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To learn more about pointers I wrote just a simple test function which creates a pointer, allocates space ande after some output on the shell the space shall be freed.

void main() {
  int *p = (int*) malloc(sizeof(int));
  int a = 42;
  printf("p: %x\n", p);
  printf("*p: %d\n", *p);
  printf("a: %s\n", a);
  printf("&a: %x\n", &a);
  p = &a;
  printf("p: %x\n", p);
  printf("*p: %x\n", *p);

  //Until now everything works as expected
  free(p); //ERROR
  // printf("p: %x\n", p); // Shall cause error, because space was freed
  // printf("*p: %x\n", *p); // Shall cause error, because space was freed
}

At first runs everything was ok. free(p) caused no error. Then I tried the same test with a struct and got the double free error. "Ok, maybe I do something wrong, lets go to the start", I thought and Ctrl+Z everything to this one above, like it was before. Now I still get this error. Why? Thats a newbie question, I know. The code above you can find everywhere in the web as a simple demonstration of malloc and free. Like here: http://www.cplusplus.com/reference/cstdlib/free/ Hope, someone can tell me what I do wrong.

Phantomas Phantom
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2 Answers2

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The pointer passed to free() must be returned from a previous call to malloc() (or calloc() or realloc()). This is not the case as p has the address of a after:

p = &a;

and p is then passed to free() causing undefined behaviour.

Note that in the example on page linked in the question nowhere is the pointer being reassigned after the call to any of the dynamic allocation functions.

The comment "Shall cause error, because space was freed" in the following code snippet is incorrect:

free(p); //ERROR
// printf("p: %x\n", p); // Shall cause error, because space was freed

as it is safe to print the address of a pointer, regardless of whether the memory it has been associated with has been free() or not. It is an error if you attempt to access memory that has been free()d, so this comment is correct:

// printf("*p: %x\n", *p); // Shall cause error, because space was freed

because p is being deferenced and an attempt to read free()d memory is occurring.

This is an error:

printf("a: %s\n", a); /* Must be '%d' as 'a' is an 'int'. */

Note that:

printf("p: %x\n", p);

is also undefined behaviour: use %p format specifier for pointers. See Correct format specifier to print pointer (address)?

Do I cast the result of malloc?

Community
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hmjd
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  • Ah, that's what I overseen completely. So free tries to deallocate the space of a, which is on the stack and not in the heap. And moreover the original allocated space is still there and not freed and I can't point to him, right? – Phantomas Phantom May 04 '13 at 19:35
  • @PhantomasPhantom, correct. The assignment causes a memory leak also. – hmjd May 04 '13 at 19:36
  • Thats something what we didn't learn explicitely. Thanks a lot for helping. I've been testing this for hours. Now I understand. – Phantomas Phantom May 04 '13 at 19:38
  • @PhantomasPhantom, no. `free()` does not (and cannot) set the pointer to `NULL`. It just releases the memory the pointer was pointing to. – hmjd May 04 '13 at 19:44
  • (And don't set the pointer to `NULL` yourself after `free`ing, unless you have some specific code in mind that will be checking it. That's a *very* bad habit to get into.) – David Schwartz May 04 '13 at 19:45
  • Once again. Thanks. I've learned here more about pointers as in the lessons. – Phantomas Phantom May 04 '13 at 19:57
  • Well, a bit strange. If I free a struct, the old data is still there, what is normal because it wasnt overidden by something else. But when I used the int pointer the value after freeing was 0 and not the original one. – Phantomas Phantom May 04 '13 at 20:21
  • @PhantomasPhantom, it is [undefined behaviour](http://en.wikipedia.org/wiki/Undefined_behavior). – hmjd May 04 '13 at 20:23
  • Oh, one more of C's inconsequences^^ But after freeing the content of free memory isn't relevant. But nice to know – Phantomas Phantom May 04 '13 at 20:33
  • @hmjd Didnt knew that it chooses the answer that shall be the main one. thought of it as kinda like-button :D and clicked in the second answer too^^ – Phantomas Phantom May 04 '13 at 20:42
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void main() {
  int *p = (int*) malloc(sizeof(int));   // here is error number one.
                                         // you casted your fresh malloc'd int
                                         // to an int*
                                         // which is not what you want to do
                                         // get rid of the cast you don't need it.

   // here is where your problem continues.
  int a = 42;   // this variable A was created on the stack not the heap!!
                //you cant free this, which is what you are doing below.
  printf("p: %x\n", p);
  printf("*p: %d\n", *p);
  printf("a: %s\n", a);
  printf("&a: %x\n", &a);

  p = &a; // this is a correct statement. 
          // you getting the address of var a
          // and making pointer p point to A's contents.
  printf("p: %x\n", p);
  printf("*p: %x\n", *p);

  //Until now everything works as expected
  free(p); //ERROR: here you are trying to free var A 
           // but you cant because it wasnt dynamically allocated.
  // printf("p: %x\n", p); // Shall cause error, because space was freed
  // printf("*p: %x\n", *p); // Shall cause error, because space was freed
}

hope this helps a bit. I highly recommend that you read CH 5 on pointer in K&R(Kernegie and Richie) C programming language book. its really short and you can probly find it free online somewhere. in fact i recommond that you read the whole book. best c book out there. remember you can only free what you malloced (remember each malloc gives you blocks in memory from the heap)

m33k
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    The casting seems to be kinda thing of taste. Some say "no, never cast", others say "cast!" https://www.securecoding.cert.org/confluence/display/seccode/MEM02-C.+Immediately+cast+the+result+of+a+memory+allocation+function+call+into+a+pointer+to+the+allocated+type – Phantomas Phantom May 04 '13 at 20:48