Difference between revisions of "CSC231 Homework 7 Fall 2012"

From dftwiki3
Jump to: navigation, search
(→‎Submission)
 
(7 intermediate revisions by the same user not shown)
Line 11: Line 11:
 
   x = 305441741 = 0x1234ABCD
 
   x = 305441741 = 0x1234ABCD
  
 +
where 305441741 is printed by a call to '''print_int''', and 0x1234ABCD is printed by your program.
  
 
==Submission==
 
==Submission==
Line 30: Line 31:
 
<br />
 
<br />
  
−
Concentrate on the hexadecimal code for each letter.  You will notice that 'A' through 'O' have the same upper nybble.  'A' is 0x'''4'''1 and 'O' is 0x'''4'''F.  So, if I wanted to ''squeeze'' a string made of upper-case letters between 'A' and 'O', I would only need to store the lower nybble, and I'd save some memory.
+
Concentrate on the hexadecimal code for each letter.  You will notice that 'A' through 'O' have the same upper nybble.  'A' is 0x'''4'''1 and 'O' is 0x'''4'''F.  So, if I wanted to ''squeeze'' a string made of upper-case letters between 'A' and 'O', I could keep only the lower nybbles, pack them,  and I'd save some memory!
  
 
The purpose of this problem is for you to take a string created using this method, decode it, and print it on the screen.
 
The purpose of this problem is for you to take a string created using this method, decode it, and print it on the screen.
−
 
−
Let's look at an example.  Look at the 2 variables below.  '''xLen''' is a double word, and contains 18.  That represents the number of bytes containing the characters of the original string. Since each byte of x contains 2 nybbles that are the lower parts of two different characters, then '''xLen''' is the number of pairs of nybbles. 
 
−
 
−
Let's take a look at '''x''' now.  The first byte of '''x''' is 0xd1, which means that the first two characters of the original string are 0x4'''d''' and 0x4'''1''', or 'M' and 'A'.  So your program will output 'M' then 'A' when it decodes the string.
 
  
 
   
 
   
Line 42: Line 39:
 
  x        db    0xd1, 0x41, 0xd5, 0x03, 0x81, 0xee, 0x5c, 0x0d
 
  x        db    0xd1, 0x41, 0xd5, 0x03, 0x81, 0xee, 0x5c, 0x0d
 
           db    0x55, 0x40, 0x21, 0x40, 0xd1, 0xe0, 0x9e, 0x0c, 0x9d, 0x10
 
           db    0x55, 0x40, 0x21, 0x40, 0xd1, 0xe0, 0x9e, 0x0c, 0x9d, 0x10
−
   
+
 
 +
 
 +
Let's look at an example. Look at the 2 variables above.  '''xLen''' is a double word, and contains 18.  18 represents the '''number of bytes''' or the '''number of characters''' in the original string. Since each byte of '''x''' contains 2 nybbles that are the lower parts of two consecutive characters, then '''xLen''' is the number of pairs of nybbles. 
 +
 
 +
Let's take a look at '''x''' now.  The first byte of '''x''' is 0xd1, which means that the first two characters of the original string are 0x4'''d''' and 0x4'''1''', or 'M' and 'A'.  So your program will output 'M' then 'A' as the first two characters when it decodes the string.
 +
 
 +
==Your Assignment==
  
 
Your program must decode the two strings listed below and print them on the screen.  Your program should replace the character '@' with a space when it prints it.
 
Your program must decode the two strings listed below and print them on the screen.  Your program should replace the character '@' with a space when it prints it.
Line 49: Line 52:
 
  y        db    0xd5, 0x54, 0x00, 0x31, 0xe4, 0x94, 0x0f, 0x21, 0xd1
 
  y        db    0xd5, 0x54, 0x00, 0x31, 0xe4, 0x94, 0x0f, 0x21, 0xd1
 
           db    0x01, 0xe4, 0x02, 0xfe, 0x90, 0xf6, 0x07, 0x1e, 0x10
 
           db    0x01, 0xe4, 0x02, 0xfe, 0x90, 0xf6, 0x07, 0x1e, 0x10
−
  xLen     dd    18
+
 
−
  x       db    0xd1, 0x41, 0xd5, 0x03, 0x81, 0xee, 0x5c, 0x0d
+
  zLen     dd    24
−
           db    0x55, 0x40, 0x21, 0x40, 0xd1, 0xe0, 0x9e, 0x0c, 0x9d, 0x10
+
  z       db    0xef, 0x07, 0xff, 0x40, 0xd1, 0x41, 0xd5, 0x03, 0x81  
 +
          db    0xee, 0x5c, 0x01, 0xe4, 0x02, 0x14, 0x0d, 0x1e, 0x0d
 +
           db    0xf1, 0xe0, 0x9e, 0x0c, 0x9d, 0x10
  
 
<br />
 
<br />
Line 66: Line 71:
 
=Python Program=
 
=Python Program=
 
<source lang="python">
 
<source lang="python">
−
 
+
text = "no good madame channel and bad man moan in lima"
−
text = "madame channel meet bad man in lima"
+
#text = "madame channel meet bad man in lima"
−
text = "meet  candid obama and boni of gana"
+
#text = "meet  candid obama and boni of gana"
 
text = text.upper()
 
text = text.upper()
 
print( len(text), text )
 
print( len(text), text )
−
 
+
−
 
+
 
lastNybble = ' '
 
lastNybble = ' '
 
out = []
 
out = []
−
 
+
 
if len( text )%2 == 1:
 
if len( text )%2 == 1:
 
     text += "@"
 
     text += "@"
−
 
+
 
count = 0
 
count = 0
 
for i, c in enumerate( text ):
 
for i, c in enumerate( text ):
Line 90: Line 95:
 
         count += 1
 
         count += 1
 
     lastNybble = nybble
 
     lastNybble = nybble
 +
 +
 +
print( "xLen    dd    ", count  )
 +
print( "x        db    ", ', '.join( out )  )
  
