Free C++ Institute CPA-21-02 Exam Dumps Questions & Answers

What happens when you attempt to compile and run the following code?
#include <iostream>
using namespace std;
int main (int argc, const char * argv[])
{
int a = 30, b = 1, c = 5, i=10;
i = b < a < c;
cout << i;
return 0;
}

• A.It prints: 1
• B.It prints: 10
• C.compilation fails
• D.It prints: 0

What happens when you attempt to compile and run the following code?
#include <iostream>
using namespace std;
class First
{
public:
void Print(){ cout<<"from First";}
};
class Second:public First
{
public:
void Print(){ cout<< "from Second";}
};
void fun(First *obj);
int main()
{
First FirstObject;
fun(&FirstObject);
Second SecondObject;
fun(&SecondObject);
}
void fun(First *obj)
{
obj?>Print();
}

• A.It prints: from First
• B.It prints: from Firstfrom First
• C.It prints: from Firstfrom Second
• D.It prints: from Secondfrom Second

What is not inherited from the base class?

• A.operator+()
• B.operator=()
• C.constructor
• D.destructor

What happens when you attempt to compile and run the following code?
#include <iostream>
using namespace std;
class A {
public :
void print() {
cout << "A ";
}
};
class B {
public :
void print() {
cout << "B ";
}
};
int main() {
B sc[2];
B *bc = (B*)sc;
for (int i=0; i<2;i++)
(bc++)->print();
return 0;
}

• A.It prints: B A
• B.It prints: A A
• C.It prints: A B
• D.It prints: B B

What happens when you attempt to compile and run the following code?
#include <iostream>
#include <string>
using namespace std;
class Base
{
string s;
public:
Base() { s="Sample text";}
Base(string s) { this?>s=s; }
void Print() { cout << s; }
};
int main()
{
Base *o = new Base();
o?>Print();
}

• A.It prints: text
• B.None of these
• C.It prints: Sample text
• D.It prints: Sample