Midterm #2 With Solutions solutions in blue marking notes in red

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1 Question 1 (10 points) Midterm #2 With Solutions solutions in blue marking notes in red There are quite a few English words, mostly slang, that consist of the same word repeated twice, separated by a hyphen. For example, "boo-boo", "ta-ta" and "nicely-nicely". Write a sed command that will replace all such double words with the word "slang". For example, your script should change the line: Making a boo-boo is a no-no. into: Making a slang is a slang. Details: 1. You may assume that double words in the input will consist entirely of lower-case letters. So you don't have to worry about catching something like "Ta-ta". 2. Each half of the double word should be a complete word. So your script should not change things like "boo-boos" or "gno-no". My Solution: s/\<\([a-z]\+\)-\1\>/slang/g The above was all that was required; I wanted the command you would give sed. In hindsight, I realize that "sed command" could be taken to mean an entire command you would give to the shell. If you wrote something like sed -e 's/\<\([a-z]\+\)\>-\<\1\>/slang/g' That was just fine too. I gave full marks if the actual substitute command was in there, even if the format of the call to sed had some minor problems, like forgetting to quote the command. Variations were possible. You could match end of word before the hyphen and beginning of word after the hyphen not necessary, but no harm. You could include beginning/ending word anchors inside the parenthesis or leave them outside. Most people did well on this question The most common error I saw was using.* instead of [az]\+. Even surrounded by \<..\>, this can match multiple words, which means your pattern would replace something like "one two-one two". General Marking Scheme: 6 points for the regular expression to match double words 4 point for putting in a substitute command, including the "g" flag

2 Question 2 (8 points) In each of the boxes below, write the output of this C program If any of the answers below don't make sense to you, see the explanations on the next page. #include <stdlib.h> #include <stdio.h> #include <string.h> struct twonums { int a; int b; }; /* end twonums */ int main() { int x = 5; int *p = &x; int *q = p; int y = *q + 1; printf("%d, %d\n", x, y); x = 6; y = 8; int *p1 = malloc(sizeof(int)); /* assume heap wasn't empty */ *p1 = 10; int *p2 = &x; int *p3 = &y; p3 = p1; p1 = p2; p2 = p3; *p3 = 12; x = 14; y = 16; printf("%d, %d, %d\n", *p1, *p2, *p3); 14, 12, 12 int nums[3]; int i; for (i = 0; i < 3; i++) nums[i] = i+10; p = nums; 10 printf("%d\n", *p); 5, 6 char name1[20] = "Gandalf"; printf("%d\n", strlen(name1)); 7 char name2[20] = "Frodo"; strcpy(name1, name2); printf("%s, %s\n", name1, name2); Frodo, Frodo char name3[20] = "Peregrin"; name3[3] = '\0'; printf("%s\n", name3); Per char name4[] = "one ring"; char *end = &name4[4]; strcpy(end, "movie"); printf("%s\n", name4); one movie struct twonums r = {1,2}; struct twonums s = {3,4}; r = s; r.a = 5; r.b = 6; printf("%d, %d\n", s.a, s.b); 3, 4 return EXIT_SUCCESS; } /* end main */ Page 2 of 7

3 Each of these questions was short and worth just one point, so I didn't feel it was necessary to give any partial credit. None of the boxes depended on the answers to previous boxes, so there was no chance of being penalized twice for the same error. Following is a detailed explanation of each answer. I didn't pay attention to spaces and commas in the output, just the actual values. int main() { int x = 5; int *p = &x; p = address of x int *q = p; q = address of x int y = *q + 1; *q is the contents of the location q is pointing to in other words, the value of x, or 5. So y is being set to 6 printf("%d, %d\n", x, y); The output is 5, 6. x = 6; y = 8; int *p1 = malloc(sizeof(int)); /* assume heap wasn't empty */ p1 = address of an anonymous integer on the heap (I'll refer to this as "anon" in the following) *p1 = 10; anon now contains the value 10 int *p2 = &x; int *p3 = &y; p2 contains the address of x and p3 contains the address of y p3 = p1; p1 contains the address of anon, so p3 now contains the address of anon p1 = p2; p2 contains the address of x, so p1 now contains the address of x p2 = p3; p3 contains the address of anon, so p2 now contains the address of anon In summary, after these three pointer assignments p1 points to x and p2 and p3 point to anon *p3 = 12; p3 points to anon, so anon now contains 12 x = 14; y = 16; printf("%d, %d, %d\n", *p1, *p2, *p3); p1 points to x and x contains 14, so the first number output is 14 p2 points to anon and anon contains 12, so the second number output is 12 similarly, p3 points to anon so the third number is 12 int nums[3]; int i; for (i = 0; i < 3; i++) nums[i] = i+10; Now nums[0] = 10, nums[1] = 11, and nums[2] = 12. p = nums; The value of nums, without any indexes, is the address of the start of the array in other words, the address of nums[0]. printf("%d\n", *p); p is the address of nums[0], so *p is the contents of nums[0], or 10. char name1[20] = "Gandalf"; printf("%d\n", strlen(name1)); The strlen function returns the number of characters in a string before the null, not counting the null. It's not the number of characters allocated; there's no way for C to check that. So the answer is 7. Page 3 of 7

