Exam (elaborations)

COS1512 Assignment (DETAILED ANSWERS) 4 2023 - DUE 18 September 2022

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Course
Introduction to Programming II

Institution
University Of South Africa

COS1512 Assignment (DETAILED ANSWERS) 4 2023 - DUE 18 September 2022 100% TRUSTED workings, explanations and solutions. For assistance call or us on . Question 1 The program below contains an incomplete recursive function raised_to_power(). The function returns the value of the first paramet...

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cos1512 assignment 4 2023

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COS1512
ASSIGNMENT 4 SEMESTER 2 2023

, COS1512 ASSIGNMENT 4 2023

TUTORIAL MATTER: Chapters 14, 15 and 17 of the Study Guide
(Tutorial Letter 102 available under Additional
Resources on the COS1512 website on
myUnisa )
Chapters 14 (excluding section 14.3), 15
(excluding sections 15.2 and 15.3) and 17 of
Savitch

WEIGHT: 25%
QUIZ AVAILABLE: 26 August 2023
DUE DATE: 18 September 2022
CUT-OFF DATE 18 September 2022

NB: This assignment consists of two parts:

• a part where you write and implement program code (this part) and
• an MCQ part where you answer questions on the code you have written,
and the material covered in this assignment.
The MCQ part of the assignment will be available in the Assessment Shell for
Assignment 4 on the myModules site for COS1512.
You will not be able to do the MCQ part unless you have completed the
coding part.

Question 1
The program below contains an incomplete recursive functionraised_to_power().
The function returns the value of the first parameternumber of type float raised to
the value of the second parameterpower of type int for all values of power greater
than or equal to 0.

The algorithm used in this question to write a recursive function to raise a float
value number to a positive power uses repeated multiplication as follows:
numberpower =1 if power = 0

= number x numberpower-1 otherwise

In other words, number raised to power gives 1 if power is 0;

Open Rubric

, and otherwise numberpower can be calculated with the formula:
number x numberpower-1

1. #include <iostream>using namespace std;
2. float raised_to_power( )
3. {
4. if (power < 0)
5. {
6. cout << "\nError - can't raise to a negative power\n";
7. exit(1);
9. }
10. else if ( )
11. return ( );
12. else
13. return (number * raised_to_power(number, power - 1));
14. }
15. main()
16.
17. float answer = raised_to_power(4.0,3);
18. cout << answer;
19. return 0;
20.}

(a) Complete the function header in line 3.
(b) Using the fact that any value raised to the power of 0 is 1, complete the base
case in line 10 and 11.
(c) Why do we need a base case in a recursive function?

(d) What is the purpose of the general case?

Question 2
Examine the code fragment below and answer the questions that follow:

1: #include <iostream>
2: using namespace std;
3:
4: //
5:
6: class A
7: {
8: private:
9: int x;
10: protected:
11: int getX();
12: public:
13: void setX();
14: };

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Exam (elaborations)

COS1512 Assignment (DETAILED ANSWERS) 4 2023 - DUE 18 September 2022

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