CS202 Midterm - March 9, 2023 - Answers and Grading

James S. Plank

Question 3

A. Animal is a class that exhibits polymorphism: True. The method ggg() has two implementations that differ based on the types of their arguments.
B. Animal defines a customized copy constructor: True. That is the following line in Animal.cpp:
Animal(const Animal &a);
C. Animal defines a customized assignment overload: False. Please see the lecture notes for how you define that.
D. Please tell me the value of s1 just before main() returns. s1 is set to "simba", and then it is the first parameter to ggg(). That is not a reference parameter, so it is copied, and s1 is unchanged. The answer is "simba".
E. Please tell me the value of s2 just before main() returns. s2 is set to "jaba", and then it is the first parameter to hhh(). That is not a reference parameter, so it is copied, and s2 is unchanged. The answer is "jaba".
F. Please tell me the value of t1 just before main() returns. t1 is set to "tony", and then it is the second parameter to ggg(). That is a reference parameter, so ggg() could change it. It could be anything. The answer is "multiple".
G. Please tell me the value of t2 just before main() returns. t2 is set to "speedy", and then it is the second parameter to hhh(). That is a const reference parameter, so and t2 is unchanged. The answer is "speedy".
H. Please tell me the value of i1 just before main() returns. i1 is set to 75, and then it is the parameter to ggg(). It is a reference parameter, so its value could by anything afterwards. The answer is "multiple".
I. Please tell me the value of i2 just before main() returns. i2 is set to 85, and then it is the parameter to iii(). It is a const reference parameter, so its value cannot be changed. The answer is 85.
J. Please tell me the value of a1.tiger just before main() returns. a1.tiger is set to "sherekhan", and then a1.ggg() is called. Since ggg() is const, a1.tiger won't be changed. The answer is "sherekhan".
K. Please tell me the value of a2.tiger just before main() returns. a2 is copied from a1, and then a2.ggg() is called. Since ggg() is const, a2.tiger won't be changed. The answer is "sherekhan".
L. Please tell me the value of a3.tiger just before main() returns. a3.tiger is set to "speedy", and then a3.hhh() is called. Since hhh() is not const, a3.tiger can be anything afterwards. The answer is "multiple".
M. Please tell me the value of a4.tiger just before main() returns. a4.tiger is set to "hangry", and then a4.iii() is called. Since iii() is not const, a4.tiger can be anything afterwards. The answer is "multiple".
N. Please tell me the value of av.at(0).tiger just before main() returns. av.at(0).tiger is set to "lazy", and then av is resized. However, that resizing can't change av.at(0), so The answer is "lazy".
O. Please tell me the value of av.at(1).tiger just before main() returns. When av is resized to 4, av[1] will be initialized by the copy constructor. Since we don't know how that is implemented, av.at(1).tiger can be anything. The answer is "multiple".

Here's a table of the correct answers:

A B C D E F G H I J K L M N O
Fred True. True. False. spot elsie multiple jenner multiple 85 rex rex multiple multiple moomoo multiple
Barn True. True. False. sweater fran multiple bruce multiple 85 baba baba multiple multiple wooly multiple
Animal True. True. False. simba jaba multiple speedy multiple 85 sherekhan sherekhan multiple multiple lazy multiple
Hoof True. True. False. flo venison multiple hartford multiple 85 bambi bambi multiple multiple bleh multiple
Bug True. True. False. dots blinky multiple buzz multiple 85 watt watt multiple multiple henry multiple

Grading

Parts A throuugh C were two points each. The rest were 1.5 points.

Question 4

A: 369033 hashes to 13. Indices 13, 14, 15, 16 are taken, so the answer is 17.
B: 1077784 hashes to 4 and the second hash will increment the value by 5. Indices 4, 9, 14 and 19 are taken, so there is no answer: -1.
C: 62794 hashes to 4 and the second hash will increment the value by 19. Indices 4 and 3 are taken, so the answer is 2.
D: 1320609 hashes to 9. Index 9 is taken, so the answer is 10.
E: 342232 hashes to 12. Indices 12, 13 and 16 are taken, so the answer is (12+9) mod 20 = 1.
F: 418294 hashes to 14. Indices 14, 15 are taken, so the answer is is (14+4) mod 20 = 18.
G: 1573160 hashes to 0. The answer is 0
H: 413252 hashes to 12 and the second hash will increment the value by 8. Index 12 is taken, so the answer is (12+8) mod 20 = 0.
I: 1108415 hashes to 15. Indices 15, 16, 19 and 4 are taken, so the answer is (15+16) mod 20 = 11.

Here are all of the answers sorted by the value being hashed:

21216        0
62794       11
117156       0
148026      -1
342232       1
369033      17
396148      12
413252       0
418294      18
442954       3
446936      -1
565904      13
607907       3
611833      13
683442      18
694279       1
711201       2
824760       0
874614      18
875579       0
903535       4
920434       5
939140       4
945008       8
950336       3
1077784     -1
1090745      9
1104489     -1
1108415     11
1227626      8
1242237      1
1316015     11
1320609     10
1335465      1
1410088     12
1521186      7
1557280      4
1573160      0
1584886      7
1591584      8
1623214      0
1642628      8
1670764     10
1822849     -1
1903030     19

Grading

Each question was worth 2 points. If your starting index was off by ten, but you got the problem correct with that starting index, you got 0.8 points.

