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points worth attention:1. The function to be called has been determined at the compilation stage and is called directly using the function address.2. The call of member functions of the class adopts the "this" Call constraint, and the implicit this pointer is stored in the exc register. The second statement:; 31: cbase bobj = dobj; Lea eax, dword ptr _ dobj $ [EBP]Push eaxLea ECx, dword ptr _ bobj $ [EBP]Call ?? 0cbase @ Qae @ abv0 @ Z From the syntax point of view, a base class object is cons

area ,. data is a readable and writable data area, while. BSS is a readable and writable data zone without initialization. Code and data zones are collectively called sections in elf. You can use other standard sections or add custom sections as needed, but at least one elf executable program should have one. text section. The following is our first assembler, In the att assembly language format: Example 1. att format # Hello. s. Data # Data Segment DeclarationMSG:. String "Hello, world! // N

example, the relationship between pixel RGB values is r> G> B, and the maximum monochrome ratio is r red, the compound color ratio is the compound color yellow formed by the maximum R and the middle G. It is not complicated to use the program code to implement the above grayscale calculation formula. The difficulty is to select the corresponding monochrome and compound Color Ratio Based on the pixel RGB relationship. An additional coloring function is added to the black/white adjustment functio

) in the memory unit opened by SWAp in the stack as the form parameter X and Y. This can be seen from the following Assembly Code (the author adds the annotation ): 22: void main () 23 :{ ...... ...... 13: int A = 1, B = 2; 00401088 mov dword ptr [ebp-4], 1 00401_f mov dword ptr [ebp-8], 2 14: int * P1 = ; 00401096 Lea eax, [ebp-4] 00401099 mov dword ptr [ebp-0Ch], eax 15: int * P2 = B; 00401_c Lea ECx, [ebp-8] 0040366f mov dword ptr [ebp-10h], ECx 16: swap (P1, P2 ); 004010a2 mov

8. Offset is the number of immediately. For example, [EBP + EDX * 8 + 200] is a valid address expression. Of course, in most cases, it is not necessary to have such a complex relationship as inter-address, proportional factor, or offset. The basic unit of memory is byte ). Each byte has eight binary bits. Therefore, the maximum number of characters that each word can save is 11111111, that is, 255 in decimal format. In general, it is more convenient

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*)

, 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;