explanation on simple c++ inheritance -

#include <iostream>  using namespace std; class { public:     void m1(){ cout << 'a'; }     virtual void m2(){ cout << 'b'; }     virtual void m3(){ cout << 'c'; } };  class b: public { public:     void m1(){ cout << 'd'; }     void m2(){ cout << 'e'; } };  class c: public b { public:     void m3(){ cout << 'f'; } };  int main() {     cout << "hello world!" << endl;     a* = new b();     a->m1();     a->m2();     a->m3();     return 0; } 

what output? thought "d e c" after running program "a e c"

could 1 elaborate going on behind line of code:

a* = new b(); 

virtual member functions dispatched based on dynamic (run-time) type of object. non-virtual member functions dispatched based on static (compile-time) type of object.

a *a = new b(); 

a points object dynamic type b. static type of a a*, however, means static type of *a a.

virtual functions (m2 , m3) dispatched based on dynamic type, b::m2 , b::m3 called.

non-virtual functions dispatched based on static type. static type of *a a, a::m1 called.

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what going on in new line? new object of type b created dynamically, , new expression returns pointer object (of type b*). then, derived-to-base conversion applied pointer convert a*, used initialise variable a.

in pseudo-code, showing intermediary steps:

b *tmp_b = new b(); // allocate , initialise b object *tmp_a = convert_derived_to_base(tmp_b); *a = tmp_a;