Can you explain the process of this code.(i thought that answer would be [ X, 65 ] but real answer i get as a output [A, 65] ).

Question

#include <iostream>
using namespace std;
union mytype{
char c;
unsigned int i;
} u;
int main(){
u.c='X';
u.i=65;
cout <<u.c<<" "<<u.i;
return 0;
}

Answer 1

A union shares the same memory for all its members. If you change c, then you change i. Likewise, if you change i, then you might also change c (depending on which byte of i that you change).
Be certain to understand that the data types are different sizes. So c overlaps the same memory with i only on the first byte of i.
Also beware that on another type of computer and another compiler they might overlap on the high byte of i or a middle byte (big-endian versus little-endian, different integer size).