Difference between s and *s in pointer

Question

This code print Only first letter i.e 'H'

#include <stdio.h>

int main(void)
{
    char *s = "Hello";
    printf("%c\n", *s);
}

But this return 'Hello'

#include <stdio.h>

int main(void)
{
    char *s = "Hello";
    printf("%s\n", s);
}

changing %c to %s will return an error in the first code
changing %s to %c will return an error in the second code

In my understating *s is dereferencing operator i.e what value we find when you go to that address and s is what store in s(address of 'H'). Please, someone explains this.

Answer 1

printf("%c\n", *s); is the same as printf("%c\n", s[0]);

By definition of printf, the %c format specifier in printf expects a character. So, *s says go to the address stored in s, and give me the character there. The square bracket notation says "get the address of s and add 0 to it, then dereference it to get its value."

printf("%s\n", s);

By definition of printf, the %s format specifier expects a pointer to a string. That's why you give it s as s is the pointer.