1. Assignment statement

mov Instructions

C++ Code

gi = 0x12345678;

Assembly instruction

mov   dword ptr ds:[001202DCh],12345678h

Machine instructions

00115718 C7 05 DC 02 12 00 78 56 34 12

Raise questions

• Memory 00AB02DCh It's really gi The address of ？
• mov Instructions are really stored in the address 00115718 In memory ？

Found that regular

The integer represented on the computer is Store in reverse byte order Of

Knowledge point ： Storage mode of the large and small end （Intel Computers are small terminals ）

Conclusion ： Instructions are combinations of bytes

Raise questions ： We can build our own instructions to execute ？

#include <iostream>
#include <Windows.h>
int gi{
    };
void* address;
void* buildCode();
int main() {
    
	void* codeAdr = buildCode();
	_asm mov address, offset _adr;
	DWORD oldProtect{
    };
	VirtualProtect(codeAdr, 16, PAGE_EXECUTE_READWRITE, &oldProtect);
	gi = 1;
	printf("gi = %d\n", gi);
	_asm jmp codeAdr;
	gi = 13;
_adr:
	printf("gi = %d\n", gi);
	getchar();
}

void* buildCode()
{
    
	char* code = (char*)malloc(16);
	char* pMov = (char*)code;
	char* pJmp = (char*)code + 10;
	char* pAddress = pMov + 2;
	//C7 05 B0 00 94 00 01 00 00 00
	pMov[0] = 0xC7;
	pMov[1] = 0x05;
	*((int*)pAddress) = (int)&gi;
	*((int*)(pAddress + 4)) = 2;
	//FF 25 88 01 7B 00
	pJmp[0] = 0xFF;
	pJmp[1] = 0x25;
	*((int*)(&pJmp[2])) = (int)&address;
	return code;
}