Notes on the 2016 Exam

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Marylou Lynch
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1 Notes on the 2016 Exam The stuff inside borders is all that you needed to write. The rest is commentary. Please also read notes on previous years exams especially the 2005 set of notes. Note that up to 2008 each question was worth 20 marks (for a total exam mark out of 80). So the relative marks per piece of solution will be very different. Question 1 This required a sentinel controlled loop. Marks as follows 10 for correct handling of input (the sentinel controlled loop); 10 for correct computing of grades (that s 5 marks for noting the marks less than 30 and for counting the marks less than 40; 2 marks for correctly computing the average mark; 3 marks for correct grade based on average and number of marks less than 30 and 40); 10 for the counting of each grade and the bar chart. 3 for any attempt at a function; either for the bars of the graph or for turning a mark into a letter grade. There were a number of ways to handle the <30 / <40 grades. I have chosen one of them here. Be careful not to output twice that the same student failed (because of a <30 and a low average for example) or to count a failure twice. In my solution I output the grade in one place only and count up that grade at that point only. const int NUM_EXAMS = 12; int main() string student_no, letter_grade; double average, tot; int exam_mark; bool fail; //true if student has failed one <30 or two <40 bool onelessthan40; //true if student has failed at least one <40 int countd = 0, countp1 = 0, countp2 = 0, countp3 = 0, countf = 0; //number of students getting each overall grade cout << fixed << setprecision(2); cin >> student_no ;

2 while (student_no!= "XXX") //initialise variables for this student tot = 0; fail = false; onelessthan40 = false; cout << student_no << "\t"; for (int i=0; i < NUM_EXAMS; i++) cin >> exam_mark; //update variables for computing average tot = tot + exam_mark; //handle fails if (exam_mark < 30) fail = true; else if (exam_mark < 40) if (onelessthan40) fail = true; //if this is the second mark <40 then failed! else onelessthan40 = true; //else not a failed exam //end FOR loop average = tot/num_exams; //Debug message: // cout << endl << "Average:" << average << "\tgrade: "; if (fail) //if one <30 or two <40 letter_grade = "F"; countf++; else if (average < 40) letter_grade = "F"; countf++; //failed on the average mark else if (average < 50) letter_grade = "P3"; countp3++; else if (average < 60) letter_grade = "P2"; countp2++; else if (average < 70) letter_grade = "P1"; countp1++; else letter_grade = "D"; countd++; cout << letter_grade << endl; //next record... cin >> student_no; //end while student_no!= 999 //now print the bar chart cout << " D ";

3 for (int i=1; i<=countd; i++) cout << "*"; cout << endl; cout << "P1 "; for (int i=1; i<=countp1; i++) cout << "*"; cout << endl; cout << "P2 "; for (int i=1; i<=countp2; i++) cout << "*"; cout << endl; cout << "P3 "; for (int i=1; i<=countp3; i++) cout << "*"; cout << endl; cout << " F "; for (int i=1; i<=countf; i++) cout << "*"; cout << endl; return 0; The declaration of the functions you might have used are string letter_grade (double mark); void printnstars (int n); You can find these functions among the code on the web page. Question 2 This question is testing your ability to use functions. Therefore there were marks for designing, declaring, defining, and calling functions properly. Note that none of these functions has any cin or cout statement!! Values come to the function through the parameters and the answer is explicitly returned to the calling program or calling function using return, or values computed by the function are made available to the calling program/function via call-by-reference parameters (&). Always think carefully before putting a cin or cout statement in a function does the function explicitly require reading (i.e. its job is to read things in from the user) or printing (i.e. its job is to display stuff on screen for the user). If not, then get the values your fnction needs from the parameters list, and either return the answer or (if there are multiple answers ) use &s and update those parameters. (a) sort2 10 marks This question checked whether you knew about call-by-reference parameters. There were 3 marks for the &s in the declaration. These say that what is passed to the function is a reference to a place where an integer is

4 stored. The function uses the reference to access the location and update it. Without the & beside x1, x1 = x2 would have no effect on the value of the variable passed as the first parameter to the function the u of the first call in the question, or the w in the second call. void sort2 (int& x1, int& x2) if (x1 > x2) int temp; temp = x1; x1 = x2; x2 = temp; 1 mark for void. 4 marks for the swapping of the numbers, mostly for realizing that you need a temporary variable to swap two numbers. 2 for using IF properly. Note that you don t need an else part if x1 is not greater than x2 then you do nothing at all. (b) This question was largely based on a practical. (i) 6 marks 3 marks for the declaration and the return statement i.e. the machinery of functions. 3 marks for the sums. double distance(double x1, double y1, double x2, double y2) double x = pow((x1-x2),2); double y = pow((y1-y2),2); double result = sqrt(x+y); return result; (ii) 5 marks A lot of marks for very little. You lost one if you had if blah return True; else return False; since this is equivalent to blah. This function returns a boolean value. Fabs returns the absolute value (i.e. negative numbers are switched to positive) of a double. bool veryclose (double s1, double s2) return (fabs(s1-s2) < );

