAREAutoSoftMultRun2004017DF 68 02000080 push 80000002004017E4 FF15 0C604200 call dword ptr ds: [......004017FE FF15 08604200 call dword ptr

Let's first look at the static compilation result of a simple code: #include "stdafx.h"int _tmain(int argc, _TCHAR* argv[]){01041380 55 push ebp 01041381 8B EC mov ebp,esp 01041383 81 EC C0 00 00 00 sub esp,0C0h 01041389 53 push ebx 0104138A 56 push esi 0104138B 57 push

Stdcall: If you see EBP + a few, you have to ret 4 * n unless the proc uses parameter defined by the pseudo command, the parameter C add ESP, 8 after caller call addtewo 1. Why sometimes an error occurs when RET is returned: stdcall Convention subprogram addtwo needs to clear parameters Description of RET 8 (mov ESP, add ESP after EBP ret, 8 clear parameters)