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计算机期中代考 CS 341代写 C语言代写

2023-03-07 17:03 星期二 所属: C++/C代写,c语言代写代考-100%安全,包过 浏览:272

计算机期中代考

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 = ___

 

计算机期中代考
计算机期中代考

 

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

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