5 (iii) 8 marks 2 marks for the declaration and return statement. 1 for the call to distance function 3 marks for the logic the ifs, (or) and && (and). 2 marks reserved for using veryclose function correctly. It is not correct to just count veryclose pairs No cout statements this function returns an integer to the calling program. The calling program decides what it wants to do with that information. int triangletype(double x1, double y1, double x2, double y2, double x3, double y3) int type; double side1, side2, side3; side1 = distance(x1,y1,x2,y2); side2 = distance(x1,y1,x3,y3); side3 = distance(x2,y2,x3,y3); if (veryclose(side1,side2) && veryclose(side2,side3)) type = 3; else if(veryclose(side1,side2) veryclose(side1,side3) veryclose(side2, side3)) type = 2; else type = 0; return type; Note that you don t need to say if (veryclose(x,y) == True) since that is the same as if (veryclose(x,y)). (iv) 4 marks 1 mark for some way of giving values to the vertices (cin or hardcoded) 1.5 marks for call to triangletype 1.5 for using an if statement to output textual answers, not just 3, 2, or 1. double x1, y1, x2, y2, x3, y3; cout << "Enter x1, y1, x2, y2, x3, y3:"; cin >> x1 >> y1 >> x2 >> y2 >> x3 >> y3; int type = triangletype(x1, y1, x2, y2, x3, y3);

6 cout<<"the vertices ("<<x1<<","<<y1<<"), ("<<x2<<","<<y2<<"), ("<<x3<<","<<y3<<") make up a "; if (type == 3) cout<<"equilateral triangle"<<endl; else if (type == 2) cout<<"isoceles triangle"<<endl; else cout<<"scalene triangle"<<endl; return 0; Question 3 This question was also testing your ability to define and use functions. Many of the marks related to definition, declaration and use of read, print and percentage_gain. Also important to use daynumber and share_price functions correctly. This was an easy question for 33 marks! (a) 3 marks double percentage_gain (int old_value, int new_value) double change = ( * (new_value - old_value)) / old_value; return change; (b) 5 marks, 3 of them for &s void read_details(string& company, int& num, int& day, int& month, int& year) cin >> company >> num >> day >> month >> year; (c) 8 marks (very generous!! ) You were asked to print out value of the shares, not unit rpice. void print_details(string company, int num, int old_price, int new_price) cout << company << <<num cout << << old_price * num; cou << << new_price * num; cout<< << calculate_gain (old_price, new_price) << "%"<<endl; (d) 17 marks total; 1 for declaring and initializing variables and constants; 3 for using daynumber to compute a number for today; 3 for calling read_details; 3 for using daynumber to compute the number for the purchase date; 4 for the two calls to share_price; 3 for calling print_details.

7 //global constants for today's date const int THIS_DAY = 27; const int THIS_MONTH = 5; const int THIS_YEAR = 2016; //functions provided// int daynumber (int day, int month, int year); int share_price (string company, int daynumber); int main () int num_held, d, m, y; int cost_price, current_price; string company; int today = daynumber(this_day,this_month,this_year); read_details (company, num_held, d, m, y); int day_bought = daynumber(d,m,y); cost_price = share_price (company, day_bought); current_price = share_price (company, today); print_details(company, num_held, cost_price, current_price); return 0; Question 4 (a) (i) 4 marks This question does not require fill_up, which is designed to read an unknown number of values. Here the function is given the number of values to read. Note that it is reading words, which are strings (without spaces). Use a for loop because you know how many times to repeat. void readnwords (string a[], int n) int i; for (i=0; i<n; i++) cin >> a[i]; (ii) 6 marks This is a bool function. There are two ways to organise it. If you use a for loop, if you reach a mismatch, return false immediately, effectively abandoning the loop. If the loop completes, no mismatch was found, so return true. Note that the const keywords before the parameters are not required but what they say is that this function does not affect the values in the arrays passed in.

8 bool match (const string a[], const string b[], int size) int i; for (i=0; i<size; i++) if (a[i]!= b[i]) return false; //if we reach here all items matched return true; A while loop more naturally allows you to keep going till you find a mismatch. But you also need to watch out for the end of the array. And after the loop ends, you have to check which condition caused the loop to end the end of the array (i.e. i==size) or a mismatch. So it ends up a bit more complicated. //Alternative form bool match (const string a[], const string b[], int size) int i=0; while (i < size && a[i] == b[i]) i++; if (i == size) return true; else return false; (b) 23 marks Part (a) should have helped you to think about array functions, but you did have to make some changes. The read function for part (b) needs to read into a char array, not a string array (2 marks for making this change). And the match function of part (a) was not directly useful for part (b) (though one or two strong students did make suitable use of it), but might have steered you towards reading answers into an array and then calling a function to take the solution and an actual answer set and compute the score. The problem can be solved without using an array for the answers you would have to read answers one at a time and compare them with the correpsonding entry in the solution array, and score as you read. Marks were awarded as follows: 8 marks for the score function (roughly 3 for its declaration and the return, 2 for the for loop, 3 for the if statement / sums) only 2 marks for read since it was almost identical to Q4(a)(i). 13 marks for main, broken down roughly as 5 marks for the sentinel loop, 2 for declaring arrays, 3 for reading solution and answers (i.e. calls to readanswers), 2 for calling grade function correctly, and 1 for output. Note that if you used no functions, you lost 6 marks. #include <iostream> using namespace std;

