Question

If I create an object on the stack and push it into a list, then the object loses scope (outside of the for loop in the example below) will the object still exist in the list? If the list still holds the object, is that data now invalid/possibly corrupt?

Please let me know, and please explain the reasoning..

Thanks,
jbu

class SomeObject{
public:
   AnotherObject x;
}

//And then...
void someMethod()
{
   std::list<SomeObject> my_list;
   for(int i = 0; i < SOME_NUMBER; i++)
   {
      SomeObject tmp;
      my_list.push_back(tmp);

      //after the for loop iteration, tmp loses scope
   }

   my_list.front(); //at this point will my_list be full of valid SomeObjects or will the SomeObjects no longer be valid, even if they still point to dirty data
}

EDIT: so what if it were a std::list<SomeObject*> my_list; instead of list…in that case would it be invalid?

Answer 1

All containers make a copy of what they store. It’s a requirement that an object be copy-constructible and assignable, if it is to be used in a container.

So yes, vector, list, etc. all make a copy of your object.

---

An even shorter example:

struct foo {};
std::vector<foo> v;

v.push_back(foo()); 
// makes a copy of the temporary, which dies at the semicolon.

If it didn’t make a copy, the above code would be bad.

---

The following code is not ok:

struct foo {};
std::vector<foo*> v;

{
    foo f;
    v.push_back(&f); // fine, but...
} // ...now f stops existing and...

v.front(); // ...points to a non-existent object.