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.

 

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. The second line of output is printed by statement “b->fun();”.

　　

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　　2013-11-27　　15:56:12