C pointer always contains its own memory address?

Question

Why does the pointer p always point to its own memory address as an integer in the following example. I can't see where it is initialized and would guess that it would be a garbage value. Can someone show me why it is not a garbage value. By the way I am compiling this in gcc with -std set to c99.

#include <stdio.h>

int main() {

    int *p; int a = 4;
    p = &a;
    *p++;
    printf("%d %u\n", *p, p);
}

Your code contains undefined behavior. Trying to explain the output of the program is a pointless exercise. See [this](https://stackoverflow.com/a/33797630/3386109), [this](https://stackoverflow.com/questions/2397984), and [this](https://stackoverflow.com/questions/4176328) for more information. You can however, use a debugger to step through the assembly code. — user3386109, Sep 12 '16 at 18:46
thank you. I would accept your answer if I could. My question was really about why the undefined behavior seemed deterministic, but I'm sure the assembly code will help. — Luke Murray, Sep 12 '16 at 18:48
Can you clarify? You can't see were it's initialised? What about "p = &a;"? — Steve, Sep 12 '16 at 18:52
@Steve Note that in C, `p = &a;` is _assignment_. `int a = 4;` is _initialization_. `p` is not _initialized_. — chux - Reinstate Monica, Sep 12 '16 at 21:18

Riley · Accepted Answer · 2016-09-12T19:28:17.480

4

Your problem (as the other answers point out) is with *p++;. What that says to do is dereference p then increment the address in p.

From what you are seeing, we can assume p comes directly after a in memory

_________________________________________
|something |  a  |  p  | something else |
-----------------------------------------

So what ends up happening is p points to a, then is incremented so it points to itself (or more specifically: p stores the address that p is at).

edited Sep 12 '16 at 19:28

answered Sep 12 '16 at 18:49

Riley

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Ohhhhh. Makes perfect sense. Thank you! – Luke Murray Sep 12 '16 at 18:51
Nit: `*p++` evaluates to the value stored at the current value of `p`, *then* increments `p`. So `*p++` should evaluate to `4`. – John Bode Sep 12 '16 at 19:18
@JohnBode Very true. Thanks! – Riley Sep 12 '16 at 19:28

score 0 · Answer 2 · answered Sep 12 '16 at 18:33

0

First you need to print a pointer value with %p, and your code has undefined behavior. You move the pointer one place after a and dereference it.

Your code doesn't illustrate the point you (it seems) wanted, the following will:

#include <stdio.h>

int main() {

    int *p; int a = 4;
    p = &a;
    printf("%d %p %p\n", *p, p, &p);
}

It produces something like:

4 0x7fff5c17da44 0x7fff5c17da48

p points to a then *p is the value of a. The value of p is 0x7fff5c17da44 which is the adresse of a and the address of p (&p) is 0x7fff5c17da48.

answered Sep 12 '16 at 18:33

Jean-Baptiste Yunès

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could you try inserting the line *p++; after the line p=&a. I used %u to illustrate the fact that the memory addresses are the same, but you'll see that this causes the output to change. After incrementing the pointer, the pointer points to a memory address, which contains an integer representation of itself. My question is why that integer representation is in memory there, when it doesn't appear to ever have been assigned. – Luke Murray Sep 12 '16 at 18:37
I told you that it will produce an "undefined behavior": pointing just after an existing variable is authorized, but reading the value here is "undefined behavior". – Jean-Baptiste Yunès Sep 12 '16 at 18:39
No you didn't illustrate that the addresses are the same. – Jean-Baptiste Yunès Sep 12 '16 at 18:39

C pointer always contains its own memory address?

2 Answers2