tags:

views:

229

answers:

8

Hi, I am have this simple C++ code, but I dont know how to use the destructor.

class date {

public:
    int day;
    date(int m)
    {
        day =m;
    }

    ~date(){
    cout << "I wish you have entered the year \n" << day;    
    }
};


int main()
{
  date ob2(12);
  ob2.~date();
  cout << ob2.day;
  return 0;
}

the question is,what should I write in my destructor code, that after calling the destructor, then it will delete the "day" variable. ???

+8  A: 

You do not need to call the destructor explicitly. This is done automatically at the end of the scope of the object ob2, i.e. at the end of the main function.

Furthermore, since the object has automatic storage, its storage doesn’t have to be deleted. This, too, is done automatically at the end of the function.

Calling destructors manually is almost never needed (only in low-level library code) and deleting memory manually is only needed (and only a valid operation) when the memory was previously acquired using new (when you’re working with pointers).

Since manual memory management is prone to leaks, modern C++ code tries not to use new and delete explicitly at all. When it’s really necessary to use new, then a so-called “smart pointer” is used instead of a regular pointer.

Konrad Rudolph
This is all good and well, but I believe the poster has a much more fundamental question.
Marius
Slight clarification on your third paragraph: explicitly calling destructors can be acceptable (albeit ugly and often avoidable) when using placement new.
Sydius
@Sydius: that’s why I wrote “only in low-level library code”. Let’s not confuse the OP further by throwing even more advanced notations around.
Konrad Rudolph
+13  A: 

Rarely do you ever need to call the destructor explicitly. Instead, the destructor is called when an object is destroyed.

For an object like ob2 that is a local variable, it is destroyed when it goes out of scope:

int main() 
{ 
    date ob2(12); 

} // ob2.~date() is called here, automatically!

If you dynamically allocate an object using new, its destructor is called when the object is destroyed using delete. If you have a static object, its destructor is called when the program terminates (if the program terminates normally).

Unless you create something dynamically using new, you don't need to do anything explicit to clean it up (so, for example, when ob2 is destroyed, all of its member variables, including day, are destroyed). If you create something dynamically, you need to ensure it gets destroyed when you are done with it; the best practice is to use what is called a "smart pointer" to ensure this cleanup is handled automatically.

James McNellis
+3  A: 

As you day variable is of int type (i.e. it lives in the stack) you don't need to do anything to delete it in your destructor.

If you have any pointers (heap objects) as variables in your class you should delete each of them in your destructor.

http://www-numi.fnal.gov/offline_software/srt_public_context/WebDocs/Companion/first_steps/stack_and_heap.html

Limo Wan Kenobi
It's also a good idea to set each pointer to NULL after its object is deleted, just in case you have dangling pointers to your class object laying around.
Loadmaster
The `date::day` member variable _does_ have type `int`, but that doesn't mean "it lives on the stack." It exists wherever the `date` object of which it is a member exists. In the OP's code, it does exist on the stack because `ob2` is an automatic variable. If a `date` was allocated dynamically, then it wouldn't exist on the stack.
James McNellis
The `day` member variable does not necessarily live in the stack. It could also live in modifiable memory, static memory, or in the heap. It depends on how the `date` object was instantiated.
Loadmaster
Oops, My bad, of course you are correct James McNellis.
Limo Wan Kenobi
A: 

In this case your destructor does not need to delete the day variable.

You only need to call delete on memory that you have allocated with new.

Here's how your code would look if you were using new and delete to trigger invoking the destructor

class date {

  public: int* day; 
  date(int m) { 
      day = new int;
      *day = m; 
  }

  ~date(){ 
      delete day;
      cout << "now the destructor get's called explicitly";
  } 
};

int main() { 
  date *ob2 = new date(12); 
  delete ob2;
  return 0; 
}
Doug Beleznay
Now you’re accessing `ob2` after deleting it, and you’re using `.` to access members instead of `->`. Apart from that: please don’t show code like this, it’s not good C++ and already far too many beginners’ C++ books promote this style.
Konrad Rudolph
Besides the fact that the code is allocating memory on the heap in order to store an integer, what else is wrong with the style?
Doug Beleznay
@Doug: that’s what I meant. But additionally (as I pointed out in my posting below), any usage of `new` in conjunction with raw pointers is somewhat deprecated in modern C++ because it leads to so much trouble. Certainly for beginners.
Konrad Rudolph
The code has the added problem of not having a properly defined copy constructor and copy assignment operator, which is an extremely common pitfall for beginners.
James McNellis
+1  A: 

Only in very specific circumstances you need to call the destructor directly. By default the destructor will be called by the system when you create a variable of automatic storage and it falls out of scope or when a an object dynamically allocated with new is destroyed with delete.

struct test {
   test( int value ) : value( value ) {}
   ~test() { std::cout << "~test: " << value << std::endl; }
   int value;
};
int main()
{
   test t(1);
   test *d = new t(2);
   delete d;           // prints: ~test: 2
}                      // prints: ~test: 1 (t falls out of scope)

For completeness, (this should not be used in general) the syntax to call the destructor is similar to a method. After the destructor is run, the memory is no longer an object of that type (should be handled as raw memory):

int main()
{
   test t( 1 );
   t.~test();            // prints: ~test: 1
                         // after this instruction 't' is no longer a 'test' object
   new (&t) test(2);     // recreate a new test object in place
}                        // test falls out of scope, prints: ~test: 2

Note: after calling the destructor on t, that memory location is no longer a test, that is the reason for recreation of the object by means of the placement new.

