Converting address to int

Question

So the goal was to convert a hexadecimal address to a decimal int but i get an error along the lines of loses precision could someone explain why what im doing isnt possible yet int x = 0xff is?
Code:
#include <iostream>
using namespace std;

struct Test{
Test* data = this;
int key = (int)data;
};

int main() {
Test l;
cout<<&l<<"\n"<<l.data<<"\n"<<l.key;
return 0;
}

Answer 1

sizeof(Test*) gives 8 bytes, while sizeof(int) gives 4. You are trying to cast a value represented using 8 bytes, to a container which stores 4 bytes. The compiler does not allow you to do so and will warn you of lossy conversion.

Instead, try using intptr_t, which is 8 bytes in size.

struct Test{
Test* data = this;
int key = (intptr_t)data;
};

int main() {
Test l;
cout<<&l<<"\n"<<l.data<<"\n"<<l.key;
return 0;
}

intptr_t ref: http://www.cplusplus.com/reference/cstdint/