Why cout the pointer can change the data - c++

Question

#include <iostream>
using namespace std;
void reset(int *a, int *b){
int sum = *a + *b;
*a = (sum / 2.0 - sum / 2) >= 0.5 ? sum / 2 + 1 : sum / 2;
*b = *a;
cout << hex << (void *)a<<endl;
cout << hex << b<<endl;
}
int main(){
int a, b;
cin >> a >> b;
reset(&a, &b);
cout << a <<' '<< b << endl;
}
I use the code to reset two variable,but after I print the pointers of variable,the data changed.When I commente the two cout statement.It works.
Looks like this:

Answer 1

You should return to dec, because you changed the output to hex
#include <iostream>
using namespace std;
void reset(int *a, int *b){
int sum = *a + *b;
*a = (sum / 2.0 - sum / 2) >= 0.5 ? sum / 2 + 1 : sum / 2;
*b = *a;
cout << hex; //set hex for output stream
cout << (void *)a<<endl;
cout << b<<endl;
cout << dec; // return to dec system
}
int main(){
int a, b;
cin >> a >> b;
reset(&a, &b);
cout << a <<' '<< b << endl;
}

Answer 2

Variables are not modified, you change basefield format flag after using std::hex here:
cout << hex << (void *)a << endl;
cout << hex << b << endl;
That's why commenting this out "fixes" this "problem".
You can uncomment this call, but modify you reset() function, so it will restore original stream state:
void reset(int *a, int *b)
{
int sum = *a + *b;
*a = (sum / 2.0 - sum / 2) >= 0.5 ? sum / 2 + 1 : sum / 2;
*b = *a;
cout.setf(std::ios::hex); //Print as hexadecimal numbers
cout << (void *)a << endl;
cout << b << endl;
cout.setf(std::ios::dec); //Restore to decimal
}
I modified your reset to set hex only once instead of doing << hex every time you print something (although it is not necessary, because, as you noticed, stream is modified permanently - that's why I prefer explicit functions call).
Further read: std::ios_base::setf.