Question

Question 1

Given the program below, what are the errors.

#include <iostream>

using namespace std;

int main() {

int ind_square(int &);

int x, *p;

x=15;

p = &x;

ind_square(*p);

}

int ind_square(int &p)

{

*p = *p **p;

}

Question 1 options:

	C. No return for non-void function
	A. Indirection for the variables in ind_square requires pointer operand
	A and B
	B. Invalided use of address (&) symbol as a parameter for ind_squared
	A and C

Question 2

Which statement is false about what Data Types defines

Question 2 options:

	What value a variable will hold?
	What values a variable cannot hold?
	How much memory will be reserved for the variable?
	How the program will use the data type?

Question 3

Use the correct operator to concatenate two strings:

string firstName = "John ";

string lastName = "Doe";

cout << firstName

lastName;

Question 4

The delete operator deallocates memory that was allocated.

Question 4 options:

	True
	False

Question 5

The address operator is not needed to assign an arrays address to a pointer.

Question 5 options:

	True
	False

Question 6

Given the structure, assignment and statement below. Which of the following statements about accessing the members of structure is incorrect?
struct Employee{
    string emp_id;
    string emp_name;
    string emp_sex;
};

Employee Emp, *pEmp;
pEmp=&Emp;

Question 6 options:

	Emp->emp_id;
	pEmp->emp_id;
	Emp.emp_id;

Question 7

In the program below, what are the errors?

#include <iostream>

using namespace std;

int main() {

int ivar;

int *iptr = &ivar;

ivar =10;

cout << "The value of ivar is: "<< iptr<<endl;

int nvar, nptr = &nvar;

float fvar;

int *lptr = &fvar;

int nums[50], *pptr = nums;

cout << "The address of nums is: "<< *pptr<<endl;

int *zptr = &ivar;

cout << "The value of zptr is: "<< *zptr<<endl;

int zvar;

cout << "The value of zvar is: "<< zvar<<endl;

}//

Question 7 options:

	A. The first cout prints the address of iptr
	B. lptr is an integer not a float
	C. nvar is not a pointer
	D. The second cout doesn’t print the assigned the address of nums
	a and b
	a and c
	c and b
	a, b, c and d

Question 8

Fill in the missing part to create a variable of type string called “city” and assign it the value of “Chicago”

;

Question 9

Given the code snippet below, what prints?

int main()

{

    int a[5] = {1,2,3,4,5};

    int *ptr = (int*)(&a+1);

    cout << *(a+1)<< *(ptr-5);

    return 0;

}

Question 9 options:

	2 1
	compiler error
	runtime error
	2 5

Question 10

To compare or print their value a pointer is pointing to, use the “ &” operator.

Question 10 options:

	True
	False

Answer 1

1) For the given code there are two error... one is in the function ind_square(). here

*p = *p **p; statement is weong... it should be *p = *p *(*p); there must be a parenthesis...

another one is the invalid use of & symbol....

Note: No return for non-void function will give a warning, but no error is there

so the correct option is A and B

2) the false statement is "How the program will use the data type"... How the program will use the data type? is totally depends on the user/programer...

For the other option, variables defines

What value a variable will hold : it is true

What values a variable cannot hold : it is also true

How much memory will be reserved for the variable : it is also true

3) the correct way is

string firstName = "John ";

string lastName = "Doe";

cout << firstName<<lastName;

"<<" operator is used to concatinate two string

4) the statement is True...

the syntax of delete keyword is

delete pointer_variable;

5) the given statement is true...

the syntax of assigning an array address to a pointer is

int arr[4] = {1,2,3,4};

int *ptr = arr; so there is no need of '&' for assigning.

6) The wrong option is Emp->emp_id;

Because Emp is a structure variable, not a pointer .. so we cannot use '->' operator to a veriable... it can only be used for pointers... the statement should be " Emp.emp_id; " so a dot(.) operator is needed instead of '->'

the other options are correct because '->' is used for pointer and ' . ' is used for variable..

7) For this problem, all the statements are correct...

The first cout prints the address of iptr : true because iptr is a pointer

lptr is an integer not a float : true because int pointer cannot store float variable address

nvar is not a pointer : true because nvar is an integer variable

The second cout doesn’t print the assigned the address of nums: true because it will print a garbage value

so all are correct... i.e. a,b,c and d

8) the syntax is Datatype variable_name = value;

so answer is, string city = " Chicago ";

9) The correct answer is " 2 1 "

because, *ptr holds the next address of the array a.

now, *(a+1) means value at a[1]. it will print 2

and *(p-5) means value at p[-5] which is equivalent to a[0].. so it will print 1

10) to compare, we use <,>,<=,>=,== operators and for printing the value a pointer is pointing to, we use ' * ' operator.

eg. cout<<*p; it will print the value which is at the address p.

so the statement is False