Writing a function to free a pointer and the assigning it NULL

Question

I was asked this in a recent interview, basically writing a function to combine the functionality of free and assigning null. I answered in the following manner:

void main()
{
      int *ptr;
      ptr = new int;
      ptr = newdelete(ptr);
}

(int*) newdelete (int *ptr)
{
      delete(ptr);
      return NULL;
}

So after execution, the ptr local to main will hold the null value as I am returning it from the newdelete function. If I had just assigned NULL in the newdelete function, the ptr local to newdelete would be nulled and not the ptr local to main.

I think my solution was correct, the interviewer accepted it too. However, he was expecting some other answer. He was insisting I do not return the NULL from the function and still achieve the desired result.

Is there any way to accomplish that? All I can think of is passing another argument which is the pointer to the pointer ptr local to main, but I don't see why it's better than what I did!

Answer 1

Is there any way to accomplish that??

template <typename T> void safeDelete(T*& p){
    delete p;
    p = 0;
}

int main(int argc, char** arv){
    int * i = new int;
    safeDelete(i);
    return 0;
}

Answer 2

I guess he was expecting something like:

void reset(int*& ptr)
{
      delete(ptr);
      ptr = NULL;
}

An even cleaner solution would have been to use a boost::shared_ptr<> and to simply call ptr.reset(). However, I suppose this wasn't an option.

Answer 3

If the requrement wasn't to write a function, you could always write a macro that would do it for you as well:

#define my_delete(x) { delete x; x = NULL; }

Of course, calling it like this will get you into all sorts of trouble:

my_delete(ptr++)

So, I think I prefer the non-macro way.