Question 1 - 10 points

Argv[1] is "Curious-George". The first thing I'd do is put each letter over its array index:

Ok, now s and argv[1] are the same: 0xbfffedd8. V will be a pointer to the first occurrence of 'G', which is 8 characters past s: 0xbfffede0. That 'G' is turned into a 'g', and then we move v forward 5 characters to 0xbfffede5, which is a pointer to the final 'e'. That gets changed to an 'i'.

Since u was created with strdup(), it has a copy of "Curious-George", which is unaffected by v. The strstr() statement will return a pointer to the "ge" in u, which starts at the 12th character from the beginning of u. Thus, the strstr() returns 0x3000fc.

Here's the output:

UNIX> pex Curious-George Prince-George Eddie-George
0xbfffedd8 0x3000f0
argc: 4
argv[1]: 0xbfffedd8 Curious-georgi
s: 0xbfffedd8 Curious-georgi
u: 0x3000f0 Curious-George
v: 0xbfffede5 i
strstr: 0x3000fc ge

Part 2: One problem is that argc is not checked -- if you call pex will no arguments, the strchr() statement will probably generate a seg fault or bus error.

The second problem is if argv[1] contains a 'G' within five characters of the end of the string -- then the "*v = 'i'; statement will stomp over memory that is not part of the string.

Grading

Histogram of scores for Question 1

Question 2 - 10 points

List *new_list()
{
  List *l;

  l = (List *) malloc(sizeof(List));
  if (l == NULL) { perror("malloc"); exit(1); }
  l->front = NULL;
  l->back = NULL;
  l->size = 0;
  return l;
}

void list_prepend(List *l, Jval v)
{
  Node *n;

  n = (Node *) malloc(sizeof(Node));
  if (n == NULL) { perror("malloc"); exit(1); }
  n->val = v;
  n->link = l->front;
  l->front = n->link;
  l->size++;
  if (l->size == 1) l->back = n;
}

void free_list(List *l) 
{
  Node *n, *n2;

  n = l->front;
  while (n != NULL) {
    n2 = n;
    n = n->link;
    free(n2);
  }
  free(l);
}

Grading

• new_list(): 1 point for malloc/testing the return value. 1 point for setting front and back to NULL, 1 point for setting size to zero and returning the new list.

• list_prepend(): 1 point for malloc/testing the return value. 1 point for hooking the new node onto the front of the list properly, 1 point for setting size and val correctly, and one point for setting back when the list was empty.

• list_free(): 1 point for a proper while() loop, 1 point for freeing individual nodes and the list itself, and 1 point for not getting your pointers mixed up and using a pointer that you have already freed. If you made some mistakes that didn't fit this framework, I simply took off points that reflected how much I thought your solution made sense.

Histogram of scores for Question 2

Question 3 - 5 points

Grading

• Two points for talking about setting up prototypes or typedefs to (void *) in the header file.
• One point for saying you put the real definition into the source file that implements the header file.
• One point for saying that the source will cast to and from the (void *).
• One point for saying that in this way the users can't mess with the data.

I was pretty flexible on where I took off points here -- as with free_list(), I basically took off enough points so that your score reflected how good your answer was, and tried to take them off in the parts that showed where your answer was lacking.

Histogram of scores for Question 3

Question 4 - 10 points

This is a straightforward program. It reads from standard input until it reads a blank line or EOF, at which point it exits. For all the non-blank lines that it reads, it pushes the characters of the first word onto a stack, onto a queue, and to the beginning and end of a linked list. This means that when we pop the characters off the stack, we will get them in reverse order. When we dequeue them, we will get them in the proper order. And if we look at the list, it will contain the characters in reverse order followed by the characters in their proper order.

So, on the first line of standard input, the queue will hold '1', the stack will hold '1' and the list will hold '1','1'. The first line of output will be "1111".

On the second line of standard input, we will have:

Therefore, line two of output is:

Fdde rerF erFr dFed

Therefore, line three of output is:

Eeei didd ddEE iddd eEie

Therefore the output is:

1111
Fdde rerF erFr dFed
Eeei didd ddEE iddd eEie

Grading

• If it looked like you had some clue about what was going on: 1 point
• If you gave three lines of output: 1 point
• If the letters of your output were composed only of the letters from the first word of each line of input: 1 point
• If each line contained as many words as there were letters on the first word of that line: 1 point
• If each word had exactly four characters: 1 point
• If I saw "Fred" and "Eddie" somewhere coherent: 1 point
• If I saw "derF" and "eiddE" somewhere coherent: 1 point
• If you got the other details correct: 3 points

The problem came when you got caught on the fact that the program is putting characters on the various data structures, and not whole words. That led to the worst scores, but of course, that shows that you're making a fundamental mistake, as opposed to, for example, writing five lines of output rather than three, which is simply a missed detail.

Histogram of scores for Question 4