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Home/ Questions/Q 6875363
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Editorial Team
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Editorial Team
Asked: May 27, 20262026-05-27T04:18:47+00:00 2026-05-27T04:18:47+00:00

I’m trying to understand the following (Lets pretend MyStorageClass is huge) : class MyStorageClass

  • 0

I’m trying to understand the following (Lets pretend MyStorageClass is huge) :

class MyStorageClass
{
public:
  string x;
  string y;
  string z;
};

class storage
{
public:
  storage();
  ~storage()
  {
    vector<MyStorageClass *>::iterator it = myVec.begin(); it != myVec.end(); it++)
    {
      delete *it;
    }

  vector<MyStorageClass*> getItems()
  {
    for(int i = 0; i < 10; ++i)
    { 
      v.push_back(new MyStorageClass());
    }
    return v;
  }

private:
  vector<MyStorageClass*> v;

};



main()
{
  storage s;
  vector<MyStorageClass*> vm = s.getItems();
}

From my understading, when s returns the vector and is assigned to vm this is done as a copy (by value). So if s went out of scope and calls it destructor, vm has its own copy and its structure is not affected. However, passing by value is not efficient. So, if you change it to pass by reference:

vector<MyStorageClass*>& getItems()
{
  for(int i = 0; i < 10; ++i)
  { 
    v.push_back(new MyStorageClass());
  }
  return v;
}

You pass the memory location of v (in class Storage). But you still assign a copy of it using the = operator to the vector vm in the Main class. So, vm is independent of v, and if Storage destructor is called vm unaffected.

Finally, if getItems returned a reference and in main you had the following:

main()
{
  storage s;
  vector<MyStorageClass*> &vm = s.getItems();
}

Now, vm holds the address of v. So it is affected by the Storage destructor.

Question: Is what I’ve stated above true?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-27T04:18:48+00:00Added an answer on May 27, 2026 at 4:18 am

    Yes, you have understood this correctly. Now, if you want to take it to the next level, start by finding reasons why this is bad:

    1. Returning pointer or references to class internals breaks encapsulation.
    2. The reference you get will become a dangling reference once the corresponding s-object is destroyed, thereby breaking the invariant that references are always valid.
    3. By returning a vector of pointers you leave the caller wondering weather he has to delete those pointers or not.

    A better solution would be for storage to expose methods begin and end that return the corresponding iterators from s. Alternatively, you can use the Visitor-pattern for algorithms that need to operate on s.

    Also, in your case it seems like the vector s is supposed to own the objects it contains. That would be a good indicator for use of boost::ptr_vector.

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