Question

I know STL containers copy the objects. So say I have a

list<SampleClass> l;

each time when I do

SampleClass t(...);
l.push_back(t);

a copy of t will be made. If SampleClass is large, then it will be very costly.

But if I declare l as a container of references,

list<SampleClass&> l;

When I do

l.push_back(t);

Will it avoid copying the objects?

Answer 1

Sadly no, it won’t compile (with stlport at least). But the alternative, which is to store pointers to your objects in the container, will compile perfectly fine.

This will leave you with a bit of extra syntactic noise around your code – you’ll have to new things in order to insert them into your container.

std::list<class_type*> l;
l.push_back(new class_type);

However though the objects now won’t be copied, they also won’t be automatically cleaned up for you when the list is destructed. Smart pointers will solve this for you, but at the cost of even more syntactic noise. And since you can’t put std::auto_ptr’s in standard containers because they can’t be copied, you have to use their slightly heavier-weight boost cousins, shared pointers.

std::list<boost::shared_ptr<class_type> > l;
l.push_back(boost::shared_ptr<class_type>(new class_type));

Shared pointed do incur some extra overhead, but it is minimal.