ILP C++ Final Quiz

1. A special library function which is called automatically when no handler at any level catches the exception

Exit()

Terminate()

Quit()

Ellipse

The terminate() function is a special library function that is called automatically when no exception handler at any level catches the exception. It is responsible for terminating the program and cleaning up any resources that were allocated.

Explanation

The terminate() function is a special library function that is called automatically when no exception handler at any level catches the exception. It is responsible for terminating the program and cleaning up any resources that were allocated.

2. int main() { int a=2,b=3; if ((a==2) || (b++==4)) cout<<"success"; return 0; } What will be the values of a and b ?

2,3

2,4

2,5

None of the above

The values of a and b will be 2 and 3 respectively. In the if statement, the first condition (a==2) is true, so the second condition (b++==4) is not evaluated. Therefore, the value of b remains unchanged. Since the if statement is true, the statement "success" is printed.

Explanation

The values of a and b will be 2 and 3 respectively. In the if statement, the first condition (a==2) is true, so the second condition (b++==4) is not evaluated. Therefore, the value of b remains unchanged. Since the if statement is true, the statement "success" is printed.

3. Class VirtualClass { int x,y; public: VirtualClass() { x=1,y=2; } virtual void operator<<(ostream&); }; // end of class void VirtualClass::operator<<(ostream &os) { os<<this->x<<"\t"<<this->y; } int main() { VirtualClass vcObj; vcObj<<cout; return 0; }

1 2

Compilation error

Run time error

None of the above

The code defines a class called VirtualClass with two integer variables x and y. The constructor initializes x to 1 and y to 2. The class also overloads the

Explanation

The code defines a class called VirtualClass with two integer variables x and y. The constructor initializes x to 1 and y to 2. The class also overloads the

4. What is the output of the following code int main() { char s1[20]="Hello"; char s2[20]="Hai"; s2=s1; cout<<"String1 : "<<s1; cout<<" String2 : "<<s2; return 0; }

String1 : Hello String2 : Hai

String1 : Hello String2 : Hello

Error : forbidden assignment of arrays

Error: incompatible types in assignment

The given code tries to assign one array to another, which is not allowed in C++. This results in a compilation error stating "forbidden assignment of arrays". Arrays cannot be assigned directly to each other in C++, instead, you need to use functions like strcpy or strncpy to copy the contents of one array to another.

Explanation

The given code tries to assign one array to another, which is not allowed in C++. This results in a compilation error stating "forbidden assignment of arrays". Arrays cannot be assigned directly to each other in C++, instead, you need to use functions like strcpy or strncpy to copy the contents of one array to another.

5. What is the output of the below code int main(){ int i=0; for(i=0;i<20;i++) switch(i) { case 0:i+=5; case 1:i+=2; case 5:i+=5; default: i+=4; } cout<<i; return 0; }

84

5

22

16

The code starts with initializing the variable i to 0. Then, it enters a for loop with the condition i
When i is 0, it adds 5 to i and continues to the next case.
When i is 1, it adds 2 to i and continues to the next case.
When i is 5, it adds 5 to i and continues to the next case.
If none of the above cases match, it goes to the default case and adds 4 to i.

After the switch statement, the value of i is printed using cout.

In this code, the loop runs until i becomes 20. The switch statement modifies the value of i based on the cases.

The final value of i is 22, which is the output of the code.

Explanation

The code starts with initializing the variable i to 0. Then, it enters a for loop with the condition i
When i is 0, it adds 5 to i and continues to the next case.
When i is 1, it adds 2 to i and continues to the next case.
When i is 5, it adds 5 to i and continues to the next case.
If none of the above cases match, it goes to the default case and adds 4 to i.

After the switch statement, the value of i is printed using cout.

In this code, the loop runs until i becomes 20. The switch statement modifies the value of i based on the cases.

The final value of i is 22, which is the output of the code.

6. What is the output of the following code class base{ public : virtual void func1()=0; void func2() { cout<<"\n In base, function2"; } }; //end of class class derived : public base{ public : void fun3() { cout<<"\n In derived , function3"; } };//end of class int main(){ derived d; d.fun3(); return 0; }

In derived, function1

In derived , function3

Compilation error

In base, function2

The code will result in a compilation error because the derived class is not implementing the pure virtual function func1() from the base class. Since func1() is declared as a pure virtual function in the base class, it must be implemented in any derived class that inherits from the base class. Since the derived class does not implement func1(), the code will not compile.

Explanation

The code will result in a compilation error because the derived class is not implementing the pure virtual function func1() from the base class. Since func1() is declared as a pure virtual function in the base class, it must be implemented in any derived class that inherits from the base class. Since the derived class does not implement func1(), the code will not compile.

7. What is the output of the following code class base{ public : virtual void display() { cout<<" \n I am in base-display ";} void virtual show() { cout<<" \n I am in base-show "; } virtual void show1(){cout<<"In base show1 ";} }; class derived : public base {}; int main(){ derived d; cout<<sizeof(d)<<"\n";}

4

8

10

16

The output of the code is 8. This is because the sizeof operator is used to determine the size of an object or data type in bytes. In this case, the object "d" is of type "derived", which is derived from the base class. Since the derived class does not have any additional member variables or functions, its size is the same as the base class, which is 8 bytes.

Explanation

The output of the code is 8. This is because the sizeof operator is used to determine the size of an object or data type in bytes. In this case, the object "d" is of type "derived", which is derived from the base class. Since the derived class does not have any additional member variables or functions, its size is the same as the base class, which is 8 bytes.

8. class Test{ public: ~Test() { cout<<" Destructor"; } }; //end of class int main() { Test t; t.~Test(); return 0; }

Run time error

Destructor Destructor

Compilation error

Destructor

The given code snippet creates an object of the Test class named "t" in the main function. Then, the destructor of the Test class is explicitly called using the syntax "t.~Test();". This results in the destructor being called twice, leading to the output "Destructor Destructor". Therefore, the correct answer is "Destructor Destructor".

Explanation

The given code snippet creates an object of the Test class named "t" in the main function. Then, the destructor of the Test class is explicitly called using the syntax "t.~Test();". This results in the destructor being called twice, leading to the output "Destructor Destructor". Therefore, the correct answer is "Destructor Destructor".

9. What is the output of the following code int main(){ int i=4,j; j=i<<2; cout<<j; return 0;}

4

8

12

16

The code assigns the value 4 to the variable i. Then, it performs a left shift operation on i by 2 bits, resulting in 16. Finally, it prints the value of j, which is 16.

Explanation

The code assigns the value 4 to the variable i. Then, it performs a left shift operation on i by 2 bits, resulting in 16. Finally, it prints the value of j, which is 16.

10. #define SQR(x) (x*x) main() { int a,b=3; a=SQR(b+2); cout<<a; return 0; }

11

25

Error

17

The correct answer is 11.

In the given code, the macro SQR(x) is defined as (x*x).

In the main function, the value of b is 3.

The expression a=SQR(b+2) is evaluated as a=(b+2)*(b+2), which becomes a=3+2*3+2, which simplifies to a=11.

Therefore, the value of a is 11, which is outputted by the cout statement.

Explanation

The correct answer is 11.

In the given code, the macro SQR(x) is defined as (x*x).

In the main function, the value of b is 3.

The expression a=SQR(b+2) is evaluated as a=(b+2)*(b+2), which becomes a=3+2*3+2, which simplifies to a=11.

Therefore, the value of a is 11, which is outputted by the cout statement.

Ilp C++ Final Quiz