Answer / lylez00

If A is a base class, and from that, B is derived, and a
dynamically allocated object of type B is deleted via a
pointer of type A, then B's destructor will not be invoked
unless A's destructor is virtual.

A *a = new B();
delete a; // won't invoke B's destructor unless A's
destructor is virtual

Answer / p govind rao

A destructor can be declare virtual. virtual destructor is
mainly useful during inheritance.

class base
{
public:
base(){}
virtual ~base(){}
};

class derv
{ char *p;
public :
derv(){ptr=nes char[2];}
~derv(){delete ptr;}

} ;

main()
{
base *baseptr=new derv();
delete baseptr;
}

If base class, and derived class, and a dynamically
allocated object of type derived is deleted via a pointer
of type base, then derived's destructor will not be invoked
unless base's destructor is virtual.

base *baseptr = new derv();
delete baseptr; // won't invoke B's destructor unless A's
destructor is virtual

this is to swap to strings....but in output the whole strings are swapped leaving first as it is...why it is so #include<iostream.h> int main() { char a[]="ajeet"; char b[]="singh"; long x=*a; long y=*b; cout<<x<<":"<<y; x=x+y; y=x-y; x=x-y; *a=x; *b=y; cout<<x<<":"<<y; cout<<&a<<endl; cout<<&b<<endl; }

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