jmp anova

Address instruction or data content ... ... 020 Cla 0→AC, accumulator AC zeroing 021 ADD 30 (AC) + (30)? AC, the value of the accumulator AC is added to the data in main memory address 30, and the result is stored in the accumulator AC 022 STA 40 (AC) 40, the value of the accumulator AC is stored in the main memory address 40 023 NOP Empty operation,

must be four bytes. Therefore, this method is not feasible. It seems that you can't steal a lazy.We need to take a path. So, what else is associated with this address on the stack? The closest thing to the stack is the ebp,esp of the two registers, which, as you probably already know. Yes, when the function ret instruction returns, after taking away the saved EIP, the location that the ESP points to is the initial address of the shellcode.Figure 31So, to get the program to execute Shellcode, ju

windows. incinclude unzip masm32.incinclude Debug. incincludelib unzip Debug. Lib. Data ddsource DD 11 h, 22 h, 33 H dddest dd lengthof ddsource DUP (?). Codemain proc Lea ESI, ddsource Lea EDI, dddest mov ECx, lengthof ddsource CLD rep movsd dumpmem offset dddest, sizeof dddest; 11 00 00 00-22 00 00 00-33 00 00 00 retmain endpend main Movsw example: ; Mongoflat, stdcallinclude windows. incinclude unzip masm32.incinclude Debug. incincludelib unzip Debug. Lib. Data ddsource DW 11 h,

function _ exit: 0x800034c 0x800034d 0x8000350 0x8000355 0x8000358 0x800035a 0x800035d 0x800035f 0x8000361 0x8000363 Movl limit x1, % eax; 1 System Call movl 0, % ebx; ebx is exit parameter 0 int limit x80; System Call To sum up, the merged assembly code is: movl $ execve system call number, % eax movl "bin/sh \ 0" Address, % ebx Movl name array address, % ecx movl name [n-1] address, % edx int limit x80; execve) Movl limit x1, % eax; 1 System Call movl 0, % ebx; ebx is the exit param

Fragmented records:The 8086CPU registers are 16-bit and can hold two bytes; The 4 registers, AX, BX, CX, DX, are commonly used to hold general data, known as general-purpose registers.Ax:In 8086CPU, the four registers can be divided into two independent use of 8-bit registers, with the-H and-L for the left and right eight bits;Instruction Mov,add: Left operand = left operand and right operand operation result.Physical Address method in 8086CPU: Two 16-bit addresses synthesize a 20-bit physical a

Storage of words in memoryThe main meaning of this passage is: a word =2b=16bit,the CPU is a two memory units stored in a word (if the 0 address to store the font data, is to get its high-byte 0+1 bit and low byte 0 bits data, the data is read by the high address bit to the low address bit)Questions:(1) What is the byte-type data stored in the 0 address unit? # 20H(2) What is the font data stored in the 0 address word cell? # 4e20h(3) What is the byte-type data stored in the 2 address word cell?

oriented to assembly language programming, my posts are machine-oriented and disassembly, hoping to play a complementary role. With the previous two foundation, the assembly language book on most of the instructions should be able to understand, understand. Here are a few more common and more complicated instructions. I am talking here about the hard instructions of the machine, not for any assembler. Unconditional Transfer Instruction JMP: There a

large scope is determined, we can use a small script in metasploit to complete this task:/opt/metasploit/apps/pro/msf3/tools/pattern_create.rb Generate a pattern string (200-2100 = 1900) with a length of 200. We will send such a packet: STOR +/../+ A * 1900 + pattern PCMAN collapsed after receiving the data. In windbg, we can see that the eip is 0x41346441: Return to kali and use pattern_offset to calculate the offset: The offset is 102, that is, the value 0x41346441 appears in 200 of the 10

finally obtain the physical address corresponding to CreateProcessW, using \ phymem, we can avoid the copy-on-write mechanism and directly rewrite CreateProcessW. 　　 3. achieve the goal through the most common means, first enumerate all processes in the system, then modify the page attributes of each process through functions such as VirtualQueryEx, VirtualAllocEx, and VirtualProtectEx, and allocate new space. Finally, we use WriteProcessMemory to write our code to the space of each process, an

, and find that there is a layer of Aspack in the layer. We have read Telock0.98 in the advanced Article, which is troublesome. resumable data cannot be stored in disorder, for example, normal Int3 breakpoint data and hardware breakpoint data. However, it can still be interrupted through the memory breakpoint. Telock0.96 is weaker than Telock0.98. The OD loader does not ignore invalid command exceptions and memory exceptions. First, the fire detection Telock has a memory exception and two invali

test ax, AX0000004b JE shadowse.00000010800426051 mov ecx, dword ptr ss: [ESP + C]00426055 PUSH EDI00426056/cmp ax, 30; number?0000005a | jb short shadowse.0000006c; 003665c | cmp ax, 3900426060 | ja short shadowse.0020.6c;> "9" indicates the hop.00426062 | and eax, 0 FFFF00426067 | sub eax, 30;-0x300000006a | jmp short shadowse.00000096003666c | cmp ax, 41; uppercase letters?00426070 | jb short shadowse.00000082; 00426072 | cmp ax, 5A00426076 | ja s

the user event processing step-by-step.Therefore, we need to find the language to give users a place to distribute messages.Tracking down:00411688 |.FFD0 call eax; Ewnd. Processnotifysys (Int,ulong,ulong)One of the steps to process the message, go ahead.004116b4 |. E8 77c7ff call Ckrnlapp::notify (Int,ulong,ulong)To continue:0040e36d |> \8b4424 mov eax, dword ptr ss:[arg2]; Case 7D8 of Switch ewnd.40de4f0040E371 |. 8BCE mov ecx, esi0040E373 |.-Push eax; /ARG1 = [ARG2]0040E374 |.E8 67a600 call C

2016 "Anheng Cup" National University Network Information Security management operation and maintenance Challenge competition _re20064 bits. NET ProgramRun:Parameter Check Reflector analysis: f= "incorrect"----loc.2i= "parameter error"----loc.3p= "? #?? # #? #?? #?##? ?# ?#?#?#? #?# ?#?? # #?# "l_0005 ... l_000a stloc.0 l_0046 ldloc.0l_0047 .... l_004c ldc.i4.8l_004d beq.s l_0056 Check whether the number of parameters is 8, equal skip output "parameter error" STR6 hold numeric characters, revers

Eax,eax00031017 5D POP EBP00031018 C3 RETN00031019 B8 249b0300 MOV eax,test.00039b240003101E E9 00000000 JMP test.0003102300031023 PUSH ECX00031024-PUSH EDX00031025 PUSH EAX00031026 1c7a0300 PUSH Test.00037a1c0003102B E8 0E000000 Call test.0003103e00031030 5A POP EDX00031031 ECX POP00031032 FFE0 JMP EAX00031034 B8 209b0300 MOV eax,test.00039b2000031039 E9 e5ffffff JMP