I want to receive a variable of type DWORD which will contain address pointer float *.
If I write this:
float *Object = 5;
std::cout << &Object;
It gives me exactly the value that I want (i.e: 0F235C1A).
Can you tell me how to assign this value to DWORD for using it in my Memory Write function?
Trying:
DWORD ObjAddress = &Object;
I got compiler error: cannot convert from 'float *' to 'DWORD'
I don't understand your object definition:
float *Object = 5;
The 5 should be an address? a value?
See following code, maybe it is what you want.
float Object = 5;
int ObjAddress = (int)&Object;
cout << "object adr1: "<< hex << &Object << endl;
cout << "object adr2: "<< hex << ObjAddress << endl;
Output
object adr1: 0x28ff08
object adr2: 28ff08
Or when you really need the object as a pointer
float value = 5;
float *Object = &value;
int ObjAddress = (int)&Object;
cout << "object adr1: "<< hex << &Object << endl;
You just need to cast your &Object;
DWORD ObjAddress = (DWORD)&Object;
And I think that will solve your problem.
I think you can use reinterpret_cast like this (replace unsigned long with DWORD)
#include <iostream>
int main(void)
{
float *object = new float;
*object = 5;
unsigned long objAddress = reinterpret_cast<unsigned long>(&object);
std::cout<<objAddress<<std::endl;
return 0;
}
DWORD?DWORDis a microsoft typedef'dunsigned longfloat *Object = 5;shouldn't compile, for one thing. What compiler are you using?