COS 126: Midterm Exam

NAME:

Midterm Exam

COS 126 Solutions, Spring 2012

Expressions

Type Value
String "10099"
int 3
String "199A"
boolean false
double 1.0
double 1.5
boolean false
Number Systems

Decimal Binary in Memory Hex on display
7 0000 0000 0000 0111 0007
-7 1111 1111 1111 1001 FFF9
162 0000 0000 1010 0010 00A2
-275 1111 1110 1110 1101 FEED
Methods and I/O
1. ```
2
0
0
```
2. ```
5
0
0
```
Loops and Arrays
1. 0 1 2 3 4 5 6 7 8 9
  1
  
  1 1
  
  1 1 1
  
  1 1 1 1
  
  1 1 1 1 1
  
  1 1 2 1 1
  
  1 1 3 1 1
2. ```
0 1 2 2 2 3 9 
```
3. ```
a blank line
```

Debugging

```
6 double x = StdIn.readDouble();
```
```
3 int N = Integer.parseInt(args[0]);
```
```
5 for (int i = 0; i < N; i++) {
```

TOY
1. ```
 
R[6] <-- 01
R[4] <-- R[4] - R[6]
```
2. 6 times.
3. The program computes the sum of all integers from 1 to N
4. Overflow.
  When R[4] is 0100 or more, the sum of the numbers is larger than 7FFF, the largest possible positive integer.

Recursion

The rightmost picture.
4 squares are uncovered.

There were many correct answers. This is one:

draw( x - s, y - s, s/2.2, level-1, k ); // bottom left
draw( x + s, y - s, s/2.2, level-1, k ); // bottom right

Sorting and Analysis
1. 1-3 hours
2. Insertion Sort
3. Linear. The second call has a sorted array so only the comparisons between adjacent elements occur.