edx mba

operation is directly applied to the memory, instead of using registers. The register is generally used as a required command or used to make the program run faster,("Sidt % 0/N": "M" (LOC); indicates that the input variable loc does not pass the Register Mathing (DIGIT) ConstraintsWhen a variable is used in input and output, "0", "1", "2"... (I think so)ASM ("incl % 0": "= A" (VAR): "0" (VAR ));Store VaR in the register % eax first, add 1 to the register, and then send it to the VaR Common Emb

Storage program computer (the stored programs computer) Memory store directives and data CPU Interpretation Instructions X86 implementation (X86 implementation)Registers (Register) Universal register (general-purpose registers) Segment Register (Segment registers) Flag Register (EFlags register) x86_64 Register (x86_64 registers) Memory MOVL%eax,%edx　　

a link ): Https://raw.githubusercontent.com/CodingLi/test/master/exp_buffer Enable ie in Windbg Then g runs the program and an exception occurs. The exception is as follows: 0:000>g(830.fb8):Accessviolation-codec0000005(firstchance)Firstchanceexceptionsarereportedbeforeanyexceptionhandling.Thisexceptionmaybeexpectedandhandled.eax=0000004aebx=022cde82ecx=0013e140edx=00140000esi=0013df00edi=0013e140eip=77c12332esp=0013dea0ebp=0013deaciopl=0nvupeiplnznaponccs=001bss=0023ds=0023es=0023fs=003bgs=0

0x00310000Two, custom structureConclusion: A custom structure can be thought of as an array.Custom structure, as a parameter, will put all member variables, one by one into the stackIf you pass a custom structure pointer, only the address is passed.Global custom struct-body variables, and global fixed-length array classes swab.The program loads, like code, already in memory, into the static zone.Uninitialized put 00 data,The name of the variable appears in the code and is replaced by the addres

= dword ptr 4 . TEXT:004E8EF0 Arg_4 = dword ptr 8 . TEXT:004E8EF0 Arg_8 = dword ptr 0Ch . text:004e8ef0 . TEXT:004E8EF0 mov ecx, [esp+arg_0]; Could be Len address of length . TEXT:004E8EF4 Push EBX . TEXT:004E8EF5 mov eax, [esp+4+arg_4]; Cache Address . TEXT:004E8EF9 push ESI . TEXT:004E8EFA mov esi, ecx . TEXT:004E8EFC Push EDI . TEXT:004E8EFD mov edi, [esp+0ch+arg_8]; Encrypt key Address . Text:004e8f01 and ESI, 3; the remainder that corresponds to the cache length divided by 4 . text:004e8f0

], 10; (int*) PST = ten mov edx, a mov eax, [ebp+var_4]; eax = PST mov [eax+4], DX ; (word*) (pst+4) = a mov ecx, [ebp+var_4]; ecx = PST mov byte ptr [ecx+6], 30; (byte*) (pst+6) = 30, verified that the second is word byte mov edx, [ebp+var_4]; edx = PST mov dword ptr [edx+8], 40; (int*)

receiving the advice given by the teacher, I thought twice and finally decided to go to the Graduate School of Management of mcgiori, majoring in MBA. "Frankly speaking, at the beginning, I was concerned about the technology itself, and I was concerned about the technical logic. As for its purpose, I think this should be something someone else should consider. Later, especially during my MBA period, I expa

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 $0x31c67,0xc (%ESP)980483A4:xx TenLong Long MULNL =2

, at least I have never met either: DThe above is a conclusion, that is, if we get any of the two addresses, We can get another one, as long as you know the offset between them.The first step is to get this address, but the address in the memory is dynamically changed and useless. Here I will teach you to change it to static, so that it will never change! I will continue to take "Chu Liuxiang xinchuan" as an example. If you have this tour, you will do it with me. It doesn't matter if you don't h

, this software cannot obtain the required sensitive characters in strings and memory. It is difficult to break down. Fortunately, you can find the ASCII "all" through the plug-in and take it off. //////////////////////////////////////// //////////////////////////////////////// //////////////= Check the number of users =00434CA4 |. 50 push eax;/Arg1 = 0012F78000434CA5 |. E8 AAE2FFFF call un_SDE3S.00432F54; un_SDE3S.00432F5400434CAA |. 59 pop ecx00434CAB |. 83C0 04 add eax, 400434CAE |. 8945 D4 m

