Is the conclusion to this code correct?

Question

#include <iostream>
using namespace std;

int main() {
int a[]={1,4,9,24,5};
int *p=a;
cout<<"the array is: ";
for(int i=0;i<5;i++)
{cout<<a[i]<<" ";

};

//cout<<++*p<<endl; this increments the value at pointer first and keeps the pointer pointing at the same position
//cout<<*p; output:2 2

cout<<"\n on doing *p++, it shows 1 and takes pointer to next element, which is 4: "<<*p++<<endl;//this showcases the value at pointer i.e. 1 and icrements it and points to the next element
cout<<" a[0]: to show the value at a[0]remains the same: "<<a[0]<<endl;
cout<<"*p shows the element which pointer points at presently: "<<*p<<endl; //this is now pointing to next element
cout<<"*p-- will show the show the value of 4 and decrement the pointer to a[0] from a[1]: "<<*p--<<endl;
cout<<" the pointer is now pointing at: "<<*p<<endl;
cout<<"--*p this will reduce the value of a[0] by 1: "<<--*p<<endl;
cout<<"where is pointer pointing now?? "<<*p<<endl;
cout<<"now we show the value again and shift the pointer to a[1] by *p++: "<<*p++<<endl;
cout<<"where is pointer pointing now?? "<<*p<<endl;

cout<<"the array is: ";
for(int i=0;i<5;i++)
{cout<<a[i]<<" ";};
cout<<"\n conclusion is: prefix operators work on the present value of the pointer \n and dont shift the pointer,\n whereas the postfix show the same value \n and shift the pointer";
return 0;
}

Answer 1

*p++ == *(p++) changes pointer.
(*p)++ changes value the pointer points to.
++*p == (++*p) changes the value of the pointer points to.
*++p == *(++p) changes pointer.