CS140 Midterm Exam - October 11, 2005 - Answers

Jim Plank

Question 1

The variable p holds the address of array element a[2]. The first printf() statement says that this is 0xbffff024. Therefore, a starts at 0xbffff024-8 = 0xbffff01c. p is at address 0xbffff044. Therefore, this is what memory looks like just after that printf() statement:

Now, b() is called with two parameters: p, which is 0xbffff024, and &p which is 0xbffff044. Therefore, ip = 0xbffff024 and ipp = 0xbffff044. The first printf() statement in b() prints the address of ip, which is 0xbffffeff8. Therefore, this is what we know about memory when b() starts. Note, I don't put ipp in a memory slot because we don't know its address:

The first printf statement will print 0xbffff024 and 0xbffff044. Next, we execute *ipp += 2, which means that the value in address 0xbffff044 will have eight added to it. This means that p in main() has its value changed to 0xbffff02c. Here is the change:

The next statement is **ipp = 1. Since *ipp is 0xbffff02c, this changes array element a[4] to be 1:

Now we perform *ip = 17. Since ip is 0xbffff024, this sets array element a[2] to be 17:

Now, we return, and the p++ statement adds four to p:

Finally, we set *p to be three. Since p is 0xbffff030, this sets array element a[5] to be three:

Therefore, the final printouts will be: 

0xbffff01c
0xbffff044
0xbffff030
15 14 17 12 1 3 9 8 7 6
So -- the output to this program is: 

0xbffff024 0xbffff044
0xbfffeff8
0xbffff024 0xbffff044.
0xbffff01c
0xbffff044
0xbffff030
15 14 17 12 1 3 9 8 7 6

Grading - 9 points

Question 2

The correct answers are D, E and F. Here are pictures of each implementation, starting just after the newn->val call.

Grading - 9 points

Question 3

When i is even, the first word on the line gets appended to a list. When i is odd, the last word on the line gets prepended to the list. So, the building of the list is:

i Action d
0
1
2
3
4
5
6
7
 Append
Prepend
Append
Prepend
Append
Prepend
Append
Prepend
 "A"
"top", "A"
"top", "A", "Is"
"flame", "top", "A", "Is"
"flame", "top", "A", "Is", "The"
"love", "flame", "top", "A", "Is", "The"
"love", "flame", "top", "A", "Is", "The", "And"
"name", "love", "flame", "top", "A", "Is", "The", "And"

The output is: 

name
love
flame
top
A
Is
The
And

Grading - 6 points

Question 4

When i is even, the first word on the line is enqueued onto a queue. When i is odd, the first word on the line is pushed onto a stack. Then the queue is printed out, and the stack is printed out. The output: 

A
Is
The
And
is
beauty
silver
the

Grading - 6 points

Question 5

For each line, x points to the first occurrence of the character 'o'. If there is no 'o', then it points to the first character on the line. Variable y points to the first occurrence of the character 's'. If there is no 's', then it points to the first character on the line. The program then prints the difference between y and x. Although these are pointers, that's ok -- chars are one byte each, so this will print out the number of characters between the pointers.

Line number x y y - x
1 &(is->text1[3]) &(is->text1[7]) 4
2 &(is->text1[5]) &(is->text1[0]) -5
3 &(is->text1[0]) &(is->text1[1]) 1
4 &(is->text1[0]) &(is->text1[0]) 0
5 &(is->text1[11]) &(is->text1[4]) -7
6 &(is->text1[12]) &(is->text1[0]) -12
7 &(is->text1[0]) &(is->text1[8]) 8
8 &(is->text1[0]) &(is->text1[1]) 1

The output is therefore 

4
-5
1
0
-7
-12
8
1

Grading - 6 points