Pointer with same memory address with different value

Question

I casted the memory address from double to an integer . Even though they point to the same address why the values are different ?

#include<iostream>
using namespace std;

int main()
{
    double d = 2.5;
    auto p = (int*)&d;
    auto q = &d;
    cout<<p<<endl; // prints memory address 0x7fff5fbff660
    cout<<q<<endl; // print memory address  0x7fff5fbff660
    cout<<*p<<endl; //prints 0
    cout<<*q<<endl; // prints 2.5
    return 0;

}

But why the value are different

0x7fff5fbff660
0x7fff5fbff660
0
2.5
Program ended with exit code: 0

Answer 1

Suppose you have "11" written on a piece of paper. That is eleven if it's decimal digits. That is two if there's one mark for each value. That's three if it's binary. How you interpret stored information affects the value you understand it to be storing.

Answer 2

double d = 2.5;
auto p = (int*)&d;
auto q = &d;

p and q are created pointing to the same memory location. The memory holds a double (usually 8 bytes)

When you create

auto p = (int*)&d;

you are telling the compiler ( reintepret_cast< int*> ( &d) ) that the value in d was an integer.

So the values of the pointers are the same, but the types are not.

When you print out

cout<<*q<<endl; // prints 2.5

You are displaying the correct value - as it came in and out through that.

when you print out

cout<<*p<<endl; //prints 0

You are looking at 4 (typically) bytes of the 8 byte memory, and interpreting them as an integer.

These happen to be 0x00, 0x00, 0x00, 0x00