Why does this code print 5 4 3 2 1 instead of 1 2 3 4 5?

Question

#include <iostream>
using namespace std;

int main(){
int x[5] = {1,2,3,4,5};

int* p = x;
int i;

for (i = 0; i < 2; i++){

int temp = *(p + i);
*(p + i) = *(p + 4 -i);
*(p + 4 - i) = temp;

}

for (i = 0 ; i < 5; i++)
cout << x[i] << " ";
return 0;
}

Answer 1

For i=0 ,
temp = 1;
*(p+i)=5;
*(p+4-i)=temp;

suppose p has a address of 102064 it is now being increased by 0 bytes as 0*2 =0 so new memory address is 102064 which is pointing to number 1 in array x ,so what *(p+i)=*(p+4-i) is telling to compiler is to store the value located at (p+4-i)(equates to 5 after dereferencing(* is used for dereferencing)) address to
to the address *(p+i) which is 102064

and so this is how 1 and 5 are being swapped ,

Hopefully it's clear enough or otherwise let me know

Answer 2

You are swapping the array, so you are getting 5, 4, 3, 2, 1 instead of 1, 2, 3, 4, 5.
You have taken array of 1, 2, 3, 4, 5 but in first for loop you are swapping first two elements with last two elements. So, it is printing in reverse.