EEN118 22nd November These are my answers. Extra credit for anyone who spots a mistike. Except for that one. I did it on purpise.

Transcription

1 EEN118 22nd November 2011 These are my answers. Extra credit for anyone who spots a mistike. Except for that one. I did it on purpise.

2 5. An automatic animal monitoring device sits in the african savannah monitoring all the animals that pass within range. Throughout the day it creates a large text file that records all of its observations. At the end of the day that file is uploaded to a research base and analysed. The file has one line for each animal observed. Each line has the same format: first the time (in 24 hour form), then the species of the animal (a single word, using only lower case letters), then the direction it was heading (N, S, E, or W), then an estimate of its weight (in pounds), and finally an estimate of its speed (in miles per hour). Here is a sample from a typical file gorilla E gorilla E gorilla E zebra W lion W penguin S gorilla W It seems to show a family of gorillas strolling towards the rising sun, then shortly later a zebra chased by a lion going in the opposite direction, and then a small penguin flying south and one of the gorillas coming back. The name of the file is animals.txt. Write a function that processes this file and does the following things: + It must create a new file called gorillas.txt containing all the records of gorilla observations. + It must calculate the average weight of all the lions observed. + It must note the speed of the slowest gorilla seen heading east in the morning (time between 0000 and 1200). So if the sample data given above were the whole file, your function would create a gorillas.txt file containing 0530 gorilla E gorilla E gorilla E gorilla W And it would print for the user to see: Average lion weight pounds Slowest east-bound morning gorilla 2.8 mph

3 void main() { // a ifstream infile("animals.txt"); // b if (infile.fail()) { cout << "Can't read animals.txt\n"; exit(1); } // c ofstream gorfile("gorillas.txt"); // b again if (gorfile.fail()) { cout << "Can't write gorillas.txt\n"; exit(1); } // d double impossible_speed = ; double lion_sum = 0.0, slowest = impossible_speed; int num_lions = 0; // e while (true) { int time; string species, direction; double weight, speed; // f infile >> time >> species >> direction >> weight >> speed; // g if (infile.fail()) break; if (species=="gorilla") { // h gorfile << time << " " << species << " " << direction << " " << weight << " " << speed << "\n"; // j if (direction=="e" && time<1200 && speed<slowest) slowest = speed; } // i if (species=="lion") { lion_sum += weight; num_lions += 1; } } // k infile.close(); gorfile.close(); // m if (num_lions!= 0) // l cout << "Average lion weight " << lion_sum/num_lions << " pounds\n"; cout << "Average lion weight n/a because no lions were seen\n"; if (slowest!= impossible_speed) cout << "Slowest east-bound morning gorrilla " << slowest << " mph\n"; cout << "Slowest east-bound morning gorrilla n/a because none seen\n"; }

4 6. A struct (or object) to represent dates is to be implemented. It must contain a string for the day of the week, and three numbers for the day of the month, the month, and the year. a. Give the struct definition, along with a suitable set function for initialising a date, and a print function that prints one nicely. Using your definitions, I should be able to write this: date today, st_swithins_day; set(today, Tuesday, 22, 11, 2011); set(st_swithins_day, Friday, 15, 7, 2011); print(today); and when run, that code should print something very close to this Tuesday 22nd November 2011 or this Tuesday November 22nd 2011 b. Define a function called latest, which takes two dates as parameters, and returns as its result the one that is latest (i.e. comes second). So for example, this date x = latest(today, st_swithins_day); print(x) should also print today s date. c. Define a function called next, which takes one date as its parameter, and modifies that date by moving it on to the next day, so this would print next(today); print(today); next(today); print(today); Wednesday November 23rd 2011 Thursday November 24th 2011

