Output of C++ Program | Set 10

Predict the output of following C++ programs.

Question 1

#include<iostream> #include<string.h> using namespace std;   class String {     char *p;     int len; public:     String(const char *a); };   String::String(const char *a) {     int length = strlen(a);     p = new char[length +1];     strcpy(p, a);     cout << "Constructor Called " << endl; }   int main() {     String s1("Geeks");     const char *name = "forGeeks";     s1 = name;     return 0; }

Output:

Constructor called
Constructor called

The first line of output is printed by statement “String s1(“Geeks”);” and the second line is printed by statement “s1 = name;”. The reason for the second call is, a single parameter constructor also works as a conversion operator (See this and this for deatils).

Question 2

#include<iostream>   using namespace std;   class A {     public:     virtual void fun() {cout << "A" << endl ;} }; class B: public A {     public:     virtual void fun() {cout << "B" << endl;} }; class C: public B {     public:     virtual void fun() {cout << "C" << endl;} };   int main() {     A *a = new C;     A *b = new B;     a->fun();     b->fun();     return 0; }

Output:

C
B

A base class pointer can point to objects of children classes. A base class pointer can also point to objects of grandchildren classes. Therefor, the line “A *a = new C;” is valid. The line “a->fun();” prints “C” because the object pointed is of class C and fun() is declared virtual in both A and B (See this for details). The second line of output is printed by statement “b->fun();”.

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