Double pointer in C

Question

Anyone please elaborate what is happining here?

int main()
{
    int **p = 0;
//p=?  and why| *p=?  and why|**p=? and why

    ++p;
//p=?  and why| *p=?  and why|**p=? and why

    printf("%d\n", p);
return 1;
}

output:-

4 (why?)

@Konrad: I think the code comments confused the code author. — Lightness Races in Orbit, Sep 15 '11 at 09:44
Yo Kuntal, we put a pointer in your pointer, so you can point while you point. — Luchian Grigore, Sep 15 '11 at 10:01

Simon · Accepted Answer · 2012-01-24T21:24:05.990

7

First of all, p is a pointer to a pointer-to-integer.

int **p = 0;

p = 0, *p = nothing, **p = less than nothing.

++p;

Same as p = p + 1. Means the size of one pointer to a pointer-to-int further. A pointer is basically, at least on your OS, 32 bits length (4 bytes). p now points 4 bytes after 0. The value of p is 4.

edited Jan 24 '12 at 21:24

answered Sep 15 '11 at 09:47

Simon

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Yes this is the answer I was looking for. Most of the other answers are more than C language. thanks – Kuntal Basu Sep 15 '11 at 09:59
@NeXuS: please don't edit answers to say something different, it's very confusing.. – Simon Sep 16 '11 at 08:56
Sorry: your answer was almost ok, but not completely. I thought editing would be the right solution. I beg pardon if this was not the case. – NeXuS Sep 16 '11 at 15:12

score 6 · Answer 2 · answered Sep 15 '11 at 09:43

p is a pointer to a pointer-to-int. It's being initialised to 0, i.e. it's a null pointer.

It's then being incremented to point at the next consecutive pointer-to-int in memory.^* The next pointer will be at address 4, because on your platform the size of a pointer is 4 bytes.

Then printf interprets the pointer value as an integer, and so displays "4".

_{* Note, however, that this is now undefined behaviour.}

score 4 · Answer 3 · answered Sep 15 '11 at 09:42

4

It is clear. You have a pointer to a pointer to int (int **p means a pointer to a pointer to int), that actually holds the address 0). A pointer in itself, in your architecture, is 32 bits (4 bytes) long, so incrementing p gives you p+4, that is, 0+4 = 4.

Go get a nice C book and learn about pointer arithmetic. You'll be glad the rest of your life! :)

answered Sep 15 '11 at 09:42

Diego Sevilla

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Hey!! This stackoverflow is not only for those who have good concept. May be I have less knowledge, but doesn't mean you think yourself all rounder and comment like this.I am talking about (Go get a nice C book and learn about pointer arithmetic. You'll be glad the rest of your life!) – Kuntal Basu Sep 15 '11 at 10:27
I still don't see where is the problem in learning from a good C book about pointer arithmetic. From your question one can infer you don't know enough on the subject. Sorry if this sounded rude? Wasn't my intention. – Diego Sevilla Sep 15 '11 at 10:33
So you think I am not trying to learn C. Then why do I ask questions.. must be there is a confusion about that,thats why I asked. – Kuntal Basu Sep 15 '11 at 10:39
@Kuntal: Diego is recommending that you get a C book and read it thoroughly before attempting to code in the language, because it is clear from your question that you're not quite ready to start yet. The answer to your question will come from the understanding that you will get from reading the book. I join Diego in recommending this approach to learning. – Lightness Races in Orbit Sep 15 '11 at 10:43
Thanks for your suggestion.But I am reading "C By Dennis Ritchie" and think its the right book for learning C.Now if I have confusion on something,then shouldn't I ask here? Thats all I did. – Kuntal Basu Sep 15 '11 at 10:49
I think the major problem is that you need a thicker skin. There's no problem with asking questions - and no one said that there is. Asking questions is a perfectly legitimate way to learn. Diego suggested, in good faith and with the intention of helping, that you also do a bit of reading on the subject. I see no intention to be rude from his post, and his comment clarified that. – Tom Sep 15 '11 at 13:31
@Tom yes thats the thing I was trying to say, that asking questions is one of the best way to learn. Then why marking -ve on my question. Might be its not usefull for all but many people can still find it usefull. – Kuntal Basu Sep 16 '11 at 04:58

Steve Jessop · Answer 4 · 2011-09-15T09:49:12.533

++p is actually undefined behaviour, but what appears to have happened on your implementation is that sizeof(int*) is 4, and a null pointer is address 0. Recall that pointer increment, when it's not UB, adds a number of bytes to the address equal to the size of the referand type. So it's not all that surprising that when you take a null pointer of type int** (hence the referand type is int*) and increment it, you end up at the address 4. It's just not guaranteed.

Passing a pointer when the %d format expects an int is also undefined behavior, but it appears that the representation of int and int** are sufficiently compatible, and the varargs calling convention on your implementation treats them sufficiently similarly, that it has successfully printed 4. That's also not very surprising for implementations where sizeof(int) == sizeof(int**), but also isn't guaranteed.

Of course since it's undefined behavior, there are other possible explanations for what you see.

Since there are two undefined behaviors, that's really only +0.5 each. — Steve Jessop, Sep 15 '11 at 09:52

score 0 · Answer 5 · answered Sep 15 '11 at 09:43

0

p is a pointer to pointer to int. And it's initialized to 0, i.e. NULL.

When you increment it, it now points to next pointer to int, which, on 32-bit systems, happens to be 4.

answered Sep 15 '11 at 09:43

cyco130

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Double pointer in C

5 Answers5