Questions 5 through 10

You can compile and run these yourself. Here are the example answers:

5: qncaode
6: 25
7: 2915
8: 3821945067
9: 691732994025
10: fcebd

And here are all of the answers with the cpp filename as a key.

Question 5 aft: qncaode dark: hrsfiez liz: mdgnslf pond: trpimhz wive: tlganev
Question 6 lit: 23 nigh: 22 soon: 24 teal: 25 weep: 24
Question 7 mung: 2567 page: 2412 puny: 2691 sal: 2859 shin: 2915
Question 8 fowl: 0561243789 joey: 1795468320 sat: 3821945067 viz: 5130296748 zeus: 5084317629
Question 9 aile: 544529389468 bare: 62551787734 bee: 471726706031 jeep: 691732994025 push: 369023481054
Question 10 deck: fcebd gyp: fcebd prey: fcebd shaw: fcebd soar: fcebd

Grading

Two points each for questions 5 through 7. Three for questions 8 through 10. On question 8, I gave:

Q8: 0.5 points for the right starting letter and < 4 chars.
Q8: 1 point for the right starting letter and >= 4 chars.
Q8: 1 point for having exactly 10 characters.
Q9: 1 point for the correct starting number
Q10: Gave partial credit to answers like "a5a2a4a1a3". push_back() is pushing back a character and not a string.

Question 11

We first read in *(p[i]):

*(p[0]) = 44
*(p[1]) = 69
*(p[2]) = 24
*(p[3]) = 88

We then read in four values of n and use them to set q and a:

q[0] points to p[2], whose value is 24. So a[0] becomes 24.
q[1] points to p[0], whose value is 44. So a[1] becomes 44.
q[2] points to p[0], whose value is 44. So a[2] becomes 44.
q[3] points to p[1], whose value is 69. So a[3] becomes 69.

We then read in four values of n and use them to increment *(p[i]):

*(p[0]) is incremented by 1, becoming 45. That means *q[1] and *q[2] become 45.
*(p[1]) is incremented by 8, becoming 77. That means *q[3] becomes 77.
*(p[2]) is incremented by 3, becoming 27. That means *q[0] becomes 27.
*(p[3]) is incremented by 5, becoming 93.

Finally, we read in four values of n and use them to increment a[i]:

a[0] is incremented by 10, becoming 34.
a[1] is incremented by 6, becoming 50.
a[2] is incremented by 9, becoming 53.
a[3] is incremented by 2, becoming 71.

So here are the answers:

1st line: 45:77:27:93
2nd line: 27:45:45:77
3rd line: 34:50:53:71
4th line: 34:50:53:71 -- (*dp) is equal to a.

Grading

3 points per line. I gave some partial credit to small arithmetic errors, and some minor pointer confusion. On line 4, you got the full three points if lines 3 and 4 matched. Here are answers of the various banks:

Input: 44 69 24 88 2 0 0 1 1 8 3 5 10 6 9 2 Output: 45:77:27:93: 27:45:45:77: 34:50:53:71: 34:50:53:71:

Input: 89 21 65 45 2 1 0 0 2 7 1 5 6 8 9 3 Output: 91:28:66:50: 66:28:91:91: 71:29:98:92: 71:29:98:92:

Input: 61 48 26 81 2 0 0 1 10 8 7 2 6 5 4 3 Output: 71:56:33:83: 33:71:71:56: 32:66:65:51: 32:66:65:51:

Input: 82 45 21 68 2 1 0 0 3 4 5 10 7 9 8 6 Output: 85:49:26:78: 26:49:85:85: 28:54:90:88: 28:54:90:88:

Input: 63 85 45 23 2 1 0 0 10 6 9 1 3 5 7 2 Output: 73:91:54:24: 54:91:73:73: 48:90:70:65: 48:90:70:65:

Question 12

This is a straightforward vector and list program:

#include <vector>
#include <list>
#include <iostream>
using namespace std;

vector < list <string> > ltovl(const list <string> &l)
{
  list <string>::const_iterator lit;
  vector < list <string> > rv;

  rv.resize(2);
  for (lit = l.begin(); lit != l.end(); lit++) {
    if ((*lit)[0] >= 'A' && (*lit)[0] <= 'Z') {
      rv[0].push_back(*lit);
    } else {
      rv[1].push_back(*lit);
    }
  }
  return rv;
}

  
int main()
{
  list <string> v;
  vector < list <string> > rv;
  list <string>::iterator lit;
  string s;
  int i;

  while (cin >> s) v.push_back(s);
  rv = ltovl(v);
  
  for (i = 0; i < rv.size(); i++) {
    cout << i;
    for (lit = rv[i].begin(); lit != rv[i].end(); lit++) cout << " " << *lit;
    cout << endl;
  }
  return 0;
}

Grading

Typically you started with 15 points and had deductions. If your code was too confused or sparse, then you got some points along with "Please see the answer."

The most common deductions involved the proper use of iterators for the lists. You can iterate through lists with integers. Also, the following tests whether c is a capital letter:

if (c >= 'A' && c <= 'Z')

You do not need to memorize ASCII values.

You lost a half a point if you created two lists in ltovl(), and then you pushed them onto a vector. That's making an unnecessary copy of each of the lists.