David Rodríguez - dribeas
Usual way of allocating memory to be used with placement new is with malloc (or new with an array). This is the 1st example I see with the stack variable. btw in the 1st example, it would be test *d=new test(2);
VJo
@VJo: I started to write an example with an stack allocated memory buffer, but that required the use of placement new *before* calling the destructor (small issue) and did not show the potential issue of multiple calls to the destructor. There is also another *auto* allocated example in the new standard: with the lesser requirements placed on the members of an union, it allows for objects with non-trivial constructor/destructors. In that case, it is up to the user to explicitly call the constructor and destructor of the active member of the union.
David Rodríguez - dribeas
+1  A: 

You do not call your constructor explicitly.

When you create your object on the stack (like you did) all you need is:

int main()
{
  date ob2(12);
  // ob2.day holds 12
  return 0; // ob2's destructor will get called here, after which it's memory is freed
}

When you create your object on the heap, you kinda need to delete your class before its destructor is called and memory is freed:

int main()
{
  date* ob2 = new date(12);
  // ob2->day holds 12
  delete ob2; // ob2's destructor will get called here, after which it's memory is freed
  return 0;   // ob2 is invalid at this point.
}

(Failing to call delete on this last example will result in memory loss.)

Both ways have their advantages and disadvantages. The stack way is VERY fast with allocating the memory the object will occupy and you do not need to explicitly delete it, but the stack has limited space and you cannot move those objects around easily, fast and cleanly.

The heap is the preferred way of doing it, but when it comes to performance it is slow to allocate and you have to deal with pointers. But you have much more flexibility with what you do with your object, it's way faster to work with pointers further and you have more control over the object's lifetime.

Marius
“The heap is the preferred way of doing it” – doing *what*? Storing objects? Because that would be wrong: the stack is absolutely the preferred way of doing that in C++.
Konrad Rudolph
Working with libraries and passing objects between them it is preferable (nay, vital) to have them allocated on the heap. Sure the stack is quick and dirty and in some cases make sense. But when you have objects that you want to use, pass around, move, copy and reference count, then the stack is no place for them. Until C++0x move semantics become commonplace that is...
Marius
@Marius: sorry but this is wrong. For argument-passing, use (const) references. For return values – simply copy. Nothing bad will happen. See e.g. http://cpp-next.com/archive/2009/08/want-speed-pass-by-value/ Libraries should never require heap-allocated arguments. This would be a **very** bad library design. And there is nothing quick and dirty about stack allocation.
Konrad Rudolph
Exactly all right; I agree fully, but this really does fail to answer the fundamental question the poster had. I believe my answer will give him a better idea of how to 'use' a destructor (since that is the fundamental and simple concept he is lacking) and how it works. It may not be reference-correct and will not stand up to language-lawyers, but I think he will get a better better idea of object lifetime with *simple* examples. Which is why I think James' post is still ostensibly perfect.
Marius
A: 

Even though the destructor seems like something you need to call to get rid of or "destroy" your object when you are done using it, you aren't supposed to use it that way.

The destructor is something that is automatically called when your object goes out of scope, that is, when the computer leaves the "curly braces" that you instantiated your object in. In this case, when you leave main(). You don't want to call it yourself.

sheepsimulator
A: 

You may be confused by undefined behavior here. The C++ standard has no rules as to what happens if you use an object after its destructor has been run, as that's undefined behavior, and therefore the implementation can do anything it likes. Typically, compiler designers don't do anything special for undefined behavior, and so what happens is an artifact of what other design decisions were made. (This can cause really weird results sometimes.)

Therefore, once you've run the destructor, the compiler has no further obligation regarding that object. If you don't refer to it again, it doesn't matter. If you do refer to it, that's undefined behavior, and from the Standard's point of view the behavior doesn't matter, and since the Standard says nothing most compiler designers will not worry about what the program does.

In this case, the easiest thing to do is to leave the object untouched, since it isn't holding on to resources, and its storage was allocated as part of starting up the function and will not be reclaimed until the function exits. Therefore, the value of the data member will remain the same. The natural thing for the compiler to do when it reads ob2.day is to access the memory location.

Like any other example of undefined behavior, the results could change under any change in circumstances, but in this case they probably won't. It would be nice if compilers would catch more cases of undefined behavior and issue diagnostics, but it isn't possible for compilers to detect all undefined behavior (some occurs at runtime) and often they don't check for behavior they don't think likely.

David Thornley