自己做memcpy？

Question

所以，我想知道为什么我不能为自己做memcpy。这是可行的代码并获得正确的结果：

unsigned int VTableAddress = FindPattern( VTABLE_PATTERN, VTABLE_MASK );

unsigned int *p_VTable = NULL;

WriteMemory( &p_VTable, ( void* ) ( VTableAddress + 2 ), 4 );

//....

void D3DX9Interface::WriteMemory( void *address, void *bytes, int byteSize )
{
    DWORD NewProtection;

    VirtualProtect( address, byteSize, PAGE_EXECUTE_READWRITE, &NewProtection );
    memcpy( address, bytes, byteSize );
    VirtualProtect( address, byteSize, NewProtection, &NewProtection );
}

因此，根据我的理解，WriteMemory基本上为内存地址设置了读/写保护，然后简单地将字节复制到地址中。要了解事情是如何运作的，我已经尝试使用此代码：

//Get the address of the vtable
unsigned int VTableAddress = FindPattern( VTABLE_PATTERN, VTABLE_MASK );

unsigned int *p_VTable = NULL;

CopyWithRWPrivileges( p_VTable, (unsigned int*)( VTableAddress + 2 ) );

//...

void D3DX9Interface::CopyWithRWPrivileges( unsigned int *p_Destination, unsigned int *p_Source )
{
    DWORD Protection( 0 );

    VirtualProtect( reinterpret_cast< LPVOID >( p_Destination ), 4, PAGE_EXECUTE_READWRITE, &Protection );

    p_Destination = p_Source;

    VirtualProtect( reinterpret_cast< LPVOID >( p_Destination ), 4, Protection, &Protection );
}

但由于某种原因，最后一个代码给了我一个NULL指针。但为什么？

Answer 1

在UnholySheep的帮助下，我找到了解决问题的方法。首先，指针被复制而不是作为引用指针传递。第二，p_Source也需要作为指针处理，所以使用这段代码它可以正常工作：

void D3DX9Interface::CopyWithRWPrivileges( unsigned int *&p_Destination, unsigned int *p_Source )
{
    DWORD Protection( 0 );

    VirtualProtect( reinterpret_cast< LPVOID >( p_Destination ), 4, PAGE_EXECUTE_READWRITE, &Protection );

    p_Destination = *(unsigned int**) p_Source;

    VirtualProtect( reinterpret_cast< LPVOID >( p_Destination ), 4, Protection, &Protection );
}