tableau edx

first set breakpointsThen, two times shift+f9.Then, cancel the breakpoint. Alt+f9 and executes to user codeThen step away, and you'll reach Oep.5.1, after loading the program, you will find the first line has an address0040a86d > B8 74de4500 mov eax,qqspirit.0045de742, at the address of the next breakpoint BP 0045de743. Then shift+f9 run, and cancel breakpoint4, we are in the Retn next line set breakpoints, and then shift+f9 Run, and cancel the breakpoint045de74 B8 F9CB45F0 mov eax,f045cbf90045

. The source program entry is _start. As follows:Cpuid2.s # Cpuid2.s file.section. DataOutput: . Asciz "CPUID is '%s ' \ n". Section. BSS . Lcomm Buffer, 12. Section. Text.globl _start_start: NOP movl $,%eax cpuid movl $buffer,%edi movl%ebx, (%edi) MOVL%edx, 4 (%edi) movl%ecx, 8 (%edi) pushl $buffer pushl $output printf Addl $8,%esp PUSHL $ exit In the following three cases where the entry is _start,main,xx

If the assembly language to achieve the following C language functions, the compilation environment Ubuntu14.04 (32-bit).#include voidSwapint*p,int*q) { intTMP = *p; *p = *Q; *q =tmp;}intMain () {intA[] = {3,0,5,1,4,6,2,9,8,7}; intI, J; for(i =0; I Ten; i++) { for(j = i +1; J Ten; j + +) { if(A[i] >A[j]) {Swap (a[i], A[j]); } } } intK; for(k =0; K Ten; k++) {printf ("%d\n", A[k]); } return 0;} Assembly Code SORT.S . Section.

Linux Kernel Analysis Course summaryName: Wang ZhaoxianStudy No.: 20135114Note: Original works reproduced please specify the source + "Linux kernel analysis" MOOC course http://mooc.study.163.com/course/USTC-1000029000First, how the computer works 个人理解：计算机就是通过和用户进行交互，执行用户的指令，这些指令存放在内存中，通过寄存器存储，堆栈变化，来一步步顺序执行。Second, the storage program computer working model1.冯诺依曼体系结构—存储程序计算机 硬件角度（主板）：通过cpu中IP寄存器指向一个代码段运行某些指令；寄存区，指向内存的某一块区域（代码段） 程序员角度：将cpu抽象为一个for循环，只是执行下一条指令，从内存中取到下一条指令的内容。内存保存指令和数据，cpu

Introduction EFS Web server is a software that can manage server files over a Web side, and sending a GET request too long can trigger a buffer overflow vulnerabilityAnalysis Source: https://www.exploit-db.com/exploits/39008/ Experimental Environment WinXP SP3 Chinese versionEFS Web Server7.2Immunity DebuggerWinDbgIdaMona Vulnerability Analysis Because the author uses the address of the overlay Seh program in ImageLoad.dll, no ASLR, so the use of more stable, open on the pop-up calculator We w

science lab. concerned that everyday knowledge workers in security positions found databases too hard to query and understand, the DOE/DOD asked Hanrahan to invent a visual analysis framework for analyzing information in Databases. They challenged the person who won two Academy Awards for graphics technology to overturn the database industry.More than five years later, what began as an extensive R D project inside Stanford-supported by the DOE/DoD as well by Microsoft's well known database res

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

(LOCK_PREFIX "decl (% eax) \ n" \ # as shown here, if count is first subtracted from the thread, SF is not equal to 1 (not equal to negative ), execute it later, that is, obtain the lock. then, if the thread that tries to obtain the lock executes this atomic operation (in the unlocked state), SF equals 1, so that void (*) (atomic_t *) is executed *) type Function to enter the waiting queue. therefore, this locking atomic operation can be completed with a single command. in non-SMP scenarios, a

have been learning about Windows kernel recently, write a blog for memo.The specific process of Windows system call in the Pan teacher's "Windows kernel Principle and implementation" in the 8th chapter has been written very clearly, first read the picture given in the.Take CreateFile as an example, after some parameter checking in Ring3 's CreateFile, the final call is NtCreateFile in Ntdll. There are also zwcreatefile, but their addresses point to the same area, so they are essentially the same

the ELF format is generally divided into the following parts :. text ,. data and. bss, where. text is a read-only code 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 assemb

many variables ............. (Dozens of rows)Omitted again ........ (Dozens of rows)The key point is to understand several points. www.2cto.comFirst: 00407D29 |. C785 00D4FFFF> | mov dword ptr [ebp-2C00], 0; the initial value of the loop00407D33 |> 8B85 04D4FFFF |/mov eax, dword ptr [ebp-2BFC]00407D39 |. C1E0 0C | shl eax, 0C; left shift00407D3C |. 0385 00D4FFFF | add eax, dword ptr [ebp-2C00]00407D42 |. 8D55 E8 | lea edx, dword ptr [ebp-18]00407D45

pmodule // jump directly to the program's airspace Press F10 to run the following command:................. 015F: 0048CCB1 PUSH EBX015F: 0048CCB2 mov ebx, EAX015F: 0048CCB4 xor eax, EAX015F: 0048CCB6 PUSH EBP015F: 0048CCB7 push dword 0048CD6C015F: 0048 ccbc push dword [FS: EAX]015F: 0048 ccbf mov [FS: EAX], ESP015F: 0048CCC2 lea eax, [EBP-04]015F: 0048CCC5 PUSH EAX015F: 0048CCC6 mov ecx, [EBX + 0830]015F: 0048 cccc mov edx, 0048CD80015F: 0048CCD1 mov