9 void readanswers (char a[], int size); /* reads size chars into array a */ /* used to read both solution and actual answers */ int score (const char a[], const char soln[], int size); /* scores exam in array a given that soln contains the right answers */ int main () const int SENTINEL = 999; const int TESTSIZE = 30; int studentid; int grade; char solution[testsize], answers[testsize]; readanswers(solution, TESTSIZE); cin >> studentid; while (studentid!= SENTINEL) readanswers(answers, TESTSIZE); grade = score (answers, solution, TESTSIZE); cout << studentid << " " << grade << "marks" << endl; cin >> studentid; return 0; void readanswers (char a[], int size) /* reads size chars into array a */ /* used to read both solution and actual answers */ int i; for (i=0; i<size; i++) cin >> a[i]; int score (const char a[], const char soln[], int size) /* scores exam in array a given that soln contains the right answers */ int i, grade = 0; for (i=0; i<size; i++) if (a[i] == soln[i]) //right answer grade = grade + 1; else if (a[i]!= 'x') //wrong answer grade = grade -1; //else a[i] is 'x' //no change to grade return grade; If you are going to answer the array question, focus on getting basics right: Make sure that you declare the appropriate array(s) what will you need to put in the array (ints, strings, ), how many will you need (30 in this case, but often you need lots of space only some of which will actually be used).

10 Use a loop to work through the array, whether to read elements into it, print out each element, compare each element to something, add each element, whatever. Once you know the number of elements in the array (30 in this case, but for arbitrary sized arrays the actual size will be established by the reading fill_up function check the lecture notes), most of the array processing will use a for loop for (int i=0; i<size; i++) (but do check that i isn t already being used for something else, that you do want to start at 0, that size is where you have stored the number of elements in the array, and that you do want to process all values up to (size-1).) Know how to use functions that handle arrays. Always pass the array size in (usually as the last parameter). In the function declaration, use [] to indicate that the parameter is an array. When you call the function, just pass the array, without any square brackets. Inside the function you will be referring to the elements of the passed array as a[i] or something similar. Don t forget the basic rules about functions reading and printing functions are void; if their job is to compute a value, they are not void functions, so you need to decide what kind of value they return (score returns an int), and they should return a value (e.g. return grade; ) and when called, the answer needs to be stored or printed (e.g. grade = score(.); ). Other Miscellaneous Comments that I noted while grading in They are all very useful to you if you are repeating 1E3: Sentinel Controlled Loops Template is cin >> x; while (x!= ) *** cin > x; Where the *** are you put all processing of the current valid x. The processing of x must all be inside the loop. I had many examples of cin >> x; while (x!= ) cin > x;

11 tot = tot+x; //wrong more stuff on x.//wrong But those last two lines must be inside the loop, before the cin > x statement. Obviously instead of x, you might have studentno or code or n or whatever it is that you are reading that will eventually be the sentinel instead of a valid value. Many people used the above template (with ***s in it or in it!!!!) every time they thought a while loop was needed. The template is ONLY for SENTINEL CONTROLLED input. Functions Many of these comments or similar appear in Notes for 2005 exam. Local variables and accessing a function s value: If a function looks like this int f(int x) int ans; ans = x; return ans; you can t do this z = 4; f(z); cout << ans; // wrong ans here is not related to f s ans and expect it to print the value of f(z) ans only has meaning inside f. Even if you declared ans in the calling program, that would be a separate variable from the one f uses, which is a local variable whose scope in the function f. The calling program has no access to f s ans. The correct way to call f and use the value it returns is: or cout << f(z); int x = f(z); //now use or print x Accessing parameters: With the following declaration of f int f(int x, int y)

12 the body of f has access to variables called x and y, and the values in there are the ones provided by the calling function. Therefore, f should not redeclare these variables read values for these variable initialize these variables So more than likely these are all wrong: int f(int x, int y) int x, y; //creates new local variables x and y which //effectively hides the ones passed in as parameters int f(int x, int y) cin >> x >> y; //if it succeeds in reading data it will replace // the values the calling function provided. int f(int x, int y) y=0; //if the calling function provides f with a value for y //f should use it!! Not replace it. And, unrelated to parameters, but the f declared above should not print any values, but must return its answer. So NOT int f(int x, int y) cout << (x*y)/2; return 0; but probably something like int f(int x, int y) return (x*y)/2; Good luck!

Notes on the 2008 Exam

Notes on the 2008 Exam A hastily compiled review of this year s exam. Apologies if there are errors. Please also read notes on previous years exams especially the first set of notes. Question 1 Question