CIS 2107 Computer Systems and Low-Level Programming Fall 2010 Midterm

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1 Fall 2010 Name: Page Points Score Total: 100 Instructions The exam is closed book, closed notes. You may not use a calculator, cell phone, etc. For each of the questions of this quiz, you can assume the following sizes for C data types: type bytes char 1 short 2 int 4 long 8 float 4 double 8 void* 4 i

2 For the following questions, you can assume that my home directory is the jfiore directory. 1. Unix shell stuff. jfiore home 2107 stuff.txt... prog.c assign (1 point) (a) If I m in my home directory, and I run the command gcc -c prog.c, what is the name of the resulting file and what does it contain? Solution: It s prog.o, and it s a binary file containing instructions for the particular machine for which we ve compiled the program. (1 point) (b) If I run the command gcc -E prog.c to run the preprocessor only on prog.c, what does the resulting file contain (i.e., how is it different from prog.c)? Solution: Among other things, the #include statements are replaced with the contents of the files included, and the #define statements are processed. (1 point) (c) What s the command that allows me to see what s in prog.c one page at a time? Solution: more prog.c (1 point) (d) If I m in my home directory i.e., /home/jfiore, what s the single command that I can type in order to remove the assignment 2 directory? Solution: rmdir 2107/assign/02 (1 point) (e) What does the ar command do? Solution: It s used to create a library file. You used this in the string library assignment. (1 point) 2. Why is it that when we declare a struct, functions, etc inside a.h file, we enclose it in a #ifndef #define #endif block? Solution: So that if the.h file is #included by multiple.c files, the code in the block is only inserted once. (2 points) 3. We ve described a computer s storage hierarchy as a kind of pyramid. What two things are true as we go up the pyramid? 1 of 11 exam continues...

3 Solution: The storage is faster and we have less of it (or that it s more expensive). (1 point) 4. (a) 72 mbytes =? tbits (1 point) (b) 8 bytes =? kbits (1 point) (c) 5 minutes =? microseconds (1 point) (d) 152 gbits =? gbytes (1 point) (e) 160 tbits =? kbits (a) (b) (c) (d) 19 (e) 160,000,000,000 (2 points) Convert each of the following from base 10 to base 2 and base 16 Solution: 196 = = = = = = = = = = C4 16 (2 points) out of a possible 9 2 of 11 exam continues...

4 Solution: 181 = = = = = = = = = = B5 16 (1 point) (a) Using the approximation trick that we talked about in class, about how much are each of the following? (1 point) (b) 2 22 (1 point) (c) 2 48 (2 points) 8. What is in base 2? (a) (b) (c) 128 billion 4 million 256 trillion Which makes sense, because = , = , and = , which is (2 points) 9. What is B9F AC 16 in base 16? B 9 F A C 16 E F (2 points) 10. In hex, what is the smallest integer that can be represented by an 8-bit two s complement int? 10. 0x80 (2 points) 11. In hex, what is the largest integer that can be represented by an 8-bit two s complement int? 11. 0x7F out of a possible 11 3 of 11 exam continues...

5 (2 points) 12. In hex, what is the largest integer that can be represented by an 8-bit unsigned int? (2 points) 13. In hex, what is 1 as an 8-bit two s complement int? (2 points) 14. What is printed by the following? unsigned char x=255; x+=3; printf("x=%d\n", x); 12. 0xFF 13. 0xFF 14. x=2 15. Some bit operations. If we have char i = 0xC7, j = 0x9F;, what is the result of the following operations? Your answer must be in the form of exactly two hex digits 1. (1 point) (a) i The easiest thing to do is to convert i and j to binary and then solve. 0xC7 16 = and 0x9F 16 = (a) 0x38 (1 point) (b)!!i (b) 0x01 (1 point) (c) i & 0xF0 (c) 0xC0 (1 point) (d) i j i (d) 0x9F 1 Ignore the possibility of promotion to 32-bit ints. Behave as though we re living in the land of 8-bit arithmetic. out of a possible 10 4 of 11 question 15 continues...

6 (1 point) (e) i j (e) 0x01 (1 point) (f) i<<2 (f) 0x1C (1 point) (g) i 0 (g) 0xC7 (2 points) 16. If we re on a little-endian machine, what is printed by the following code? int x=0x ; char *p=(char*)&x; int i; for (i=0; i<sizeof(x); i++, p++) printf("%x\n", (unsigned char)*p); Solution: (6 points) 17. For this question, we re doing 5-bit two s complement representation of integers. Fill in the empty boxes in the following table. Addition and subtraction should be performed based on the rules for 5-bit, two s complement arithmetic. Recall that in your book s notation, TMin is defined to be the smallest negative two s complement number that we can represent, and TMax is the largest positive one. out of a possible 11 5 of 11 exam continues...