5 a. Remember I said you could abbreviate reasonably, you would not have written all the strings. And of course there are millions of other ways of doing it, this is just the way I think is neatest. struct date { string dow; // short for day of week int d, m, y; }; void set(date & when, string dow_in, int d_in, int m_in, int y_in) { when.dow = dow_in; when.d = d_in; when.m = m_in; when.y = y_in; } string suffix[] = { "", "st", "nd", "rd", "th", "th", "th", "th", "th", "th", "th", "th", "th", "th", "th", "th", "th", "th", "th", "th", "th", "st", "nd", "rd", "th", "th", "th", "th", "th", "th", "th", "st" }; string month_name[] = { "", "January", "February", "March", "April", "May", "June", "July", "August", "September", "October", "November", "December" }; void print(date when) { cout << when.dow << " " << when.d << suffix[when.d] << " " << month_name[when.m] << " " << when.y << "\n"; } b. date latest(date a, date b) { if (a.y > b.y) return a; if (b.y > a.y) return b; if (a.m > b.m) return a; if (b.m > a.m) return b; if (a.d > b.d) return a; return b; }

6 c. As with part a, you d abbreviate and it would be nice and small. int usual_days_in[] = { 0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 }; int leap_days_in[] = { 0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 }; bool isleap(int yr) { if (yr % 4 == 0) return true; return false; } void next(date & x) { x.d += 1; if (isleap(x.y)) { if (x.d > leap_days_in[x.m]) { x.d = 1; x.m += 1; } } { if (x.d > usual_days_in[x.m]) { x.d = 1; x.m += 1; } } if (x.m > 12) { x.m = 1; x.y += 1; } if (x.dow == "Monday") x.dow = "Tuesday"; if (x.dow == "Tuesday") x.dow = "Wednesday"; if (x.dow == "Wednesday") x.dow = "Thursday"; if (x.dow == "Thursday") x.dow = "Friday"; if (x.dow == "Friday") x.dow = "Saturday"; if (x.dow == "Saturday") x.dow = "Sunday"; if (x.dow == "Sunday") x.dow = "Monday"; }

7 7. Here is an attempt at sorting an array of ints so that they will appear in ascending order. It consists of a helper function that finds the largest int in an array, and a sorting function that repeatedly finds the largest int and moves it to the end of the array. Finally, there is a main() that tests it. int find_biggest(int A[1000] A,I ) { int biggest_so_far = 0; int pos = 0; while (pos<= J 1000 I ) { if (A[pos]>biggest_so_far) C biggest_so_far = A[pos]; pos = pos + 1; } B } int L swap(int D x, E y) { int t = x; y = t; F x = y; } void sort(int A[1000] A,H ) { int end_pos = 1000 J ; while (end_pos>0) { int big = find_biggest(a H,I ); swap(a[big], A[end_pos]); G } } void main() { int data[1000]; int number; cout << How many numbers are you going to type? ; cin >> number; cout << OK, now type them all\n ; for (int i = 0; i<number; i += 1) cin >> data[i]; sort(data[1000] H,K ); cout << Here they are sorted\n ; for (int i = 0; i<number; i += 1) cout << data[i] << \n ; } There is a lot wrong with this program. Tell me all the mistakes, and what must be done to fix them and make it work.

8 To make it easier on you, I ll tell you that there is nothing wrong with the input and output part, so don t waste time looking for mistakes in the parts with grey backgrounds. For this one, I ve marked the question with large red letters where the mistakes are, to connect to the explanations here. a. Don t say the size for array parameter. Just int A[] b. It says int biggest(...), but didn t return anything. c. Look at how sort uses this function. It must return the position of the biggest. d. Must be reference parameters, need & e. Both params need types: (int & x, int & y) f. This doesn t swap. Should be t=x; x=y; y=t; g. Infinite loop, need end_pos -= 1; h. number should be a param to both functions. int find_biggest(int A[], int N) void sort(int A[], int N) and in main: sort(data, number) i. This will keep finding the same biggest thing, should be find_biggest(a, end_pos-1) j. Array has elements [0] to [999], should be while (pos<n) and int end_pos = N-1; k. data is the name of the whole array, just say sort(data). Except for what I said for item h. l. Should be void, swap doesn t produce a value.

9 8. Draw a picture of a cat with a hat, or a hat with a cat, whichever you prefer. The cat s name is Dennis. * / \ =. = / ( ) / ( )/ ( _ )