Question 1. [15 marks]

Transcription

1 Note to Students: This file contains sample solutions to the term test together with the marking scheme and comments for each question. Please read the solutions and the marking schemes and comments carefully. Make sure that you understand why the solutions given here are correct, that you understand the mistakes that you made (if any), and that you understand why your mistakes were mistakes. Remember that although you may not agree completely with the marking scheme given here it was followed the same way for all students. We will remark your test only if you clearly demonstrate that the marking scheme was not followed correctly. For all remarking requests, please submit your request in writing directly to your instructor. For all other questions, please don t hesitate to ask your instructor during office hours or by . Question 1. [15 marks] For each part of this question, use the space to the right of the code to state what will happen when we attempt to compile and run the code. Pick one of the following options: (a) does not compile in this case, circle the line(s) that contain an error and explain the error(s); (c) undefined behaviour (compiles but may crash when run) in this case, circle the line(s) that contain an error and explain the error(s); (d) runs (compiles and runs without error) in this case, write down the output produced by the code. Part (a) [1 mark] struct some { int x, y; ; struct some *thing; thing.x = -1; thing.y = 2; printf("%d, %d\n", thing.x, thing.y); Does not compile because thing is a pointer and does not have.x or.y fields. Error [0.5]: correctly identifying the error Part (b) [1 mark] struct some { int x, y; ; struct some *thing; thing->x = -1; thing->y = 2; printf("%d, %d\n", thing->x, thing->y); Undefined behaviour because thing points to an arbitrary memory location. Error [0.5]: correctly identifying the error Page 1 of 9

2 Part (c) [1 mark] struct some { int x, y; ; struct some thing; thing->x = -1; thing->y = 2; printf("%d, %d\n", thing->x, thing->y); Does not compile because thing is not a pointer (operator -> is undefined). Error [0.5]: correctly identifying the error Part (d) [1 mark] struct some { int x, y; ; struct some thing; thing.x = -1; thing.y = 2; printf("%d, %d\n", thing.x, thing.y); Runs and prints -1, 2. Output [0.5]: giving the correct output Part (e) [1 mark] int array[][2] = {{2, 3, {4, 5; printf("%d\n", array[0][1]); Runs and prints 3. Output [0.5]: giving the correct output Page 2 of 9

3 Part (f) [2 marks] int i; char *names[] = {"alice", "bob", "carol", "dan"; for (i = 0; i < 4; ++i) printf("%c", names[i][0]); printf("\n"); Runs and prints abcd. Output [1]: giving the correct output Part (g) [2 marks] int i; char *names[] = {"alice", "bob", "carol", "dan"; for (i = 0; i < 4; ++i) { names[i][0] += A - a ; printf("%s\n", names[i]); Undefined behaviour because strings stored through pointers are read-only (cannot be modified). Error [1]: correctly identifying the error Part (h) [2 marks] int i; char names[][3] = {"joe", "dan", "bob"; for (i = 0; i < 3; ++i) printf("%s\n", names[i]); Undefined behaviour because array names does not have enough room to store null characters that terminate each string. Page 3 of 9

4 Error [1]: correctly identifying the error Give full marks for students who answer that the code runs and prints joedanbob danbob bob because they ve identified the correct problem with the code and this is what reasonable compilers will produce. Part (i) [2 marks] char *str1 = "hello"; char str2[50] = "goodbye"; strcat(str2, str1); printf("%s\n", str2); Runs and prints goodbyehello. Output [1]: giving the correct output Part (j) [2 marks] char *str1 = "hello"; char str2[50] = "goodbye"; strcpy(str2, str1); printf("%s\n", str2); Runs and prints hello. Output [1]: giving the correct output Question 2. [16 marks] Part (a) [4 marks] Write a function max2d that takes in a 2-dimensional array of ints of size N N (where N is a preprocessor constant) and that returns the largest value in the array. Page 4 of 9

5 int max2d(int array[][n]) // (int array[n][n]) is also acceptable { int i, j, max = array[0][0]; for (i = 0; i < N; ++i) for (j = 0; j < N; ++j) if (array[i][j] > max) max = array[i][j]; return max; Idea [1]: correct high-level idea, independently of syntax errors Syntax [3]: correct function header (parameter type and return type), correct syntax to access multidimensional array elements, and correct C syntax overall (including variable declarations) Marker s Comments: common error [ 1 mark]: Instructor s Comment: cannot initialize max to 0 what if every element in array is negative? The marker for this question was harsh and gave 0 to students who wrote a function that worked only for single-dimensional arrays. This was not what I intented, but we ran out of time to fix it for everyone. Also, it would have made very little difference: the most you could get then would be 1 mark. So I will not remark this part for students who got 0 because of this. Instead, I will make a general adjustment to everyone s test grades at the end of the term. After all, the test was a bit long, and the average is a bit low... Marker s Comments for Parts (b) (d): common error: Many students confused the notion of lower bound on the worst-case running time with best-case running time. If you don t understand the difference, make sure you talk to the instructor! Part (b) [4 marks] Give an upper bound and a lower bound on the worst-case running time of the following code. Briefly justify each of your bounds (show how you computed them). int array[] = { /*...values omitted...*/ ; int n = sizeof(array) / sizeof(array[0]); int i, s = 0; for (i = 1; i < n; i *= 2) s += array[i]; Page 5 of 9

