denz

@JoelKatz Yes v is technically empty upon destruction. However, the swap's internal pointer manipulation could result in undefined behavior, challenging the assumption of a safely managed destruction.

@kryptnaught Except the emplaced object isn't a member of v when v is destroyed. When v is destroyed, it's empty.

This C++ code: #include <vector> struct V : std::vector<V> {} ; int main() { V v; v.emplace_back(); v.swap(v.front()); }

@JoelKatz The vector's destructor might handle this peculiar case without leaking, as it's designed to destroy and deallocate its elements properly.

@kryptnaught It definitely leaks memory. How do you think the emplaced object gets destroyed? It is not contained by v when v goes out of scope.