ECE 190 Final Exam HoChiMinh University of Technology

Tuesday 14 August 2012

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

Name: Student ID # :

Page 2 Name: ____________________________________________

Problem 1 (30 points): Two C Programs Part A (15 points): Write the rest of the C function below to count the number of pairwise matches in an array of integers. For example, if the array contains the values {0, 5, 3, 2, 3, 5}, your function should return 2 (the two 5’s match, and the two 3’s match). If the array contains the values {7, 7, 0, 7}, your function should return 3, since there are three distinct pairs of 7s in the array. Use two loops to implement your function. Do not attempt to sort the array. int count_matches (int array[], int num_values) { int N, a, b; N = for (a = ; ; a++) { } return N; }

Page 3 Name: ____________________________________________

Problem 1, continued: Parts B, C, and D ask about the C function seek below. char* seek (char* s, char* t, int n) { if ('\0' == t[n]) { return s; } if ('\0' == s[n]) { return NULL; } if (t[n] != s[n]) { return seek (s + 1, t, 0); } return seek (s, t, n + 1); } Part B (5 points): Assuming that string s starts at address x4000, what value is returned by the following call? seek ("bat bate bet!", "ate", 0); Part C (5 points): Assuming that string s starts at address x4000, what value is returned by the following call? seek ("applesauce", "Banana", 5); Part D (5 points): Briefly explain what the function does when called with n equal to 0.

Page 4 Name: ____________________________________________

Problem 2 (30 points): Structures and Stack Frames This problem focuses on the structure and function shown below. typedef struct complex_t complex_t; struct complex_t { int real; int imag; }; /* multiply complex numbers A and B; store the product in C */ void complex_mult (complex_t* A, complex_t* B, complex_t* C) { complex_t product; product.real = A->real * B->real – A->imag * B->imag; product.imag = A->real * B->imag + A->imag * B->real; *C = product; return; } int main () { complex_t one, two, three; one.real = 10; one.imag = 2; two.real = 5; two.imag = 8; complex_mult (&A, &B, &C); return 0; } Part A (10 points): Write the contents of the structures one, two, and three just before main returns in the program above. one: { real: _________ , imag: __________ } two: { real: _________ , imag: __________ } three: { real: _________ , imag: __________ }

Page 5 Name: ____________________________________________

Problem 2, continued: Part B (20 points): The diagram below shows the state of the stack just before the function complex_mult returns (before execution of the return statement) in the program on the previous page. Most of complex_mult’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 diagram below 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 complex_mult. 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.

xC675 ← R6

xC676 ← R5

xC677 previous frame pointer

xC678 x3443 return address

xC679 return value

xC67A

xC67B

xC67C

xC67D

xC67E 10

xC67F 2

xC680 (unknown) (unknown)

xC681 three.real

xC682 three.imag

xC683 two.real

xC684 two.imag

xC685 10 one.real

xC686 2 one.imag

xC687

Page 6 Name: ____________________________________________

Problem 3 (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 do_transform replaces each of the dynamically allocated objects with a transformed object. The transformed object is produced by calling transform_object on the current object and a newly allocated object. If the call to transform_object returns any value other than 0, the node is instead deleted from the linked list, and all associated objects as well as the list node must be freed. 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.

(The code is on the next page.)

Page 7 Name: ____________________________________________

Problem 3, continued: void* malloc (size_t bytes); /* dynamically allocate memory */ /* Although the object structure is not shown here, assume that the compiler has it. */ 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 do_transform (list_elt_t* head) { list_elt_t* cur_elt; object* new_obj; list_elt_t* elim_elt; tail = to; for (cur_elt = head->next; ___________________________ ; cur_elt = cur_elt->next) { new_obj = malloc ( sizeof ( *new_obj ) ); if (NULL == new_obj) { _____________________ ; } if (transform_object (cur_elt->obj, new_obj)) { cur_elt->prev->next = _____________________; cur_elt->next->prev = cur_elt->prev; elim_elt = _______________________; cur_elt = cur_elt->prev; free (elim_elt->obj); free (elim_elt); free (new_obj); continue; } ______________________________; cur_elt->obj = new_obj; } return 1; }

Page 8 Name: ____________________________________________

Problem 4 (20 points): Debugging Exercises Be concise: if your answer contains more than 10 or 15 words or a simple picture, it is probably wrong. Part A (10 points): The following code has two errors. For each of the two errors, identify the error and indicate how to fix the error. /* Returns 1 if the value is prime, or 0 otherwise. */ int is_prime (int value) { int divisor; for (divisor = 1; value > divisor; divisor++) { if (value == value * divisor / divisor) { return 0; } } return 0; } Part B (5 points): The following code has one error. Identify the error and indicate how to fix the error. /* Allocates a new player structure and initializes it. Returns * NULL on failure, or a pointer to the new structure on * success. */ player_t* create_player_structure (char* nm, char* passwd, char* profile) { player_t p; /* player_init is the version used in class; it returns 1 on success, and 0 on failure. */ if (!player_init (&p, nm, passwd, profile)) { return NULL; } return (&p); } Part C (5 points): The following code has one error. Identify the error and indicate how to fix the error. /* Returns 1 if the value is odd, or 0 otherwise. */ int is_odd (int value) { int mod_two = (value % 2); if (mod_two = 1) { return 1; } return 0; }