edi 837

Static _ always_inline void * _ memcpy (void * To, const void * From, size_t N) {int D0, D1, D2; ASM volatile ("rep; movsl \ n \ t "" movl % 4, % ECx \ n \ t "" andl $3, % ECx \ n \ t "" JZ 1f \ n \ t "" rep; movsb \ n \ t "" 1: ":" = C "(D0 ), "= D" (D1), "= S" (D2): "0" (N/4), "G" (N ), "1" (long) to), "2" (long) from): "Memory"); return to;}/** this looks uugly, but the compiler can optimize it totally, * as the Count is const Ant. */static _ always_inline void * _ constant_memcpy (void *

/**author:davidlin*date:2014-11-11pm*email: [email protected] or [email protected]*world:the City of SZ, in China*ver:000.000.001*history:editor time do1) Linpeng 2014-11-11 created this file!2)*/Linux-0.11 Memory Management module is more difficult to understand in the source code part, now the author's personal understanding publishedFirst hair Linux-0.11 kernel memory management get_free_page () function analysisHave time to write other functions or files:)/** Get Physical Address of first (a

/**author:davidlin*date:2014-11-11pm*email: [email protected] or [email protected]*world:the City of SZ, in China*ver:000.000.001*history:editor time do1) Linpeng 2014-11-11 created this file!2)*/Linux-0.11 Memory Management module is more difficult to understand in the source code part, now the author's personal understanding publishedFirst hair Linux-0.11 kernel memory management get_free_page () function analysisHave time to write other functions or files:)/** Get Physical Address of first (a

It's free. Let's take a look.ODLoad directly as follows: 00414000> $ E8 00000000 call 0041400500414005 $ 5B pop ebx // locate the code00414006.81EB05024000 sub ebx, 00400205 // The code segment length, followed by a variable0041400C. 64: 8B3D 30000000 mov edi, dword ptr fs: [30] // FS [30]-> PEB, locate kernel32.dll00414013. 8B7F 0C mov edi, dword ptr [edi + C] /

EAX register to the memory location specified by value. Use the memory location of the variable address. as follows, multi-memory specifies multiple values in one command:values:. int 10,20,30,40,50,60,70this creates a series of data values that are contiguous in memory (similar to an array of high-level languages). When referencing data in an array, you must use the system to determine which memory location you want to access. The memory location is determined by the following expression:base_

If you returned a struct object, what would the return statement do? Here is the test code #include using namespace Std;struct BIG{Char buf[100];int i;Long D;}B,B2;Big Bigfun (Big B){b.i=100;return b;}int main (){B2=bigfun (B);return 0;}To set a breakpoint at the beginning and end of main8:int Main ()19: {004012A0 Push EBP004012A1 mov Ebp,esp004012a3 Sub esp,118hPuzzled at first, and analyzed for a long timeThe original (118h-40h) remaining memory block holds two big variablesLow address put bi

Loading and execution of programs (iii)--reading notes 23And then the last time the content said.load_relocate_programthe explanation of the process is not over yet, so create a stack segment descriptor and reposition symbol table.Allocating stack space and creating a stack segment descriptor462 ; Creating a program stack segment descriptor463 movecx,[edi+0x0c]; 4KB magnification464 movEbx0x000fffff465 SubEbx,ecxTo get

__stdcall __cdecl __fastcall vc6.0:int __stdcall/__cdecl/__fastcall Add (int x, int y){return x+y;}void Main (){Add (2,3);}1.__stdcall:1:int __stdcall Add (int x, int y)2: {00401020 Push EBP00401021 mov Ebp,esp00401023 Sub esp,40h00401026 push EBX00401027 push ESI00401028 Push EDI00401029 Lea edi,[ebp-40h]0040102C mov ecx,10h00401031 mov eax,0cccccccch00401036 Rep stos dword ptr [edi]3:return X

esi, [edx + 0x3c]; Lea ESI, [edx + esi]; mov esi, [esi + 0x78]; Lea ESI, [edx + esi]; mov edi, [esi + 0x1c]; Lea EDI, [edx + edi]; MOV[EBP-0X04], EDI; mov

Everyone knows that to use COM components in a program, you must first call coinitialize. This function is mainly used to initialize the com runtime environment. But does the function scope take the thread as the unit or the process as the unit? Maybe you have figured out the answer through the test program. That's right, it's a thread. Today, we will go into a bit more detail and confirm our ideas by analyzing the specific implementation of coinitialize. Let's take a look at coinitialize compil

]; Lea ESI, [edx + esi]; mov esi, [esi + 0x78]; Lea ESI, [edx + esi]; mov edi, [esi + 0x1c]; Lea EDI, [edx + edi]; MOV[EBP-0X04], EDI; mov edi