1. Michele Weigle - COMP 14 - Spr 04
COMP 110 Midterm review
Luv Kohli
October 27, 2008
MWF 2-2:50 pm
Sitterson 014

2. Michele Weigle - COMP 14 - Spr 04
Announcements
Program 3 due Friday, October 31, 2pm

3. Michele Weigle - COMP 14 - Spr 04
Questions?
Any other questions?

4. Michele Weigle - COMP 14 - Spr 04
Today in COMP 110
Take out a sheet of paper
Write your name on it
Take about 5 minutes to write down how you think you did on the midterm, and why

5. Midterm scores
Average score: ~76

6. Q3 a. double var1 = 9 / 2; b. int var2 = (int) (6.6 / 3.0)
9 / 2 is integer division, resulting in 4
division is done before the automatic typecast to double. var1 is set equal to 4, or 4.0
b. int var2 = (int) (6.6 / 3.0)
6.6 / 3.0 is 2.2
(int) 2.2 is 2
var2 is set to 2
c. float var3 = (float) 5 / 2
(float) 5 / 2 causes 2 to be typecast to float
(float) 5 / (float) 2 is 2.5
var3 is set to 2.5

7. Q4
Accepted most reasonable things as long as they were not too far off

8. Q8 3 * 4 * 3 = 36  3 iterations  4 iterations  3 iterations
int i = 0;
while (i < 3) {
for (int j = 1; j <= 4; j++)
for (int k = 3; k > 0; k--)
System.out.println(i + “,” + j + “,” + k); }
i++;
3 * 4 * 3 = 36
 3 iterations
 4 iterations
 3 iterations

9. Q13: Tracing code the magical snail quickly eats the strange airplane
NAM NAM NAM!!!

10. Q15: Remove middle words Y o u w i l n t g ! 1 2 3 4 5 6 7 8 9 10 11
int firstSpace = str.indexOf(“ ”);
int lastSpace = str.lastIndexOf(“ ”);
String strNoMiddle = str.substring(0, firstSpace) +
str.substring(lastSpace); Y o u w i l n t g ! 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Y o u g ! 1 2 3 4 5 6

11. Q16: Determining if a string is a palindrome
Odd-length string r a c e 1 2 3 4 5 6

12. Q16: Determining if a string is a palindrome
Even-length string r e d 1 2 3 4 5

13. Q16: Determining if a string is a palindrome
‘u’ != ‘e’
Not a palindrome r u d e 1 2 3 4 5

14. Q16: Determining if a string is a palindrome
Need a loop
What is being repeated?
Character comparison
What characters are we comparing?
First half of string to reverse of second half of string
What is our loop terminating condition?
We find a pair of characters that do not match

15. Q16: using a for loop r a c e 1 2 3 4 5 6 i str.length() – 1 - i
public boolean isPalindrome(String str) {
for (int i = 0; i < str.length() / 2; i++)
if (str.charAt(i) != str.charAt(str.length() – 1 – i))
return false; }
return true; r a c e 1 2 3 4 5 6

16. Q16: Can also do it with a while loop
public boolean isPalindrome(String str) { int first = 0; int last = str.length() - 1; while (first < last) if (str.charAt(first) != str.charAt(last)) return false; } first++; last--; return true;

17. Michele Weigle - COMP 14 - Spr 04
Wednesday
Static variables and methods