Question

If I add elements to a vector using the code below, then at the time I call foo, the elements (automatic variables) of vec have been destroyed since the scope in which they are created ends.

std::vector<A> vec;
for (int i = 0; i < n; i++) {
  A a;
  vec.push_back(a);
}

foo(vec);

My question is now what the textbook solution to such a problem is

Answer 1

No, the elements in vec will be different copies of a.

However, you need to allocate the size of vec if you want to use operator[] or else use vec.push_back():

for (int i = 0; i < n; i++) vec.push_back(A());

EDIT (after question change):

Even though push_back() takes its argument as a reference, internally it will make a copy of it. It takes it argument by reference to avoid making an unnecessary copy prior to making the copy to store internally.