How to destroy a vector of pointers in c++?

Question

I have the following code in one of my methods:

vector<Base*> units;
Base *a = new A();
Base *b = new B();
units.push_back(a);
units.push_back(b);

Should I destroy the a and b pointers before I exit the method? Or should I somehow just destroy the units vector of pointers?

Edit 1:

This is another interesting case:

vector<Base*> units;
A a;
B b;        
units.push_back(&a);
units.push_back(&b);

What about this case? Now I don't have to use delete nor smart pointers.

Thanks

Answer 1

If you exit the method, units will be destroyed automatically. But not a and b. Those you need to destroy explicitly.

Alternatively, you could use std::shared_ptr to do it for you, if you have C++11.

std::vector<std::shared_ptr<Base>> units;

And you just use the vector almost as you did before, but without worrying about memory leaks when the function exists. I say almost, because you'll need to use std::make_shared to assign into the vector.

Answer 2

A rather old-fashioned solution, that works with all compilers:

for ( vector<Base*>::iterator i = units.begin(); i != units.end(); ++i )
    delete *i;

In C++11 this becomes as simple as:

for ( auto p : units )
    delete p;

Your second example doesn't require pointer deallocation; actually it would be a bad error to do it. However it does require care in ensuring that a and b remain valid at least as long as units does. For this reason I would advise against that approach.

Answer 3

You need to iterate over the vector and delete each pointer it contains. Deleting the vector will result in memory leaks, as the objects pointed to by its elements are not deleted.

TL;DR: The objects remain, the pointers are lost == memory leak.