 +
</source>
  
−
print( "xLen    dd    ", count )
+
* Other version where we create an array of the characters that are used and then pack the nybbles representing the indexes of the characters used in the string.
−
print( "x        db    ", ', '.join( out ) )
 
  
 +
<source lang="python">
 +
#text = "no good madame channel and bad man moan in lima"
 +
#text = "madame channel meet bad man in lima"
 +
#text = "meet  candid obama and boni of gana"
 +
text = "A man, a plan, a canal, Panama"
 +
text = text.upper()
 +
print( len(text), text )
 +
 +
letters = []
 +
for letter in text:
 +
    if not letter in letters:
 +
        letters.append( letter )
 +
 +
array = []
 +
for letter in letters:
 +
    array.append( letter )
 +
print(  "table    db  " + ', '.join( ["'"+k+"'" for k in array] ) )
 +
 +
lastNybble = ' '
 +
out = []
 +
 +
if len( text )%2 == 1:
 +
    text += "@"
 +
 +
count = 0
 +
for i, c in enumerate( text ):
 +
    #if c == ' ': c = '@'
 +
    index = letters.index( c )
 +
    nybble = hex( index )[-1]
 +
    #nybble = index
 +
    print( c, hex( ord( c ) ), nybble )
 +
    if  i!= 0 and i%2==1:
 +
        hexa = "0x" + lastNybble + nybble
 +
        out.append( hexa )
 +
        print( "          0x" + lastNybble + nybble )
 +
        count += 1
 +
    lastNybble = nybble
 +
 +
 +
print( "xLen    dd    ", count  )
 +
print( "x        db    ", ', '.join( out )  )
  
 
</source>
 
</source>
 +
 
</onlydft>
 
</onlydft>
  

Latest revision as of 14:15, 30 October 2012

--D. Thiebaut 13:47, 24 October 2012 (EDT)

This assignment is due the evening of Oct. 31st, at midnight (booo!). You can work on this assignment in pairs.

Problem #1 (0.5 point doc, 1.5 points program)

Write a program that displays the contents of a double-word in hexadecimal. Your program should store the value 0x1234ABCD in a variable called x and should display this information: 

  x = 305441741 = 0x1234ABCD

where 305441741 is printed by a call to print_int, and 0x1234ABCD is printed by your program.

Submission

Submit your program (called hw7a.asm) as follows: 

 rsubmit hw7 hw7a.asm

Hints

HexChars        db          "0123456789ABCDEF"

Problem #2 (0.5 point doc, 1.5 points program)

Take a look at the section of the ASCII table below:


ASCIIUpperCase.png


Concentrate on the hexadecimal code for each letter. You will notice that 'A' through 'O' have the same upper nybble. 'A' is 0x41 and 'O' is 0x4F. So, if I wanted to squeeze a string made of upper-case letters between 'A' and 'O', I could keep only the lower nybbles, pack them, and I'd save some memory!

The purpose of this problem is for you to take a string created using this method, decode it, and print it on the screen. 


xLen     dd     18
x        db     0xd1, 0x41, 0xd5, 0x03, 0x81, 0xee, 0x5c, 0x0d
         db     0x55, 0x40, 0x21, 0x40, 0xd1, 0xe0, 0x9e, 0x0c, 0x9d, 0x10

Let's look at an example. Look at the 2 variables above. xLen is a double word, and contains 18. 18 represents the number of bytes or the number of characters in the original string. Since each byte of x contains 2 nybbles that are the lower parts of two consecutive characters, then xLen is the number of pairs of nybbles.

Let's take a look at x now. The first byte of x is 0xd1, which means that the first two characters of the original string are 0x4d and 0x41, or 'M' and 'A'. So your program will output 'M' then 'A' as the first two characters when it decodes the string.

Your Assignment

Your program must decode the two strings listed below and print them on the screen. Your program should replace the character '@' with a space when it prints it. 

yLen     dd     18
y        db     0xd5, 0x54, 0x00, 0x31, 0xe4, 0x94, 0x0f, 0x21, 0xd1
         db     0x01, 0xe4, 0x02, 0xfe, 0x90, 0xf6, 0x07, 0x1e, 0x10
 
zLen     dd     24
z        db     0xef, 0x07, 0xff, 0x40, 0xd1, 0x41, 0xd5, 0x03, 0x81 
         db     0xee, 0x5c, 0x01, 0xe4, 0x02, 0x14, 0x0d, 0x1e, 0x0d
         db     0xf1, 0xe0, 0x9e, 0x0c, 0x9d, 0x10


Note: You do not need to use functions to solve this problem, but they will make your life easier. Suggestion if you are starting on this on Wednesday 10/24 or Thursday 10/25: write the code for decoding 1 string, and wait until Friday's lecture to put this code in a function that will decode any string you pass to it.


Submission

rsubmit hw7 hw7b.asm



...














Retrieved from "http://dominiquethiebaut.com/dftwiki3/index.php?title=CSC231_Homework_7_Fall_2012&oldid=15685"