CODEWITHSUNDEEP
cpointers
Will Felipe
Will Felipe

Reputation: 63

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

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.

Upvotes: 5

Views: 125

Answers (5)

user16719198
user16719198

Reputation:

the pointer of S (without the ampersand)

This is already confusing. char *s declares s as a pointer.

but when I put the ampersand I receive another pointer

That other pointer can be called pointer of (or to) s. A pointer to a pointer (to char).

address of S

"pointer of" and "address of" must mean the same. Nobody says "pointer of s" in that context above.

Upvotes: 0

William Pursell
William Pursell

Reputation: 212248

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

printf("%p\n", s);      /* address of the 'A' in the string literal "ABCDED" */
printf("%p\n", &s);     /* address of the local variable s */

printf("%p\n", y);      /* contents of y, which is the address of the local variable s */
printf("%p\n", &s[0]);  /* address of the 'A' in the string literal */

Maybe it helps to think of s[0] as *s, in which case you can rewrite &s[0] as &*s and see that the & and the * operators "cancel" (in a very non-technical sense) each other.

Upvotes: 1

SANDRUX
SANDRUX

Reputation: 31

s[0] is equivalent to (s + 0), means that the first block of that memory is dereferenced so if you write &s[0] this is equivalent to &((s + 0)) -> (s+0) That is an actual pointer to the array's first element and the double-pointer can refer to a pointer, pointers themselves have addresses because they are similar to variables that store a memory address.

Upvotes: 2

John Bode
John Bode

Reputation: 123468

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 |
                                 +---+

Upvotes: 7

Some programmer dude
Some programmer dude

Reputation: 409176

With

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

you basically have this:

+---+     +---+     +----------+
| y | --> | s | --> | "ABCDEF" |
+---+     +---+     +----------+

That is, y points to s, and s points to (the first character of) the string "ABCDEF".

Printing y will print where it is pointing, which is the location of the variable s.

Upvotes: 4

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