dword ptr[edi + 0x39], 0x000000B4; mov dword ptr[edi + 0x3D], 0x908BED75; mov dword ptr[edi + 0x41], 0x000000F8; mov dword ptr[edi + 0x45], 0x00F89189; mov dword ptr[edi + 0x49], 0x31610000; mov dword ptr[edi + 0x4D], 0x0000C3C0; mov dword ptr[edi + 0x51], e...
m, i and j. I’ll assume the target in this case is the x86 platform. By examining the generated assembly code (/FA), I notice the variable n has been allocated to the register
eaxmovedx,4//edx=systemPID(4)//Theloopistogetthe_EPROCESSofthesystemfind_sys_pid:moveax,[eax+0xb8]//Findtheprocessactivitylistsubeax,0xb8//Listtraversalcmp[eax+0xb4],edx//Determinewhether
***Stack Dump being sent to C:\Program Files (x86)\Microsoft SQL Server\MSSQL11.SA\MSSQL\LOG\SQLDump4039.txt SqlDumpExceptionHandler: Process 12 generated fatal exception c000000d EXCEPTION_INVALID_CRT_PARAMETER. SQL Serve r is terminating this process. * *** * * BEGIN STACK DUMP: * 0...
static int get_tjmax(struct cpuinfo_x86 *c, u32 id, struct device *dev) { /* The 100C is default for both mobile and non mobile CPUs */ int err; u32 eax, edx; u32 val; .. .. /* * An assumption is made for early CPUs and unreadable MSR. ...
I’ll assume the target in this case is the x86 platform. By examining the generated assembly code (/FA), I notice the variable n has been allocated to the register ESI, the variable m has been allocated to ECX, and both i and j have been allocated to...
When I am trying to activate script with my plugin or when I am trying to activate STEALTH EDIT plugin in Memory Viewer, I always get this message: What is happening? My script: script [ENABLE] stealtheditex(one,"Asphalt9_w10_x86_rtl.exe"+329B6C,1) alloc(newmem,2048) label(returnhere...
Thread 0 crashed with X86 Thread State (32-bit): eax: 0x00000000 ebx: 0x21b7b310 ecx: 0xffffffff edx: 0x18000000 edi: 0x17eb39ba esi: 0x00000000 ebp: 0xbfffd0e8 esp: 0xbfffceb0 ss: 0x0000001f efl: 0x00010213 eip: 0x93d2c59b cs: 0x00000017 ...
"add eax, ebx" instruction is used in x86 assembly language to perform an addition operation. It adds the value in the EBX register (32-bit) or BX register (16-bit) to the value in the EAX register (32-bit) or AX register (16-bit), and stores the result in the EAX ...
mov eax, 1 ; Load 1 into register EAX mov ebx, 2 ; Load 2 into register EBX add eax, ebx ; Add EAX and EBX, result in EAX This, in turn, produces the following binary. 00000000000000000000000000000011 NOTE: The above output varies on the CPU architecture. It’s just a simple example...