Explain why the two calls to function g produces different results in each case:

Question

 Explain why the two calls to function g produces different results in each case:  #include <iostream> using namespace std; class X {     int x; public:     X() { x = 10; }     void f() { cout << "f called " << x <<  endl; } };  class Y : public X {     int y; public:     Y() { y = 20; }     void g() { cout << "g called" << y << endl; } };  int main() {    X* a = new X;    Y* b = new Y;    ((Y*)a)->g();    a = b;    ((Y*)a)->g(); } 

Explanation / Answer

X* a = new X; Y* b = new Y; ((Y*)a)->g(); // Here we are just getting reference 'a' to be converted into (Y*) without creating object.... So the variable 'y' in class Y stores '0' by default... So it will be printed "g called 0" a = b; // We are assigning classY object to classX referrence ((Y*)a)->g(); // Here after assigning classY object to referrence 'a' we are calling method g().. So as we created object for class Y, its variable will be assigned with '20' in the constructor.. So it will be printed "g called 20"

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