PUSH EDI005357C6 8BF8 mov edi, EAX005357C8 33C0 xor eax, EAX005357CA 55 PUSH EBP005357CB 68 955C5300 PUSH tk.00535C95005357D0 64: FF30 push dword ptr fs: [EAX]005357D3 64: 8920 mov dword ptr fs: [EAX], ESP005357D6 8D55 D0 lea edx, dword ptr ss: [EBP-30]005357D9 8B87 F0020000 mov eax, dword ptr ds: [EDI + 2F0]005357DF E8 1464F1FF CALL tk.0044BBF8;005357E4 8B45 D0 mov eax, dword ptr ss: [EBP-30]; [EBP-30] = E005357E7 8D55 D4 lea

. _ 44;Vcresult. x = vcresult. X/vcresult. W;Vcresult. Y = vcresult. Y/vcresult. W;Vcresult. z = vcresult. Z/vcresult. W;Vcresult. W = 1.0f}Else{Float * ptrret = (float *) vcresult;_ ASM {MoV ECx, this; vectorMoV edX, M; MatrixMovss xmm0, [ECx]MoV eax, ptrret; Result VectorShufps xmm0, xmm0, 0Movss xmm1, [ECx + 4]Mulps xmm0, [edX]Shufps xmm1, xmm1, 0Movss xmm2, [ECx + 8]Mulps xmm1, [

Movl $ sys_write, % eax Movl st_filedes (% EBP), % EBX Movl st_write_buffer (% EBP), % ECx Movl $ record_size, % edX Int $ linux_syscall # Note-% eax has the return value, which we will # Give back to our calling program Popl % EBX Movl % EBP, % ESP Popl % EBP RET File write-records.s: . Include "Linux. s" . Include "record-def.s" . Section. Data # Constant data of the records we want to write # Each text data item is padded to the proper # Length

insensitive, that is, the meaning of Function rightpos (const substr, S: string): integer;VaRIPOs: integer;Tmpstr: string;BeginTmpstr: = s;IPOs: = pos (substr, tmpstr); Result: = 0;// Find the location where substr appears for the first timeWhile IPOs BeginDelete (tmpstr, 1, IPOs + Length (substr)-1 );// Delete the searched charactersResult: = Result + IPOs;IPOs: = pos (substr, tmpstr); // find the position where the substr appearsIf IPOs = 0 Then break;Result: = Result + Length (substr)-1;End

1) classic comparison, usually at the registration code (by programhunter)1MoV eax [] can be an address or another register.MoV edX [] the preceding two addresses usually store important information.Call 00 ??????Test eaxJZ (jnz)2MoV eax [] can be an address or another register.MoV edX [] the preceding two addresses usually store important information.Call 00 ??????JNE (JE)3MoV eax []MoV

, ADDR buffer1, 128 Invoke wsprintf, ADDR buffer, ADDR template, wparam Invoke lstrcmpi, ADDR buffer, ADDR buffer1 . If eax! = 0 Invoke setdlgitemtext, hdlg, idc_handle, ADDR Buffer . Endif Invoke getdlgitemtext, hdlg, idc_classname, ADDR buffer1, 128 Invoke getclassname, wparam, ADDR buffer, 128 Invoke lstrcmpi, ADDR buffer, ADDR buffer1 . If eax! = 0 Invoke setdlgitemtext, hdlg, idc_classname, ADDR Buffer . Endif Invoke getdlgitemtext, hdlg, idc_wndproc, ADDR buffer1, 128 Invoke getclasslong,

computer tests, only through their own test, the magic is not much difference in Ubuntu, the problem is said later), and then I compile, look at their assembly code, found a mysterious thing I do not understand. The following are the assembly codes for the If-else and three mesh operations respectively. 37:if (a>b) 00401079 mov ecx,dword ptr [ebp-10h] 0040107C cmp ecx,dword ptr [ebp-14h] 0040107F jle main+79h (00401089) 38:temp=a; 00401081 mov edx,dw

Different compilers may produce different codes, resulting in different results. The Code is as follows: #include Environment: win7 Compiler: GCC IDE: vc ++ 6.0/DEV-C ++ Result: q = 22 : Q = (++ j) + (++ mov eax, dword ptr [ebp-] Move J = add eax, add 1 to the Register eax value, eax = 0040103C mov dword ptr [ebp-], eax moves the register value to the variable j, j = 0040103F mov ecx, dword ptr [ebp-] Move J = add ecx, in the register, ecx +, j = mov dword ptr [ebp-], ecx moves the value on