Question

What will be printed by the code below

What will be printed by the code below

#include <iostream>
using namespace std;

class A {
public:
A() {cout << "+1";}
~A() {cout << "-1";}
};

class B: public A {
public:
B() {cout << "+2";}
~B() {cout << "-2";}
};

class C: public B {
publc:
C() {cout << "+3";}
~C() {cout << "-3";}
};

int main() {
A a1;
B b1;
A *a2 = new A;
B *b2; = new B;
delete a2;

delete b2;
return 0;
  
}

Answer 1

#include <iostream>

using namespace std;
class A {
public:
A() {cout << "+1";}
~A() {cout << "-1";}
};

class B: public A {
public:
B() {cout << "+2";}
~B() {cout << "-2";}
};

class C: public B {
public:
C() {cout << "+3";}
~C() {cout << "-3";}
};

int main() {
A a1;//due to this line +1 is printed as contructor A() is invoked is invoked internally
B b1;//due to this line +1+2 is printed as it invokes parent A() and then B()is invoked internally
A *a2 = new A; // due to this line +1 is printed as contructor A() is invoked is invoked internally
B *b2 = new B; //due to this line +1+2 is printed as it invokes parent A() and then B() is invoked internally
delete a2;//due to this line -1 is printed as destructor ~A() is invoked is invoked internally

delete b2;//due to this line -2-1 is printed as destructor ~B() and then parent destructor ~A() is invoked internally
//now a1 and b1 are not used, so garbage collector destroys them from latest to first ie b1 then a1;
//due to b1 getting destroyed -2-1 is printed as destructor ~B() and then parent destructor ~A()
//due to a1 getting destroyed -1 is printed as destructor ~A() is invoked is invoked internally

//so output is +1+1+2+1+1+2-1-2-1-2-1-1
return 0;
  
}