4 char name2[20] = "Frodo"; strcpy(name1, name2); The strcpy function copies copies from the source to the destination until it hits a null, and copies the null as well. The first parameter is the destination and the second is the source. So the function call above copies 'F','r','o','d','o', and a null character into name1. The contents of name2 doesn't change. So name1 and name2 now print as "Frodo" and the output is "Frodo, Frodo". printf("%s, %s\n", name1, name2); char name3[20] = "Peregrin"; name3[3] = '\0'; printf("%s\n", name3); The fourth character of name3 (the second 'e') is replaced by a null character not the character representing the d zero digit. So when you print name3, it prints the 'P', the 'e', and the 'r', then sees the null character and stops so the output is "Per". char name4[] = "one ring"; char *end = &name4[4]; strcpy(end, "movie"); printf("%s\n", name4); I made an error in this question. My original draft had a length in the declaration of name4, so that it would have enough room for the result. That length somehow got deleted in the final version. First, here's what would have happened if name4 had a longer length 10 or more. The literal constant "movie" is a pointer to a string that C creates for us, containing the 6 characters 'm', 'o','v','i','e', and null. The call to strcpy copies those 6 characters into the location pointed to by end and the 5 following locations. end is the address of the fourth character of name4 the 'r'. So strcpy overwrites 'r' with 'm', 'i' with 'o', 'n' with 'v', 'g' with 'i', the original null character from name4 with 'e', and the following character with a null character. So the contents of name4 are now "one movie". The problem in the program as it appear on the exam is that C will set aside just enough characters for name4 to be room for "one ring" that's 9 characters including the null. The new null will go in the 10'th position, beyond the space reserved for name4. This is the kind of bug I've been warning you about! In the case of this simple program, there wasn't anything important being stored in that byte immediately after name4, so the program ran without a problem and printed "one movie". If this block of code had appeared in a different program, we might have had a segmentation fault or other strange bugs. I gave a full point to anyone who remarked that this kind of error was possible. I also gave a point to to a few people who gave the answer "one movi". They seemed to be assuming that C would refrain from going past the end of the space allocated for name4. That would be nice, but it's not the case. But I felt they deserved credit for realizing there was a memory space problem. struct twonums r = {1,2}; struct twonums s = {3,4}; r = s; r.a = 5; r.b = 6; printf("%d, %d\n", s.a, s.b); The point of this question was discovering whether you understood what an assignment between two structures means. The assignment r = s copies the contents of s into r. There are no pointers involved and there is no lasting connection made between r and s. So the changes to r later on have no effect on s, and s keeps its original contents of 3 and 4, and the output is "3, 4". Page 4 of 7

5 Question 3 (12 points): Recall the following definitions from the Week 7 Lab: /* Maximum length for names (not counting null at end) */ #define NAME_LEN 20 /* A struct to store information about a student: their * name, id number, and their mark in the course */ struct studentstruct { char name[name_len+1]; int id; float mark; }; /* end struct */ /* a handy typedef so you don't have to type "struct" * all the time */ typedef struct studentstruct student; On the next page, write a function called a_students that takes an array of student structs and returns a new array containing just the A students from the original array (students with marks of 80 or more). Your a_students function should take 3 parameters: 1. an array of students 2. the size of parameter #1 3. an address in which to store the size of the array you are returning For example, if you've got an array called course which contains this data: name id mark Harry Fred Hermione Ron George and you do the following: int topcount; student *top = a_students(course, 5, &topcount); the variable topcount should now be equal to 3 and the top array should contain this data: name id mark Harry Hermione Ron Page 5 of 7

6 Question 3, continued. Please write your a_student function in the space below: student* a_students(student array[], int size, int *newsize) { *newsize = 0; int i; for (i = 0; i < size; i++) { if (array[i].mark >= 80) (*newsize)++; /* these parenthesis are necessary */ } /* end for */ student *newarray = malloc(*newsize * sizeof(student)); if (newarray == NULL) { fprintf(stderr, "error: heap is full\n"); exit(exit_failure); } /* end if */ int index = 0; /* index into newarray */ for (i = 0; i < size; i++) { if (array[i].mark >= 80) { newarray[index] = array[i]; index++; } /* end if */ } /* end if */ return newarray; } /* end a_students */ If instead of counting the A students first you allocated an array of the same size as the original, that was OK. It wastes some space at the the of the array, but it works and saves the time taken by counting the A students in the first loop. Quite a few people were confused between an array of students and an array of pointers to students. The question asked for "an array of student structs", but I decided I'd accept a solution that treated the first parameter and return value as arrays of pointers to student structs, as long as you followed through consistently. Most people who tried this didn't quite pull it off; they alternated between treating the elements as structs and as pointers to structs. Several people created an array as a local variable and returned it. We spent a fair amount of time in class talking about why it's not OK to return a pointer to a stack address that will be overwritten. There was another common small goof. If newarraysize is the third parameter, a pointer to the size of the new array, many people used the statement *newarraysize++; That actually won't work, since the ++ binds tighter than the *. You need parenthesis: (*newarraysize)++; This was a small thing, and one that you may never have run into, so I noted it but didn't deduct marks. Page 6 of 7

7 Marking Scheme For Question 3: created new array on heap & returned it: 3 points copied A students into new array (algorithm, not types): 3 points sent back number of A students via 3 rd parameter: 3 points array & pointer types correct: 3 points Page 7 of 7