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COMP201 Fall 20 20 Midterm Exam
Duration: 90 minutes
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This exam contains 8 multi-part questions and you have 90 minutes to earn 9 0 marks.
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Question 1 A. Integer Representations, Bitwise Operations [9 POINTS] Assume that you are working on a 1 0 - bit machine, in which
- Two’s complement arithmetic is used for signed integers,
- short integers are encoded using 5 bits.
- Sign extension is carried out when a short is casted to an int.
- All shift operations are arithmetic. Consider the following variable declarations Fill in the empty boxes in the table below. The first two rows have already been answered for you. short sa = - 7 ; int b = 2 *sa; int a = - 32 ; short sb = (short) a; unsigned short ua = a+28; Note: You need not fill in entries marked with ”–”. TMax denotes the largest positive two’s complement number and TMin denotes the minimum negative two’s complement number. Expression Decimal Representation Hexadecimal Representation Zero 0 0x^000 (short) 1 1 0x sa - 7 0x b - 14 0x3f sb 0 0x ua 28 0x1c a >> 2 - 8 0x3f ua >> 2 7 0x b << 3 - 112 0x Tmax 511 0x1ff Tmax - Tmin - 1 0x3ff
Question 3. C-Strings In the company you are employed, your boss asked you to implement the strtok() function in the standard C library that you used in your second assignment. Recall that strtok() splits a given string into a series of tokens using a set of delimiters provided as a second input (e.g. " \t\r\n"). The function returns a pointer to the first token found in the string, and if there are no tokens left, it simply returns NULL. In doing so, strtok() basically identifies both the beginning and the end of the token so that the next the function is called, it continues the search from the previous ending position. strtok() utilizes a static variable to store the pointer to the beginning of the next token (the end of the extracted token), and let’s call this pointer pNextToken. The implementation of strtok() is given below: char ∗ strtok (char ∗str, const char ∗delim) { /∗ initialize ∗/ if (!str) str = pNextToken; /∗ find start of token in str ∗/ str += strspn(str, delims ); if (∗str == ’\0’) return NULL; /∗ find end of token in text ∗/ pNextToken = str + strcspn(str, delim); /∗ insert null−terminator at end ∗/ if (∗pNextToken != ’\0’) ∗pNextToken++ = ’\0’; return str; } As you may notice, the above implementation uses the two common string functions strspn() and strcspn() and you have to implement these functions by yourself, too. To make things a little bit simpler, you first implemented an helper function named strwhere(), declared as: int strwhere(const char ∗str , const char ch); This function can be used to locate a character in a given string. (a) [4 POINTS] Using the above helper function, implement the strspn() function by writing down the missing pieces in the code below: unsigned int strspn(const char ∗ str, const char ∗ delims) { unsigned int i ; for (i = 0; str[i] != ’\0’ && strwhere(delims, str[i]) > −1; i++); return i ; } (b) [ 4 POINTS] Similarly, using the helper function strwhere(), implement the strcspn() function which computes the index of the first delimiter character in our string. unsigned int strcspn(const char ∗ str, const char ∗ delims) { unsigned int i ; for (i = 0; str[i] != ’\0’ && strwhere(delims, str[i]) == −1; i++); return i ; }
Question 4 A. Pointers and Working with Dynamic Memory In this question, you will try to debug a set of programs and identify the problems associated with them. Here are some common errors and mistakes that you can refer to in your answers:
- storage used after free,
- allocation freed repeatedly,
- insufficient space for a dynamically allocated variable,
- freeing unallocated storage,
- freeing of the stack space,
- memory leakage,
- assignment of incompatible types,
- returning (directly or via an argument) of a pointer to a local variable,
- dereference of wrong type,
- dereference of uninitialized or invalid pointer,
- incorrect use of pointer arithmetic,
