C - I'm getting two diferents pointers from the same variable

Question

Context

I was studing about pointer in C, for that I was testing some pointer's algorithm and in one of this tries I make a little change that is bugging until now.

This is the code:

char *s = "ABCDEF";
char **y = &s;

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

printf("%p\n", y);
printf("%p\n", &s[0]);

The problem is that for some reason when I ask for C to print the pointer of S (without the ampersand), I receive one pointer, but when I put the ampersand I receive another pointer differently. I think that this didn't happen because the ampersand will give me the address of S.

So to check if isn't a bug, I make a pointer with the address of S, and I received the same that the "&s" as expected, but when I use the "&s[0]" I received the value of the single "s".

So, why is C giving me two results if in both cases I'm just asking for the first character of the string?

The values that I received

0x400634
ox7ffd95c33090
ox7ffd95c33090
0x400634

I know that probably I'm forgetting some piece of logic, but I've struggled with this, so if someone can help me, I'll appreciate it.

Answer 1

The variable s stores the address of the string literal "ABDCDEF"; the variable y stores the address of the variable s.

    char **        char *        char
    +---+          +---+         +---+
 y: |   | ----> s: |   | ------> |'A'|
    +---+          +---+         +---+
                                 |'B'|
                                 +---+
                                 |'C'|
                                 +---+
                                 |'D'|
                                 +---+
                                 |'E'|
                                 +---+
                                 |'F'|
                                 +---+
                                 | 0 |
                                 +---+