c++指针练习

Pointers
在getchar处断点，断点后，调试->窗口->反汇编查看数据

main

#include <iostream>

#include <Windows.h>

/*

Player : object

Name : string

Health : integer

Coins : integer

Coordinates : object

X : float

Z : float

Y : float

Inventory : array - Array of item objects, having the item and item count.

*/

uintptr_t _Inventory[3] = { 1,2,3 };

struct _Coordinates

{

	float x = 4.0;

	float y = 2.0;

	float z = 3.0;

} coordinates;

struct Player

{

	const char* Name = "ab";

	uintptr_t Health = 6;

	uintptr_t Coins = 3;

	/*

	// 这种方法类似把coordinates直接复制到这里来

	// Padding1的偏移量将是 playerBaseAddress+4*6

	_Coordinates Coordinates = coordinates;

	float x = 4.0;

	float y = 2.0;

	float z = 3.0;

	*/

	_Coordinates* Coordinates = &coordinates;

	// uintptr_t Padding1 = 1;

	/*

	 //类似直接复制到这

	 //std::cout << "arrar[0]: " << *(uintptr_t*)(playerBaseAddress + sizeof(uintptr_t) * 4) << std::endl;

	 //std::cout << "arrar[1]: " << *(uintptr_t*)(playerBaseAddress + sizeof(uintptr_t) * 5) << std::endl;

	 //std::cout << "arrar[2]: " << *(uintptr_t*)(playerBaseAddress + sizeof(uintptr_t) * 6) << std::endl;

	 const int Inventory[3] = { 1,2,3 };

	*/

	// 数组直接返回的就是指针，所以不用&

	 uintptr_t* Inventory = _Inventory;

} player;

int main()

{

	std::cout << "playerBaseAddress: " << &player << std::endl;

	uintptr_t playerBaseAddress = (uintptr_t)&player;

	// name

	// lea stringNameAddress, [playerBaseAddress]

	uintptr_t* stringNameAddress = (uintptr_t*)(playerBaseAddress);

	// 从指针中获取值

	// mov eax, dowrd ptr [stringNameAddress]

	std::cout << "Name: " << std::hex << *(uintptr_t*)(*stringNameAddress) << std::endl;

	// get Health

	std::cout << "Health: " << *(uintptr_t*)(playerBaseAddress + sizeof(uintptr_t)) << std::endl;

	// get Coins

	std::cout << "Coins: " << *(uintptr_t*)(playerBaseAddress + sizeof(uintptr_t) * 2) << std::endl;

	// 获取Coordinates指针

	uintptr_t coordinatesAddress = *(uintptr_t*)(playerBaseAddress + sizeof(uintptr_t) * 3);

	std::cout << "CoordinatesAddress: " << coordinatesAddress << std::endl;

	std::cout << "Coordinates->x: " << *(float*)(coordinatesAddress) << std::endl;

	std::cout << "Coordinates->y: " << *(float*)(coordinatesAddress + sizeof(float)) << std::endl;

	std::cout << "Coordinates->z: " << *(float*)(coordinatesAddress + sizeof(float) * 2) << std::endl;

	// 获取Inventory指针

	uintptr_t InventoryAddress = *(uintptr_t*)(playerBaseAddress + sizeof(uintptr_t) * 4);

	std::cout << "InventoryAddress: " << InventoryAddress << std::endl;

	std::cout << "Inventory[0]: " << *(uintptr_t*)(InventoryAddress) << std::endl;

	std::cout << "Inventory[1]: " << *(uintptr_t*)(InventoryAddress + sizeof(uintptr_t)) << std::endl;

	std::cout << "Inventory[2]: " << *(uintptr_t*)(InventoryAddress + sizeof(uintptr_t) * 2) << std::endl;

	// set

	*(uintptr_t*)(playerBaseAddress + sizeof(uintptr_t)) = 4;

	*(uintptr_t*)(playerBaseAddress + sizeof(uintptr_t)*2) = 5;

	getchar();

	return 0;

}

x86打印结果:

playerBaseAddress: 0026D05C

Name: 6261

Health: 6

Coins: 3

CoordinatesAddress: 26d050

Coordinates->x: 4

Coordinates->y: 2

Coordinates->z: 3

InventoryAddress: 26d044

Inventory[0]: 1

Inventory[1]: 2

Inventory[2]: 3

x64打印结果:

playerBaseAddress: 00007FF7CC8AD028

Name: 6261

Health: 6

Coins: 3

CoordinatesAddress: 7ff7cc8ad018

Coordinates->x: 4

Coordinates->y: 2

Coordinates->z: 3

InventoryAddress: 7ff7cc8ad000

Inventory[0]: 1

Inventory[1]: 2

Inventory[2]: 3

巴特西

c++指针练习

main

x86打印结果:

x64打印结果:

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