- array index out of bounds (a) [ 6 POINTS] What is the problem with the following code? Justify your answer, otherwise your answer will not be counted. Note that there could be more than one error. void myfunc(int arr) { int p_arr = (int) malloc(2sizeof(int)); p_arr[0] = 42; p_arr[1] = 24; arr = p_arr; } int main(int argc, char *argv[]) { int *arr = NULL; myfunc(arr); printf("arr[0] = %d\n arr[1] = %d", arr[0], arr[1]); free(arr); return 0; } dereference of uninitialized or invalid pointer: arr in main is still NULL freeing unallocated storage: free(arr) (b) [ 6 POINTS] What is the problem with the following code? Justify your answer, otherwise your answer will not be counted. Note that there could be more than one error. int main(int argc, char *argv[]) { if (argc!=3) {printf("wrong number of arguments\n"); return 1; char *param1 = *argv[1]; char *param2 = *argv[2]; char *ptr; ptr = (char ) malloc(strlen(param1)+strlen(param2)+1); while ((ptr++ = *param1++) != 0) ; strcat(ptr+strlen(param1)+1, param2); printf("%s\n", ptr); ptr = NULL; return 0; } Dereference of invalid pointer: strcat could not find end of dest memory leakage: ptr = NULL
Question 6 A. Disassembly Please answer the questions below considering the following fragment of the assembly code generated by running the objdump - d command. 4004ed: b9 00 00 00 00 mov $0x0,%ecx 4004f2: eb 16 jmp 40050a <myfunc+0x1d> (a) [ 2 POINTS] The objdump has not included the corresponding suffix to the mov instruction to explicitly state the size information. That being said, you can still identify the size of the destination operand. What is the size and how do you tell? 32 bits/4 bytes (a) [ 2 POINTS] What is the value of %rip register as the CPU executes the mov instruction in the sequence of two instructions? 400 4f (b) [ 10 POINTS] Let’s now look at the entire function from which we took these two instructions. 00000000004004ed : 4004ed: b9 00 00 00 00 mov $0x0,%ecx 4004f2: eb 16 jmp 40050a <myfunc+0x1d> 4004f4: 85 ff test %edi,%edi 4004f6: 74 0b je 400503 <myfunc+0x16> 4004f8: ba 00 00 00 00 mov $0x0,%edx 4004fd: f7 f7 div %edi 4004ff: 01 c1 add %eax,%ecx 400501: eb 07 jmp 40050a <myfunc+0x1d> 400503: 8d 04 40 lea (%rax,%rax,2),%eax 40050 6: 01 f8 add %edi,%eax 400508: 01 c1 add %eax,%ecx 40050a: 8b 06 mov (%rsi),%eax 40050c: 89 c2 mov %eax,%edx 40050e: 0f af d7 imul %edi,%edx 400511: 39 ca cmp %ecx,%edx 400513: 77 df ja 4004f4 <myfunc+0x7> 400515: 89 c8 mov %ecx,%eax 400517: c3 retq Fill in the blanks of the corresponding C function: unsigned int myfunc(unsigned int a, unsigned int b) { unsigned int result = 0; while (result < a(b)) // or and equivalent expression if (a>0) result += b/a; else result += a+3(b); return result; }
Question 7A. Makefiles [1 POINTS] In your free time you are working on a large project involving multiple source files, which have the following #include dependencies:
#ifndef BEANS_H #define BEANS_H ... #endif
#ifndef COFFEE_H #define COFFEE_H #include "beans.h" ... #endif
#ifndef MILK_H #define MILK_H
#include "beans.h" ...
#include "coffee.h" #include "milk.h"
#include "coffee.h" int main(int argc, char *argv[]) { ... } You can always compile your project using gcc - Wall - g - std=c11 - o main *.c to create an executable program. But as you learned in COMP 201 that this is not a good idea as it is not necessary to recompile and relink the files each time you make a change in just one of the files. Think about the dependencies between the source files (.c), the compiled object (.o) files, and the final executable file (main) that is created by linking the object files, and write down the Makefile that you can use for the project. CC = gcc CFLAGS = - Wall - g - std=c main: beans.o coffee.o main.o $(CC) $(CFLAGS) – o main beans.o coffee.o main.o beans.o: beans.c beans.h $(CC) $(CFLAGS) – c beans.c coffee.o: coffee.c coffee.h beans.h milk.h $(CC) $(CFLAGS) – c coffee.c main.o: main.c beans.h coffee.h $(CC) $(CFLAGS) main.c Each target is of 2 points, and you lose 1 point if you don’t consider the dependencies correctly.
(c) [ 3 POINTS] What is the output of the following C program? #include <stdio.h> #define SQUARE(a) a * a #define TRIPLE(b) 3 * b int main() { int a = 2; int b = 3; int x = TRIPLE(a); int y = SQUARE(a+b); int z = TRIPLE(1+a+b); printf("%d %d %d\n", x, y, z); return 0; } 6 11 8
