Possible memory leak without a virtual destructor?

Question

#include <iostream>
using namespace std;
class base
{
   int a;
 public: 
   base() {a =0;}
 };
 class derv :public base
 {
   int b;
  public:
   derv() {b =1;}
 };
 int main()
 {
    base *pb = new derv();
    delete pb;
 }

I don't have a virtual destructor in derv class, does it delete only base part of derv object??

Answer 1

It might.

Because base does not have a virtual destructor, your code exhibits undefined behavior. Anything might happen. It might appear to work as you expect. It might leak memory. It might cause your program to crash. It might format your hard drive.

A citation was requested. C++11 §5.3.5/3 states that, for a scalar delete expression (i.e., not a delete[] expression):

if the static type of the object to be deleted is different from its dynamic type, the static type shall be a base class of the dynamic type of the object to be deleted and the static type shall have a virtual destructor or the behavior is undefined.

The static type (base) is different from the dynamic type (derv) and the static type does not have a virtual destructor, so the behavior is undefined.