esp 3250

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Implementation of the C language interpreter-run the script (6)

ESP - 4 return-address 0 0 C 4 d 8 E 12 @ 1 16 @ 2 ------------ EIP indicates the current command location when the function is called. When the function returns, we need to pop this eip and continue executing the next command of EIP.ESP indicates the starting position of the variable space of the current function when the function is called, that is, the caller's esp. When the functi

(11) Generic Programming/Function Templates/class templates

point where it is called-compile the *///template code after the call, and tell the compiler that it starts the generic programming//typename tells the compiler that the type name is T compiler you see type T don't mess up the error .... T is type, type is parameterized just templateThe g++-s 1.cpp generates 1.SAnalysis 1.s. File "1.cpp". Lcomm __zstl8__ioinit, 1, 1.def ___main;. SCL 2;. Type 32;. Endef.section.rdata, "Dr" Lc2:.ascii "hello....\0" LC3:. ASCII "Pause\0". Text. Globl _main. def _

Natural code Input Method

INT3 017f:1003d211 7c24 JL 1003d237 (NO JUMP) 017f:1003d213 0801 OR [ecx],al 017f:1003d215 0f8581010000 jnz NEAR 1003d39c 017f:1003d21b Pusha 017f:1003d21c be00a00210 MOV esi,1002a000 "R eip eip-1", "D EIP", the 017f:1003d210 place to 80H: 017f:1003d210 807c240801 CMP BYTE [esp+08],01 017f:1003d215 0f8581010000 jnz NEAR 1003d39c 017f:1003d21b Pusha 017f:1003d21c be00a00210 MOV esi,1002a000 017f:1003d221 8DBE0070FDFF LEA edi,[esi+fffd7000]

Operating system Experiment Four experiment report _ Operating system

; uint32_t tf_eflags; /* below here is crossing rings, such as from user to kernel, defined by hardware///only if the privilege level change occurs, the additional information here is guaranteed by the hardware pressure stack Save uintptr_t Tf_esp; uint16_t Tf_ss; uint16_t Tf_padding5; } struct Context {uint32_t eip; uint32_t esp; uint32_t ebx; uint32_t ecx; uint32_t edx; uint32_t ESI; uint32_t EDI; uint32_t EBP; }

Stack overflow case analysis under Linux-gdb debug Drills

4 bytes are filled with EBP, the last 4 bytes are filled with the RET address, so it is supposed that the EIP here should be 0x65656565, then why is this 0x61616161, just the value of AAAA?According to the results of single step debugging, it is found that the EIP becomes 0x61616161 after the main function exits, and the EIP becomes 0x65656565 when the overflow exits.Why does overflow return to the main function after exiting? Possible cause: The input string does not overwrite the RET address,

Remove the taskbar icon

will break down the BPX shell_policyicona breakpoint and use F12 to check if the software is called and the parameters are used! First come to the following: Here is where the software is called at startup: * Possible reference to string resource id = 00114: "CCProxy"|: 00408770 6a72 push 00000072: 00408772 51 push ECx: 00408773 c681_f0000000005 mov byte PTR [esp + 000024f4], 05: 0040877b e8c0890100 call 00421140: 00408780 83c408 add

Introduction to how computers work

instruction and the program3. Put the written program and data into memory and start the computer work.The CPU work process can be broadly divided into three steps: Take the instruction, analyze the instruction and execute the instruction. Where the staging of both the instruction data and the address occurs in the register. The main object of assembly language is register. The register of the CPU is mainly divided into general register, control register and segment register three kinds. The In

PWNIUMCTF2014-JJSN Summary

]} Access to the address in the topic, you can get a file, open after The file header is a lpck, do not know what format, but after seeing the MZ head. Delete all the parts before the MZ head, and successfully get an EXE. Requires input password. No way, just throw it into Ida ... find the key sections below: . text:00401363 mov [esp+5ch+var_5c], offset apassword; "Password:" . text:0040136a Call puts . text:0040136f Lea EAX, [

BSS segment, data segment, code snippet for target file under Linux

c28b4424 1801c28b 44241c01 d0890424. d$ .... d$.....$ 0050 e8fcffff ff8b4424 18c9c3 ... d$ ... Contents of section. Data: 0000 54000000 00000000 55000000 T .... U ... Contents of section. Rodata: 0000 68656c6c 6f20776f 726c6421 0025640a Hello world!. %d. 0010 00323232 323200.22222. Disassembly of section. Text: 00000000 0:55 Push%EBP 1:89 e5 mov%esp,%ebp 3:83 EC Sub $0x18,%esp

