ECE 190 Final Exam HoChiMinh University of Technology

Sunday 7 August 2011

• Be sure your exam booklet has 8 pages (4 sheets). • You have THREE HOURS to complete this exam. • Wherever necessary, even if not explicitly specified in the code, assume that

stdio.h and stdlib.h have been included in the code. Thus, library calls such as printf, scanf , malloc, and so forth may be used in the code.

• Write your name at the top of each page. • This is a closed book exam. • You may not use a calculator. • You are allowed THREE handwritten A4 sheets of notes (both sides). • Absolutely no interaction between students is allowed. • Show all work. State all assumptions. • Don’t panic, and good luck!

Problem 1 20 points _______________________________ Problem 2 30 points _______________________________ Problem 3 30 points _______________________________ Problem 4 20 points _______________________________ Total 100 points _______________________________

Name: Student ID # :

Page 2 Name: ____________________________________________

Problem 1 (20 points): Short Answer Questions Be concise: if your answer contains more than 10 or 15 words or a simple picture, it is probably wrong. Part A (5 points): The following code has one error. Identify the error and indicate how to fix the error. void print_digit (int value) { char* name[10] = { "zero", "one", "two", "three", "four", "five", "six", "seven", "eight", "nine" }; if (0 <= value && 10 >= value) { printf ("%s", name[value]); } } Part B (5 points): The following code has one error. Identify the error and indicate how to fix the error. void player_free (player_t* p) { free (p); free (p->name); } Part C (5 points): Assuming that our compiler uses 16-bit 2’s complement numbers to represent ints, give values for v1 and v2 such that v1 < v2, but the less_than function below returns 0. int less_than (int v1, int v2) { return (v1 – v2 < 0); }

Page 3 Name: ____________________________________________

Problem 1, continued: Part D (5 points): The following code has one error. Identify the error and indicate how to fix the error. int* split_digits (int value) { int digits[12]; int cnt = 0; if (0 <= value) { while (10 <= value) { digits[cnt++] = (value % 10); value = (value / 10); } digits[cnt++] = value; } digits[cnt] = -1; return digits; }

Page 4 Name: ____________________________________________

Problem 2 (30 points): Two C Programs Part A (15 points): Complete the function below to print a given string along the diagonal of a square of size N, where N is the length of the string in ASCII characters (not counting NUL). Examples are shown in the figure below. The rest of the squares are filled with spaces (' '). Assume that N is a positive integer. void draw_diagonal (char* s) { int N, a, b; N = strlen (s); for (a = ; ; a++) { } }

h e l l o

Y e s

Proper Output for s="Yes" (N=3)

Proper Output for s="hello" (N=5)

E C E 1 9 0

Proper Output for s="ECE190" (N=6)

Page 5 Name: ____________________________________________

Problem 2, continued: Parts B, C, and D ask about the C function count below. int count (char* s, int inside) { if ('\0' == *s) { return inside; } if ('.' == *s) { if (inside) { return (count (s + 1, 0) + 1); } return count (s + 1, 0); } return count (s + 1, 1); } Part B (5 points): What value is returned by the following call? count ("Hello!", 0); Part C (5 points): What value is returned by the following call? count ("What.does.count.count?.....", 0); Part D (5 points): What value is returned by the following call? count ("...A...big...red...fox...ate...a...chicken...", 10);

Page 6 Name: ____________________________________________

Problem 3 (30 points): Arrays and Stack Frames This problem focuses on the function shown below. int merge (int into[], int into_len, int from[], int from_len) { int i, j; for (i = into_len, j = 0; from_len > j; j++) { if (0 != from[j]) { into[i++] = from[j]; } } return i; } int main () { int base[4] = { 42, 0, 5, 10 }; int added[2] = { 0, 19 }; int sum; sum = merge (base, 2, added, 2); return 0; } Part A (10 points): Write the contents of the arrays base and added just before main returns in the program above. base: { _________ , __________ , __________ , __________ } added: { _________ , __________ }

Page 7 Name: ____________________________________________

Problem 3, continued: Part B (20 points): The diagram below shows the state of the stack just before the function merge returns (before execution of the return statement) in the program on the previous page. Most of merge’s stack frame and some of main’s stack frame have been labeled, and some of the values have been filled in for you. Fill in the rest of the stack frame for merge by writing both the values and the meanings for the remaining locations into the boxes. For example, you need to write in the values of the parameters passed to merge. If you can not know the value at an address or the meaning of the value stored there, write “unknown.” Blank boxes will not receive points.

xC771 i ← R6

xC772 j ← R5

xC773 previous frame pointer

xC774 x3443 return address

xC775 return value

xC776

xC777

xC778

xC779

xC77A

xC77B

xC77C x0013

xC77D x0002

xC77E (unknown) (unknown)

xC77F 42 base[0]

xC780 base[1]

xC781 base[2]

xC782 base[3]

xC783 0 added[0]

xC784 19 added[1]

xC785 sum

Page 8 Name: ____________________________________________

Problem 4 (20 points): Linked Lists This problem makes use of a cyclic, doubly-linked list as discussed in lecture and in discussion section. Each list element points to an object (the contents of this object are not necessary to the problem) and to the previous and next elements of the list. A sentinel (with obj field equal to NULL) serves as the starting point for a list. The function reverse_copy creates a copy of a list in reverse order. The parameter from points to the sentinel element for the original list, and the parameter to points to the sentinel element for the list being created. The new list’s elements must be allocated using malloc, and should point to the same objects as do the elements of the old list. The function should return 1 on success and 0 on failure. Fill in the five blanks to complete the function correctly. You may not add nor remove any other code. void* malloc (size_t bytes); /* dynamically allocate memory */ typedef struct object_t object_t; typedef struct list_elt_t list_elt_t; struct list_elt_t { object_t* obj; /* object referenced by list element */ list_elt_t* prev; /* previous list element */ list_elt_t* next; /* next list element */ }; int reverse_copy (list_elt_t* from, list_elt_t* to) { list_elt_t* cur_elt; list_elt_t* tail; list_elt_t* new_elt; tail = to; for (cur_elt = from->prev; ___________________________ ; cur_elt = cur_elt->prev) { new_elt = _______________________________ ; if (NULL == new_elt) { _____________________ ; } new_elt->obj = cur_elt->obj; tail->next = new_elt; new_elt->prev = tail; tail = new_elt; } tail->next = __________________; to->prev = __________________; return 1; }