Portland State University. CS201 Section 5. Midterm Exam. Fall 2018

Transcription

1 Portland State University CS201 Section 5 Midterm Exam Fall 2018 Name: This exam has 9 pages including this cover. The last page contains tables to assist you in performing binary and hexadecimal conversions Write your name on every page before you start There are 4 questions, worth a total of 100 points This is a closed book exam. A non-programmable simple calculator is allowed. No cell phone calculators allowed. No other electronics allowed. Write clearly and show your work. State any assumptions you make You have 70 minutes. Budget your time and good luck! Question Maximum Your Score Total 100

2 Question 1: Linking a) What is an Instruction Set Architecture (ISA) and what is a Microarchitecture? (8 POINTS) Abstraction layer between the software and the microarchitecture. It serves as a programming interface to the developers. Microarchitecture is the particular implementation of an ISA b) What are the two steps involved in the linking process and briefly describe each step? (5 POINTS) 1. Symbol Resolution: The linker is responsible for matching every external symbol with its corresponding global symbol by looking across all the relocatable object files to be linked 2. Relocation: Relocation entries are replaced with actual addresses in the linked executable object file. ~ 2 ~

3 c) What are the 3 types of Object Files and how do they fit in the linking process? Which ones are relocatable? (12 POINTS) Relocatable Object Files: Contains code and data in a format that can be combined with other relocatable object files to form an executable during linking Executable Object Files: Contains code and data in a form that can be copied directly into memory and then executed. They are not relocatable Shared Object File: A relocatable object file that can be loaded into memory and linked dynamically, at either load time or run-time. d) From the following C code specify where the following symbols are stored in the compiled executable (i.e Stack, Heap, TEXT or DATA segment) (10 POINTS) Global variable myglobal: DATA Segment Global variable foo: DATA Segment Local variable myarray: Stack The dynamic array referenced by myarray: Heap The function referenced by foo: TEXT Segment static double myglobal = 3.14; static int (*foo)(int a, int b) = main; void main() char* myarray; myarray= (char*) malloc(5 * sizeof(char)); ~ 3 ~

4 Question 2: C Language a) Write a function that allocates a 3-Dimensional array of doubles (double precision floating point numbers) using malloc(). You must use the Array of Pointers method. A prototype of the function has been provided below, in which array is a pointer to the 3-Dimensional array, sizex is the length of the first dimension, sizey is the length of the second dimension and sizez is the length of the third dimension. Note that the allocated array must be available to use by the calling function. Assume all the required headers (e.g. stdlib.h) are included (15 POINTS) void Allocate3DArray(double**** array, int sizex, int sizey, int sizez) int i, j; *array=(double***) malloc(sizeof(double**) * sizex); for(i=0; i < sizex; i++) (*array)[i]=(double**) malloc(sizeof(double*) * sizey); for(j=0; j < sizey; j++) (*array)[i][j]=(double*) malloc(sizeof(double) * sizez); ~ 4 ~

5 b) Write a function that de-allocates a 3-Dimensional array of doubles, just like the one you allocated in Part A. The prototype of the function has been provided below (10 POINTS) void Destroy3DArray(double*** array, int sizex, int sizey) int i, j; for(i=0; i < sizex; i++) for(j=0; j < sizey; j++) free(array[i][j]); free(array[i]); free(array); ~ 5 ~

6 Question 3: Hex and Binary Arithmetic a) Assuming a 16-bit Two s complement system. Convert the following signed numbers from decimal to binary. Show all your work. (6 POINTS) i.) = =5 5-4=1 1-1=0 i.) = = =4 4-4=0 X= ~X= ~X+1= b) Convert the following hex numbers to binary. (4 POINTS) i.) E7FC = From the table we just convert each hex numeral into a group of 4 bits ii.) A0B = ~ 6 ~

7 c) Compute the following binary operations (Subtraction 5 POINTS, Multiplication 5 POINTS. 10 POINTS TOTAL) i.) = We convert the subtrahend to a negative number in Two s Complement notation and then add both numbers X= ~X= ~X+1= = ii.) x 101 = = ~ 7 ~

8 Question 4: IEEE Floating Point a) Convert the following binary number to IEEE Single precision (32-bits). Show all your work. (20 POINTS) x2-3 The format of the IEEE Single precision number is provided below: Exponent Bias: 127 Length of Sign Field: 1 bit Length of Exponent Field: 8 bits Length of Fraction Field: 23 bits Sign Bit = 1 Normalized Binary = x2 1 Exp= IEEE Number: ~ 8 ~

9 Conversion Tables A 1010 B 1011 C 1100 D 1101 E 1110 F 1111 ~ 9 ~