Talking about how the computer works and the simple disassembly operation under Linux __linux

and other hardware. How to drive these hardware is done by software. Programming is done on the basis of the operating system through the operation of the Code and then compiled by the compiler to execute. In Linux we have a disassembly to test how C code is converted into assembly code. We do this on the lab floor, the C code is a simple three function (a main function and two called functions): Disassembly with GCC under Linux: Gcc-s-O main.s main.c-m32 You can get the assembly-level code f

Talk about what functions on the device can achieve Network Security

network security. Because VLANs can isolate broadcast, different VLANs cannot communicate with each other, so they have a certain degree of security. 5. Port isolation With the port isolation feature, you can add the ports to be controlled to an isolation group to isolate ports in the isolation group from Layer 2 and Layer 3 data, enhancing network security, it also provides users with flexible networking solutions. Currently, only one isolation group can be set up for one device. The number of

Csapp Buffer Overflow attack Experiment (top)

causes some data on the stack to be overwritten. This results in a buffer overflow crisis:int getbuf(){ char buf[12]; Gets(buf); return1;}2.2 Buffer Stack AnalysisBefore you start a real "attack". Let's start by analyzing what the stack looks like when Bufbomb calls Getbuf ().Only a comprehensive understanding of the stack structure. We were able to "attack" it at our own pace at the back of the experiment.First, through the Objdump disassembly Getbuf () function:[Email protected] bufb

Stack allocation in C Language (digging holes)

We all know that local variables are stored in the stack during the C language operation, and the space is allocated from high to low. However, recently I encountered a program that made me a little confused. First look at a program. Obviously, addresses are allocated from high to low, just as expected. Modify it a little and then run it. Obviously, from low to high !!! Clarify the problem: the stack should change the memory allocation mode for the memory occupied by local variables. Why? W

Example analysis of memory structure of C program runtime

Prior Knowledge Static variables are stored in static storage, local variables are stored in dynamic storage (stacks), code is stored in code area Register, EBP points to the bottom, esp points to the top of the stack, the EIP points to the next instruction that is executing the instruction, three registers are stored in the address, 32-bit system, the address is 4 bytes is a DWORD All statements written in the function definition are

Linux kernel Analysis (ii)--analyzing process switching principles from a simple Linux kernel __linux

switching the most important is the broker's storehouses Switch and EIP (that is, program counters) the correct jump, mymain.c in the function My_start_kernel is the first code to execute, so from this function to start analysis. The My_start_kernel function first established 4 processes and initialization, such as the allocation stack, note that at the time of the first set up only No. 0 process status is runuable, the rest are unrunnable. There is the PCB structure of the THREAP.SP, each proc

C + + virtual function compilation

Refer to this article: http://blog.csdn.net/linyt/article/details/6336762classBase { Public: Virtual voidf () {}Virtual voidg () {}}; classDerive: PublicBase { Public: Virtual voidf () {}}; intMain () {Derive D; intA =4; Base*PB; intb =5; PB= D; intc =6; PB-f (); return 0; } %ESP| ...| | | ------>Derive d| | | ------>int C| | | ------>base *PB| | | ------>int b| | | ------>int a| +. file"main.cpp". Section. Text$_zn4base1fev,"x

Linux Kernel fragmentation notes

# The lower half of the interrupt.Common_interrupt:Addl $-0x80, (% esp)/* Adjust vector into the [-256,-1] range */# Adjust the interrupt number to 0 ~ The value range is 255.SAVE_ALLTRACE_IRQS_OFFMovl % esp, % eaxCall do_IRQ # process the Interrupt Routine.Jmp ret_from_intrENDPROC (common_interrupt) Unsigned int _ irq_entry do_IRQ (struct pt_regs * regs){Struct pt_regs * old_regs = set_irq_regs (regs ); /*

Trojan virus detected by program behavior tracking

obvious. First, you must write code for every API function that you are interested in. Second, you cannot HOOK all the API functions, due to the lack of complete API call sequence for reference, some small actions may be missed during log analysis.Another idea is to use the debugging technology to set breakpoints in advance at the entrances of all introduced DLL functions, and obtain parameters through stack information during debugging. Both IDA pro and OllyDbg can be used for dynamic debuggin

[Assembly Learning] For beginners in assembly-function call stack Change Analysis (from hacker situation)

% instead of D.36 call _ Test37. Push eax; assume that the address of push eax is XXX.38 push offset sztextfmt39 call printf40 add ESP, 841 RET42_main endp43end _ main4445; /**///////////////////////////////////// //// // The stack changes are described below46 The first thing to understand is that the operation of the stack segment SS can only use ESP or EBP registers, other registers eax ebx edx and so o

Functions and assembly

Document directory 1. Function Parameters 3. Reference This article is based on Win32 compilation.1. Function Parameters The growth direction of function stack is opposite to that of address, from high address to low address. ESP points to the top of the function stack, and EBP points to the bottom of the function stack. Sub esp xxx pushes memory blocks of XXX length into the stack Add

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