why this C++ code cannot print the content of the dereference of a pointer

I have not used c++ for a long while so maybe I forget some very simple trick. Here is the code and I tried to print the content of *p and got nothing printed out.

#include <iostream>

using namespace std;

int main() {
    char hello[] = "hello";
    char *p = hello;

    while (*p) {
        *p += 1; // increase the character by one
        p += 1; // move to the next spot
    }

    cout << "content of hello is: " << hello << endl;
    cout << "content of hello is: " << *p << endl;

    return 0;
}

Good thing is I figured out by myself while writing this question. The reason is very simple: p is a pointer type for char, and the p is altered in the while loop. So this will print out i:

cout << "content of hello is: " << *(p - 5) << endl;

Just note it here.

C-strings, which you're dealing with here, are terminated by a null character ('\0'). That's why looping while *p is true works. When p points to a null terminator, dereferencing it gives us 0, which is false.

Your loop increments p until it points to the null terminator. That means at the conclusion of the loop, it must be pointing to '\0' which counts as an empty C-string. Your program then prints that empty string.