7 Name Decimal Rep. Binary Rep. Zero n/a n/a n/a TMax TMin TMin TMax If I have the following: int main(void) { int a=10; int b=20; int *p=&a; char *cp=(char*)&a;... (*p)++; cp+=2; p+=2; and memory is laid out like this: a 1000 b 1004 p 1008 q 1012 cp 1016 Solution: a b p q 1012???? cp (1 point) (a) a what do you see if you print: (a) 11 out of a possible 1 6 of 11 question 18 continues...

8 (1 point) (b) &a (1 point) (c) b (1 point) (d) &b (1 point) (e) p (1 point) (f) *p (1 point) (g) &p (1 point) (h) cp (b) 1000 (c) 20 (d) 1004 (e) 1008 (f) 1008 (g) 1008 (h) Use the following code to answer the questions. 1 #include <stdlib.h> 2 #include <stdio.h> 3 #include <string.h> 4 5 struct Poo { 6 int x; 7 int A[1]; 8 char *s; 9 ; void func(struct Poo *p); int main(void) 15 { 16 struct Poo p; 17 char *str; str=(char*)malloc(10); strcpy(str, "hi"); 22 p.x=10; 23 p.a[0]=20; 24 p.s=str; func(&p); printf("p.x=%d\n", p.x); 29 printf("p.a[0]=%d\n", p.a[0]); 30 printf("p.s=%s\n", p.s); return 0; void func(struct Poo *p) 36 { 37 p->x=111; 38 p->a[0]=222; 39 strcpy(p->s, "bye"); p=(struct Poo*)malloc(sizeof(struct Poo)); 42 p->x=1111; 43 p->a[0]=2222; 44 p->s=(char*)malloc(10); 45 strcpy(p->s, "later"); 46 (2 points) (a) How many bytes are passed to the function func( )? (a) 4 What is the value of each of the following after func( ) is called 2? out of a possible 9 7 of 11 question 19 continues...

9 (2 points) (b) p.x (2 points) (c) p.a[0] (2 points) (d) p.s (b) 111 (c) 222 (d) bye (9 points) 20. Write a function which takes as an argument an int x. The function returns an int which has the bytes of x in reverse order. Do not assume that ints are 4-byte values. Solution: There are probably lots of ways to do this. Here are two: int revbytes01(int x) { char *b, *e; char tmp; b = (char*)&x; e = b+sizeof(x)-1; while (b<e) { tmp = *b; *b = *e; *e = tmp; b++; e--; return x; int revbytes02(int x) { int i, ret=0, len=sizeof(x); ret = x&0xff; for (i=0; i<len-1; i++) { ret <<= 8; x >>= 8; ret = x&0xff; return ret; (6 points) 21. Write a function which will swap the pointers p and q. To receive any credit, it must be a function, and when the function has finished, p points to b and q points to a. int a=10; int b=20; int *p=&a; int *q=&b; out of a possible 21 8 of 11 exam continues...

10 Solution: Remember that in C, when we pass arguments to functions, we re passing copies of the arguments. If we want to change the pointers, we have to pass a pointer to a pointer. void swapptr(int **a, int **b) { int *tmp = *a; *a=*b; *b=tmp; We d call this function by doing: swapptr(&p, &q); (10 points) 22. Write a C function which is passed a C-string s, a C-string sep, an int multiplier m. The function returns a new string which is the original string duplicated m times, separated by sep. For example, if s= bora, sep=, and m=2, the function returns the string bora bora. The caller is responsible for freeing the memory allocated by your function. Solution: There was no restriction on use of string functions from the standard C library. #include <stdlib.h> #include <string.h> char *multiply(char *s, char *sep, int m) { char *ret; int lens, lensep, i; lens=strlen(s); lensep=strlen(sep); if (m<1) return NULL; /* we ll have: * m occurrences of s * m-1 occurences of the separator * one extra byte for \0 */ if ((ret=(char*)malloc(lens*m+lensep*(m-1)+1))==null) return NULL; strcpy(ret, s); for (i=1; i<m; i++) { strcat(ret, sep); strcat(ret, s); return ret; out of a possible 10 9 of 11 exam continues...

11 If you didn t realize that you could use functions in string.h and decided to write something yourself, the functions might look something like: int m_strlen(char *s) { int i=0; while (s[i]!= \0 ) i++; return i; void m_strcpy(char *d, char *s) { int i=0; while (s[i]!= \0 ) { d[i]=s[i]; i++; d[i]= \0 ; void m_strcat(char *d, char *s) { int i=0, j=0; while (d[j]!= \0 ) j++; while (s[i]!= \0 ) { d[j]=s[i]; i++; j++; d[j]= \0 ; (10 points) 23. Write a function which is passed a C string. The function capitalizes the first letter of every word in the string. Solution: #include <stdio.h> #include <ctype.h> int main(void) { char c; int inword = 0; /* if we re currently inside a word */ while ((c=getchar())!=eof) { if (isspace(c) && inword) out of a possible of 11 question 23 continues...

12 inword = 0; else if (!isspace(c) &&!inword) { inword = 1; c = toupper(c); putchar(c); return 0; out of a possible?? 11 of 11 end of exam