Freeing a character pointer

Question

I have a function which is called multiple times during the program's execution. In said function, I have a dynamic character pointer which I resize numerous times.

My question is: do I need to free this pointer before the end of the function?

void functionName()
{
 char *variable = (char *) malloc(0);
    //variable is resized with realloc x number of times

 //should free be called here?
 return;
}

I should also note that I have tried to free the pointer, however, gdb gives me warnings when I do so.

Answer 1

Yes, you have to free it or you'll leak the memory. Your code should look something like this:

void function(void) 
{
    char *variable = (char *)malloc(0);

    variable = realloc(variable, 10);
    // do something with new memory

    variable = realloc(variable, 20);
    // do something with more new memory

    free(variable);  // clean up
}

Calling malloc(0) is a little weird, I think.

Answer 2

From the man pages:

If size is 0, then malloc() returns either NULL, or a unique pointer value that can later be successfully passed to free().

So, I believe the answer is "yes" :).

Answer 3

Yes you need to call free() once to release the block of memory. You do not need to call free for the subsequent reallocs() that you are doing, even if those return a different address/pointer. The memory manager knows that the old block is not needed any longer and will free() it.

Answer 4

You should be able to just call free(variable) at the end. If realloc ever has to move the data in order to resize it, it calls free internally, you do not need to worry about that.

Also, where you initialize variable, you could just set it to NULL instead of calling malloc; realloc will work just like malloc the first time.

Answer 5

Have a look at some of the answers I have given to a couple of questions in relation to memory management:

• Why are memory leaks common?
• What's the point in malloc(0)?
• C String confusion?
• Need help in solving segmentation fault.
• C stdlib realloc issue?
• C: Creating array of strings from delimited strings.

All of the above point out the obvious thing, for every malloc there is a free, if not you have a memory leak, so it is imperative that you free the memory when you are finished with your pointer variable that is mallocd.

Hope this helps, Best regards, Tom.

Answer 6

A few points to make:

I can't see how you use realloc() in your code, but if you're using it like this, it's wrong:

variable = realloc(variable, amount);

When it's unable to allocate more memory, realloc() returns NULL but leaves the original pointer unaltered. In the above line, this means that variable is NULL and we've lost access to the memory it pointed to, but that memory hasn't been freed. The correct idiom is this:

void *tmp = realloc(variable, amount);
if(tmp)
  {
    // success! variable invalid, tmp has new allocated data
    variable = tmp;
  }
else
  {
    // failure! variable valid, but same size as before, handle error
  }

The reason you should use the second one is because, with realloc(), failure is bad, but is quite recoverable in many situations, unlike malloc() where failure usually means "stop everything and die."

This is a more contentious issue, but it is questionable whether or not you should cast the return value of malloc() and realloc() like you do. Consider:

// functionally identical in C
char *a = malloc(10);
char *b = (char *)malloc(10);

In C++, the cast must be made, because in C++ void * cannot be implicitly converted to another pointer type. (I think this is a language mistake, but it's not my place to judge.) If your code is C++, you should be using new and delete anyway. If your code is C but needs to compile with C++ compilers (for some inane reason), you have no choice but to cast. If you don't need to compile C code with C++ compilers (which is similar to having to run Ruby code in a Python interpreter), continue to the points below, which are why I think you shouldn't cast.

1. In C89, if a function is used without being declared, it will be implicitly declared as returning an int. If, say, we forgot to #include <stdlib.h> and we called malloc(), the version without a cast would cause a compiler error (implicit casts from int to char * aren't allowed), while the version with the cast would (wrongly) tell the compiler "I know this sounds crazy, but cast it anyway." Most compilers will give you a warning for implicit (or incompatable) declarations of built-in functions like malloc(), but the cast does make it harder to find.

2. Say you have some data:

   float *array = (float *)malloc(10 * sizeof(float));
   

   Later, you discover that you need more precision on your data, and have to make this a double array. In the above line, you need to change no more than 3 different places:

   double *array = (double *)malloc(10 * sizeof(double));
   

   If, on the other hand, you had written:

   float *array = malloc(10 * sizeof *array);
   

   You would only need to change float to double in 1 place. Furthermore, always using sizeof *obj instead of sizeof(type) and never using casts means that a later call to realloc() can work without any changes, while using casts and explicit type names would require finding anywhere you called realloc and change the casts and sizeofs. Also, if you forget, and do this:

   double *array = (float *)malloc(10 * sizeof(float));
   

   On most platforms, array will now only be an array of 5 elements, assuming the alignment isn't off and the compiler doesn't complain that you're assigning a float * to a double *. Some consider the warning the compiler issues to be helpful, as it points out potentially incorrect lines. However, if we avoid sizeof(type) and avoid casting, we can see that the lines won't be incorrect, so having the compiler draw attention to them is wasting time we could be using to program.