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The result of calling malloc(0) or calloc() to allocate 0 bytes (calloc(1,0), calloc(0,0), or calloc(0,1)) is undefined. From a practical standpoint, allocating 0 bytes with calloc() and malloc() can lead to programming errors with critical security implications, such as buffer overflows. This occurs because the result of allocating 0 bytes with calloc() and malloc() may not considered an error, thus the pointer returned may not be NULL. Instead, the pointer may reference a block of memory on the heap of size zero. If memory is fetched from, or stored in this a location serious error could occur.

Non-compliant Code Example 1

In this example, the user defined function calc_size() (not shown) is used to calculate the size of the string other_srting. The result of calc_size() is returned to str_size and used as the size parameter in a call to malloc(). However, if calc_size returned zero, then when the strncpy() is executed, a heap buffer overflow will occur.


int main(int argc, char *argv[]) {
  int *i_list = NULL;
  size_t size;
  
  if (argc != 2) { return -1; }

  size = atoi(argv[1]);
  if (size == 0) {
    /* Handle Error */
  }
  i_list = (int*)malloc(size);
  if (i_list == NULL) {
    /* Handle Allocation Error */
  }	
}

Compliant Code Example 1

To assure that zero is never passed as a size argument to malloc(), a check must be made on the size parameter.

int main(int argc, char *argv[]) {
  char *str = NULL;
  size_t size;

  if (argc != 2) {
    /* Handle Arguments Error */
  }
  size = strlen(argv[1])+1;
  if (size == 0) {
    /* Handle Error */
  }
  str = malloc(size);
  if (str == NULL) {
    /* Handle Allocation Error */
  }
  strcpy(str, argv[1]);
  /* Process str */
  return 0;
}

References

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