6 Upper bound: O(log n) because it takes roughly log 2 n steps for the value of i to double repeatedly until it reaches or exceeds n, and the loop executes a constant amount of work at each iteration. Lower bound: Ω(log n) because the loop iterates at least log 2 n 1 times. Upper bound [2]: Lower bound [2]: correct and well-justified upper bound correct and well-justified lower bound Part (c) [4 marks] Give an upper bound and a lower bound on the worst-case running time of the following code. Briefly justify each of your bounds (show how you computed them). int array[] = { /*...values omitted...*/ ; int n = sizeof(array) / sizeof(array[0]); int i, j, k, s = 0; for (i = 0; i < n - 10; ++i) { for (k = i; k < i + 10; ++k) s -= array[k]; for (j = i + 10; j < n; ++j) for (k = j; k > j - 10; --k) s += array[k]; Upper bound: O(n 2 ) because each loop on k performs only a constant amount of work (it iterates exactly 10 times), the nested loop on j performs at most n-10 iterations (so O(n) work), and the loop on i performs n-10 iterations (so O(n 2 ) work). Lower bound: Ω(n 2 ) because for each value of i from 0 to n/2, j iterates through at least all of the values from n/2+10 to n so it performs at least Ω(n) work, Ω(n) many times. Upper bound [2]: Lower bound [2]: correct and well-justified upper bound correct and well-justified lower bound Page 6 of 9

7 Part (d) [4 marks] Give an upper bound and a lower bound on the worst-case running time of the following code. Briefly justify each of your bounds (show how you computed them). int array[] = { /*...values omitted...*/ ; int n = sizeof(array) / sizeof(array[0]); int i, j; for (i = 0; i < n; ++i) if (array[i] % 2) for (j = i; j < n; ++j) array[j] += 1 Upper bound: O(n 2 ) because the nested loop on j performs at most n iterations (so O(n) work) in the worst-case, and the loop on i performs n iterations (so O(n 2 ) work). Lower bound: Ω(n 2 ) because if array is initialized to {1,2,3,...,n, then the condition of the if statement will be true at every iteration of the loop for i, and the code is a standard nested loop (as in the previous part). Upper bound [2]: Lower bound [2]: correct and well-justified upper bound correct and well-justified lower bound Page 7 of 9

8 Question 3. [14 marks] Write a complete program that creates a linked list of doubles, one for each command-line argument (called Program Arguments in Code::Blocks), then computes and prints their average. More precisely, your program must contain: The definition for a struct node that can be used to work with linked lists of doubles. The definition for a function average that takes a pointer to a struct node and returns the average of the values stored in the list starting at that node. It is okay for your function to work only for non-empty lists, i.e., only when its argument is non-null. A main function that creates a linked list of doubles from its array of command-line arguments (see below for how to convert from strings to doubles), then calls function average to compute the average of these values and prints the result to stdout. Assume that every command-line argument represents a valid double, i.e., don t write any code to perform error-checking for bad inputs. For example, if your code is compiled and run with command-line arguments , it should simply print the value 2.5 on stdout. Your answer must be a complete program, i.e., it must have appropriate #include directives, etc. Part of your grade will be for style and design, i.e., you must make appropriate use of helper functions, preprocessor constants, typedef, etc. Make use of the following function, declared in <stdlib.h>: /* Convert the string str into a double, and return its value. */ double atof(const char *str); Relax! This is not as difficult as it might seem. Just take it one step at a time and have a careful look at that C Reference Card if you ve forgotten how to work with command-line arguments. #include <stdio.h> #include <stdlib.h> typedef struct node { double value; struct node *next; node_t; double average(const node_t *head) { unsigned count = 0; double sum = 0.0; while (head) { ++count; sum += head->value; head = head->next; return sum / count; Page 8 of 9

9 CSC 190 H1 Midterm Test #2 Winter 2012 int main(int argc, char *argv[]) { /* Create the linked list from command-line arguments. */ node_t *first = NULL; while (argc > 1) { node_t *new = malloc(sizeof(node_t)); if (new == NULL) { fprintf(stderr, "ERROR allocating memory -- aborting!\n"); exit(exit_failure); new->value = atof(argv[--argc]); new->next = first; first = new; /* Compute and print the average. */ if (first) printf("%g\n", average(first)); /* Free the memory allocated for the list. */ while (first) { node_t *next = first->next; free(first); first = next; return EXIT_SUCCESS; Idea [4]: clear attempt to carry out the correct algorithm even if the code is terribly designed and riddled with syntax errors [2 for main, 1.5 for average, 0.5 for struct node] Design [3]: well designed code independently of the correctness or syntax Syntax [7]: correct C syntax throughout (take off only 0.5 marks for small things like missing parentheses or semicolons, but at least 1 mark for anything more serious) Marker s Comments: common error [ 1]: forgetting to check the return value of malloc common error [ 0.5]: forgetting to free the memory at the end common error [ 2]: errors creating the linked list common error [ 2]: not converting strings to doubles common error [ 1]: missing argc and argv parameters for main Page 9 of 9