Question About Scope

As a C programmer doing C++ stuff a few things are still confusing to me. One is scope. Consider the following code

Code:

---

#include <vector>
using std::vector;

class A_Class{
 public:
  A_Class(int x);
// some stuff
};

void foo(vector<A_Class> *v){
Class c(42);
v->push_back(c);
}

int main(){
 vector<A_Class> *v = new vector<A_Class>();
 foo(v);
 return 0;
}

---

What happens to c and what happens to the vector when foo exits? std::vector<T>.push_back takes as an argument a T&, so my understanding is that the vector contains a reference to c, which is located on foo's stack. When foo returns, c should get nuked, which should screw up the vector. Is that right?

c is copied into an instance that lives in v. It isn't a reference that lives in v

Quote:

---

std::vector<T>.push_back takes as an argument a T&

---

No, const T&. push_back() stores a copy of the argument.

the vector is leaked also, you really don't need to allocate it dynamically. I do realize that's outside the "scope" of the question.... thank you, thank you. I'll be here all week.

Also, just fyi about the reference, const T&, it's there to avoid having copy the argument twice, especially if it's a large one.