Using smart_ptr for user defined class objects

Question

As a C++ neophyte trying to understand smart pointers. I have written below code to check.

It did compile and run but I was expecting the destructor of my class to be invoked and print the cout's from the destructor but it didn't .

Do we need to overload any function in the user defined class so that its destructor is called when the smart_ptr object of that class gets destroyed.

Why is that it did not invoke the object destructor. What is that i am missing?

#include <iostream>
#include <cstdlib>
#include <tr1/memory> 
#include <string>

//using namespace std;

class myclass
{
public:
  myclass();
  myclass(int);
  ~myclass();
private:
  int *ptr;
  std::string *cptr;
};

myclass::myclass()
{
    std::cout << "Inside default constructor\n";
}

myclass::myclass(int a)
{
   std::cout << "Inside user defined constructor\n" ;
   ptr = new int[10];
   cptr = new std::string("AD");
}

myclass::~myclass()
{
    std::cout << "Inside destructor..\n";
    delete [] ptr;
    delete cptr;

    std::cout << "Freed memory..\n";
}

int main()
{


   int i;
   std::cin >> i;       
 std::tr1::shared_ptr<std::string> smartstr(new std::string);
 std::tr1::shared_ptr<myclass> smart_a(new myclass(i));
   if(i == 0)
   {
      std::cout << "Exiting...\n";
      exit(-1);
   }


}

Did you exit, or just allow `main` to return? Does it behave the same in both cases? — Useless, Jun 17 '13 at 14:12
I thought I'd give goldenmean a chance to spot the difference :) — Useless, Jun 17 '13 at 14:17
@Useless: I went down the exit(-1) path. But I get the point from the answers below. — goldenmean, Jun 17 '13 at 14:22
For completeness, you should replace the call to `exit()` with an exception `throw`. Then you will see the destructor gets called. — juanchopanza, Jun 17 '13 at 14:30
@juanchopanza - Yes thanks. Will ensure that aspect of the question i had - resource freeing on exceptions. — goldenmean, Jun 17 '13 at 14:37
@juanchopanza: You'll also need to handle the exception to be sure that the stack is unwound. — Mike Seymour, Jun 17 '13 at 15:25
@Mike Seymour - So one would need a try{ ... throw ; }catch(){...}. IS it correct? any specific things needed to be done while 'catching' the exception. — goldenmean, Jun 18 '13 at 09:16
@goldenmean: You don't need to do anything specific apart from catching it (and not rethrowing). It the exception isn't handled, then `std::terminate` is called; and it's implementation-defined whether or not the stack is unwound (destroying the local variables) first. — Mike Seymour, Jun 18 '13 at 10:12

score 11 · Accepted Answer · answered Jun 17 '13 at 14:14

The reason the object is never destroyed is because you are exiting the program by calling exit. This causes the program to exit before the smart pointer objects have a chance to go out of scope thus the objects they manage are never destroyed. Since you are in main use a return statement instead of calling exit.

score 4 · Answer 2 · answered Jun 17 '13 at 14:19

And, as additional information to other answers, note from the Standard:

Per §3.6.1/4:

Terminating the program without leaving the current block (e.g., by calling the function std::exit(int) (18.5)) does not destroy any objects with automatic storage duration (12.4).

score 2 · Answer 3 · answered Jun 17 '13 at 14:16

2

In the code below,

if(i == 0)
{
   std::cout << "Exiting...\n";
   exit(-1);
}

You are terminating the program by calling exit(), so the object is never destroyed. So remove the exit(-1); from the code.

answered Jun 17 '13 at 14:16

Deepu

7,592
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score -1 · Answer 4 · edited May 23 '17 at 12:05

-1

One possible solution is ensure that your buffer is flushed in your destructor. Use std::endl; in your destructor. For more information, please look here: Buffer Flushing, Stack Overflow

edited May 23 '17 at 12:05

Community

1
1

answered Jun 17 '13 at 14:13

Tyler Jandreau

4,245
1
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Using smart_ptr for user defined class objects

4 Answers4