Author: wangweilll These times have been playing CrackMe learning algorithm analysis ~, Today, I am looking for a foreign software to analyze analysis algorithms.I am not looking for any software in our country ~~~~!!!!!!!!!!!!!Because software writing is a great deal of effort ~~, It is not good to publish their algorithms ~~~Let's talk about the analysis.My username is: wangweiThe false code I entered is: 12345678900546FC3 51 push ecx00546FC4 53 push ebx00546FC5 56 push esi00546FC6 57 push ed

\ xe7 \ xff \ xe7" Put this tag in front of your shellcode: w00tw00t After debugging this command sequence in OD, you can sort out the following code: 00406030 66: 81CA FF0F or dx, 0FFF; generate an initial value of edx 00406035 42 inc edx; move EDX --- code 00406036 52 push edx; save

Recently called instruder 0-day vulnerability: http://www.exploit-db.com/exploits/18140/ Write the debugging analysis here. The level is limited, making everyone laugh. You are welcome to correct your criticism. The analysis result of the dump file is as follows: Prediction_code: (NTSTATUS) 0xc0000005-"0x % 08lx" FAULTING_IP: Win32k! ReadLayoutFile + 88 Bf89ed23 0fb75006 movzx edx, word ptr [eax + 6] TRAP_FRAME: b28068a0 -- (. trap 0xffffffffb28068

Source Very simple C language code, the function is to exchange two number:1 #include 23void swap (intint * b) { 4 *a = *a + *b-(*b = *a); 5 return ; 6 }Assembly Code Parsing executed in the GCC compilation environment, the Gcc-s-o test.s test.c command generates the relevant assembly code.1. file"test.c"2 . Text3 . Globl _swap4. def _swap;. SCL 2; . type 32; . Endef5 _swap:6 LFB6:7 . Cfi_startproc8 PUSHL%EBP9. cfi_def_cfa_offset8Ten. cfi_offset5, -8 One movl%esp,%EBP A.

This lesson mainly on the computer operating principle and assembly language is briefly introduced.Von Joyman Architecture is the storage program Computer , that is, the program written in memory, by the CPU through the bus from memory to read a program, according to the contents of the program to perform specific steps.When the CPU reads the instruction, it uses the register IP to refer to the next instruction (an EIP if it is a 32-bit system)The register of the CPU is divided into general regi

--------------------------------------------------------------------------------------------------------------- -------------------------　　This article is the MOOC course http://mooc.study.163.com/course/USTC-1000029000 of Linux kernel analysis. Homework after the first lesson. Student Name: Liu Zheng　　The main content of this paper is to write a simple C language program, convert it into assembly code, analyze the changes in the stack. If there is an error, please give your valuable comments, t

ss: [ebp + 8] add eax, 1 mov edx, dword ptr ss: [ebp + 0Ch] add edx, 100 add eax, edx pop ebp retn 8_test endp_main proc push dword ptr ds: B; in the disassembly, we can see that B is not B, but a [******] Digital dword ptr, which means we put [*****] in the ds (data segment); in the beginning, a double-character-length value is obtained to push dword ptr ds: a;

00400000 | hinst= 004000000012fb8c 00000000/lparam = NULL Windowname is "really? Really! ", This is the content to be changed. Note that do not use Ctrl + F9 or Alt + F9 to return the result. Check the first record of the stack:Call to createmediawexa from reallyre.0057d7c4 In the Assembly window, press Ctrl + G and enter 0057d7c4. You can see:0057d797 |. 50 push eax;/lparam0057d798 |. 57 push EDI; | hinst0057d799 |. 8b53 5C mov edX, dword ptr [EBX

conventions are used by function callers and function bodies, program execution errors may occur and must be considered as part of the function declaration; Ii. Common function call conventions; Function call conventions in vc6; Call Convention stack clearing parameter passing_ Cdecl the caller passes the stack from right to left_ Stdcall function is passed through Stack from right to left_ Fastcall: Use register (ECx, EDX) first, and then use StackT

Method 1 MoV eax, 0MoV EBX, 0MoV ECx, 0MoV edX, 0 Method 2 XOR eax, eaxXOR edX, EDXXOR ECx, ECxXOR edX, ECx Method 3 Hmm... yep, here the stall is partially balanced by the non-change of the source argument (a work/branch avoided by the CPU), so in this specific case, the difference is minimized. XOR eax, eaxXOR ECx, ECxMoV EBX, eaxMoV