Consider the following program printing a string to the standard output character by character:

#include <stdio.h>

int main(int argc, char **argv) {
    char *string = "hello, world!\n";

    int i = 0;
    while(string[i] != '\0')
        printf("%c", string[i++]);

    return 0;
}

Alter the loop so as to use a char * pointer as the iterator and as the way to access characters within the string for printing. The source code should contain no square bracket. The expected output is:

./pointer4
hello, world!

To check the correctness of your program, use a Linux distribution with check50 installed and write your solution in a file named pointer4.c. In a terminal, with that file in the local directory, check with this command:

check50 -l --ansi-log olivierpierre/comp26020-problems/2023-2024/week3-c-pointers-stdlib/10-pointer4