what is happening in this simple double pointer example

Question

I have the following example code:

void doAlloc(int **p) {
  *p = (int *)malloc(sizeof(int));
  **p = 80;
}

int main() {
  int *ptr2;
  doAlloc(&ptr2);
  printf("%d\n", *ptr2);
  free(ptr2);
}

If I understand correctly, ptr2 is a pointer variable, and &ptr2 is "address of" that pointer variable in memory.

My question is, since doAlloc accepts a double pointer int **p as parameter; I don't understand why &ptr is a int **.

I would expect something like this to be done instead:

int **ptr2;
doAlloc(ptr2);

How is the "address of" ptr2 considered to be a valid value to pass as a double pointer?

Any help is greatly appreciated! Thanks.

because doAlloc gets a pointer to a pointer. A pointer itself - and not its value - is just like saying the address of. — Ariel Pinchover, Nov 10 '13 at 20:59
its something usually done in C (to send address of) but i cant recall the reason, something with extra space — Ariel Pinchover, Nov 10 '13 at 21:16
OT: In C it isn't necessary nor recommended to cast the results of `malloc/calloc/realloc`: http://stackoverflow.com/a/605858/694576 — alk, Nov 10 '13 at 21:26

score 0 · Answer 1 · edited Nov 10 '13 at 21:40

0

&ptr2 is the address of ptr2.

Since pt2r is a pointer to an integer

&ptr2 is a pointer to a pointer to an integer:

int ** ptr2;

If it may help unserstandig the code above is equivalent to:

int  *doAlloc() {
  int *p;
  p = (int *)malloc(sizeof(int));
  *p = 80;
  return p; 
}

int main(void) {
  int *ptr2;
  ptr2 = doAlloc();
  printf("%d\n", *ptr2);
  free(ptr2);
}

edited Nov 10 '13 at 21:40

alk

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answered Nov 10 '13 at 21:01

Paolo

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Is there a return statement missing in your doAlloc? – Nov 10 '13 at 21:14
If there was a return line, would it be ... `return &p` ? – Nov 10 '13 at 21:22
do you mean if it was generated by the compiler? maybe – Ariel Pinchover Nov 10 '13 at 21:23
@Infested: Not returning anything from a non void function provokes undefined behaviour. Also @Paolo: In C it isn't necessary nor recommended to cast the results of `malloc/calloc/realloc`: http://stackoverflow.com/a/605858/694576 – alk Nov 10 '13 at 21:28
@alk thats what i said - maybe and perhaps. – Ariel Pinchover Nov 10 '13 at 21:30
@Infested: "*some compilers will add a return line*" I heavily doubt this. – alk Nov 10 '13 at 21:32
@alk well, i can tell you they do. – Ariel Pinchover Nov 10 '13 at 21:33
@RobertHume: `return p;` is missing. – alk Nov 10 '13 at 21:33
@Infested: How would they know what to return? – alk Nov 10 '13 at 21:34
@alk some compilers are smarter than some ;) – Ariel Pinchover Nov 10 '13 at 21:34
@Infested: Which are the smarter ones please? Btw: If it's to be considered "smarter" not complying to the C-Standard is matter to discuss. – alk Nov 10 '13 at 21:42
@alk VS, sometimes ubuntu will also generate it. – Ariel Pinchover Nov 10 '13 at 21:44
@Robert Hume: yes, return p was missing, thanks to alk to fix my code. AFAIK compilers won't add a return line. Some IDEs would show a warning as the function is not void but end without returning anything. – Paolo Nov 10 '13 at 22:54
@Infested: Hu, I didn't knew ubuntu is a compiler. And as far as I know VS is an IDE. – alk Nov 11 '13 at 15:33
@alk youre right, not the ubuntu - a compiler in ubuntu for c/c++ files. – Ariel Pinchover Nov 11 '13 at 18:05
@Infested: Ah I see, which compiler, please (name, version)? – alk Nov 11 '13 at 18:33
@alk im not that into this, i just know it works like this on some. – Ariel Pinchover Nov 11 '13 at 22:22

Arkku · Answer 2 · 2013-11-10T21:10:42.407

A pointer is an address to a variable. The type of ptr2 is int * (pointer to int), i.e., it contains the address of an int. The argument of doAlloc is of type int ** (pointer to pointer to int), i.e., it contains the address of a pointer to int. Passing &ptr2 passes the address of ptr2, and since ptr2 is a pointer to int (int *), its address is of the correct type (int **).

(The reason why a pointer to pointer is required here is because doAlloc wants to modify the ptr2 pointer of main, not just the int it points to. A way to do it without the pointer to pointer would be to have doAlloc return a pointer to the allocated memory—then it would be the responsibility of the caller to assign it to a pointer.)

Very helpful, thanks! Antonio's answer helped me figure it out conceptually, and now I understand your answer too. :) — , Nov 10 '13 at 21:29

score 0 · Accepted Answer · answered Nov 10 '13 at 21:17

0

Very simply, if you have

int a;

then

&a

Is of type int*

In your case, you have to consider an extra pointer for all your occurrences.

answered Nov 10 '13 at 21:17

Antonio

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So I have `int *ptr2`, therefore `&ptr2` is a `int **`, correct? – Nov 10 '13 at 21:20
@RobertHume Yes, it's really that simple. – Antonio Nov 10 '13 at 21:21
Can you briefly explain why doAlloc can't just accept the original `int *ptr2` as an argument? Why does it have to be a `int **` ? – Nov 10 '13 at 21:21
@RobertHume This is explained well in Arkku answer – Antonio Nov 10 '13 at 21:26
http://www.cs.usfca.edu/~wolber/SoftwareDev/C/paramPassing.htm this might show a real usage – Ariel Pinchover Nov 10 '13 at 21:26
Thanks -- your simple answer was the simple piece of info my brain was missing. :) – Nov 10 '13 at 21:27

score 0 · Answer 4 · answered Nov 10 '13 at 21:42

A pointer in 'C' - yes - points to a memory location, where to store data (lets not talk about the type of, for now)

so let us assume: you do something like this (I try to go from the basics):

 int *p = 70;
 *p=65;

then you have your 'p' pointing to the address #70 (this is my syntax here for now thinking in 8-bit bytes no modern CPU architeckture), and then you assign '65' to where your 'p' points -> #70, so how looks the memory?

#6e ---
#6f ---
#70 65
#71 ---

so where is your 'p'?

#2212 --- here starts the stack
#2213 70 // p! 
#2214 --

and then you call: doAlloc(&ptr2);

that means: doAlloc get the address of 'p' ergo: 2213 and then it puts (a new alocated buffer address(!) there - into that 2213 - lets say '96'

#2212 --- here starts the stack
#2213 96 // p! new address of buffer 
#2213 --

and then you put '80' where this thing points to: 96 ... **p = 80;

#68 ---
#69 ---
#70 65
#71 ---
...
#94 ---
#95 ---
#96 80
#97 ---

I hope to point out what basicaly happens you could (theoretically) write

void doAlloc(int *p) {
  p = (int)malloc(sizeof(int));
  *p = 80;
}

so leave all the 2nd '*' alone

the compiler will tell you that it is wrong (and it is!), but that is what happens

what is happening in this simple double pointer example

4 Answers4