CS 341: Computer Architecture and Organization
In-Class, Open Book Mid-Term Examination
计算机期中代考 All questions can be answered by one or two short sentences. Do not try to make up for a lack of understanding by providing a rambling answer.
All questions can be answered by one or two short sentences. Do not try to make up for a lack of understanding by providing a rambling answer.
Note: I give partial credit! Show all work!
1.(20 points) Short Questions 计算机期中代考
a.(2 points)What is the use of the Control Bus?
b.(2 points) If %eax = 0x100200, what is wrong with the following Intel assembly instruction?
c.(4 points)Name an advantage and a disadvantage of a processor that has more bits in its address bus and data bus over another processor that has fewer bits on these busses?
Advantage: ______________________________
Disadvantage:____________________________
d.(2 points) What is the difference between dynamic RAM and static RAM
e.(6 points) For the following program:
void main( ){ char * ptr; *ptr = 5;}
i) Which part of the memory can you find ptr?
ii) What does the program do when you run it on the linux server?
iii) What does the program do when you run it on the tutor VM?
f.(4 points) In Lab 2, why do you use the function pgm_read_byte(&a[i]) to read the EEPROM content instead of just using a[i]?
2.(20 points) Evaluations 计算机期中代考
a.(10 points) The content of 16 memory locations starting at 0x00100250 is:
00100250 01 23 45 67 58 02 10 00
89 ab cd ef 00 01 10 11
Show the content of %ecx, %eax after
executing these instructions:
b.(10 points) Show the hex value of the eax register and state of the specified condition flags after executing the instruction.
movl $0xf0f0f0f0, %eax
movl $0x0f0f0f0f, %ecx
xorl %ecx %eax
%eax = 0x 0xffffffff
CF = ___
SF = ___
ZF = ____
OF = ___
3.(20 points) Stack Operations
You are debugging a program on the tutor VM. The content of 16 memory locations starting at 0x003fffe0 is:
003fffe0 bc f0 10 00 00 bd 89 ab
00 00 e8 01 00 00 00 cc
%eax = 0x12345678
%esp = 0x003fffe4
What are the memory and register content after you execute the instruction: pushl %eax ?
0x003fffe0: ________
0x003fffe1: ________
0x003fffe2: ______
0x003fffe3: ______
0x003fffe4: ________
0x003fffe5: ________
0x003fffe6: ________
0x003fffe7: ________
%eax = ___________
%esp = ____________
4.(40 points) Assembly language program 计算机期中代考
Write a C callable assembly language function (change_case.s) to change the alphabets in a string to either upper case or lower case. The user will select an option. The change_case function will return the string with the correct case or it will return the error string if an invalid option is selected.
The function prototype of the assembly language function in C is shown as:
extern char * change_case(char option);
Your assembly language function should get the option from a C main function shown below:
/* change_casec.c: C driver for the changing case function. Users can enter an option:
option = ‘U’ or 0x55: change all to upper case
The function will return the error string if an invalid option is selected.
*/
#include <stdio.h>
extern chat * change_case(char option);
int main()
{
char option;
char *ptr;
printf("Enter case change option : ");
scanf(“%c”, &option);
ptr = change_case(option);
printf("\nThe new string is : %s\n", ptr);
return 0;
}
# Change case assembly language program
# You only have to implement the option = ‘U’
# Return the error string if option != ‘U’
# You can find the ascii code chart at the end
#
ASCII Code Chart: 计算机期中代考
Here is the ASCII Encoding, a correspondence of keyboard characters with integers from 0 to 127 (0x7F in hexadecimal, 0177 in octal)
| char | hex | oct | char | hex | oct | char | hex | oct | char | hex | oct | |||
| NUL | 00 | 000 | SP | 20 | 040 | @ | 40 | 100 | ` | 60 | 140 | |||
| SOH | 01 | 001 | ! | 21 | 041 | A | 41 | 101 | a | 61 | 141 | |||
| STX | 02 | 002 | “ | 22 | 042 | B | 42 | 102 | b | 62 | 142 | |||
| ETX | 03 | 003 | # | 23 | 043 | C | 43 | 103 | c | 63 | 143 | |||
| EOT | 04 | 004 | $ | 24 | 044 | D | 44 | 104 | d | 64 | 144 | |||
| ENQ | 05 | 005 | % | 25 | 045 | E | 45 | 105 | e | 65 | 145 | |||
| ACK | 06 | 006 | & | 26 | 046 | F | 46 | 106 | f | 66 | 146 | |||
| BEL | 07 | 007 | ‘ | 27 | 047 | G | 47 | 107 | g | 67 | 147 | |||
| BS | 08 | 010 | ( | 28 | 050 | H | 48 | 110 | h | 68 | 150 | |||
| HT | 09 | 011 | ) | 29 | 051 | I | 49 | 111 | i | 69 | 151 | |||
| NL/LF | 0A | 012 | * | 2A | 052 | J | 4A | 112 | j | 6A | 152 | |||
| VT | 0B | 013 | + | 2B | 053 | K | 4B | 113 | k | 6B | 153 | |||
NP/FF |
0C | 014 | , | 2C | 054 | L | 4C | 114 | l | 6C | 154 | |||
| CR | 0D | 015 | – | 2D | 055 | M | 4D | 115 | m | 6D | 155 | |||
| SO | 0E | 016 | . | 2E | 056 | N | 4E | 116 | n | 6E | 156 | |||
| SI | 0F | 017 | / | 2F | 057 | O | 4F | 117 | o | 6F | 157 | |||
| DLE | 10 | 020 | 0 | 30 | 060 | P | 50 | 120 | p | 70 | 160 | |||
| DC1 | 11 | 021 | 1 | 31 | 061 | Q | 51 | 121 | q | 71 | 161 | |||
| DC2 | 12 | 022 | 2 | 32 | 062 | R | 52 | 122 | r | 72 | 162 | |||
| DC3 | 13 | 023 | 3 | 33 | 063 | S | 53 | 123 | s | 73 | 163 | |||
| DC4 | 14 | 024 | 4 | 34 | 064 | T | 54 | 124 | t | 74 | 164 | |||
| NAK | 15 | 025 | 5 | 35 | 065 | U | 55 | 125 | u | 75 | 165 | |||
| SYN | 16 | 026 | 6 | 36 | 066 | V | 56 | 126 | v | 76 | 166 | |||
| ETB | 17 | 027 | 7 | 37 | 067 | W | 57 | 127 | w | 77 | 167 | |||
| CAN | 18 | 030 | 8 | 38 | 070 | X | 58 | 130 | x | 78 | 170 | |||
| EM | 19 | 031 | 9 | 39 | 071 | Y | 59 | 131 | y | 79 | 171 | |||
| SUB | 1A | 032 | : | 3A | 072 | Z | 5A | 132 | z | 7A | 172 | |||
| ESC | 1B | 033 | ; | 3B | 073 | [ | 5B | 133 | { | 7B | 173 | |||
| FS | 1C | 034 | < | 3C | 074 | \ | 5C | 134 | | | 7C | 174 | |||
| GS | 1D | 035 | = | 3D | 075 | ] | 5D | 135 | } | 7D | 175 | |||
| RS | 1E | 036 | > | 3E | 076 | ^ | 5E | 136 | ~ | 7E | 176 | |||
| VS | 1F | 037 | ? | 3F | 077 | _ | 5F | 137 | DEL | 7F | 177 |
