2.1 From the Class Notes link on the Course Home PageIndex

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• Assignment #06 - Submit midterm test corrections
• 120_CharacterEncoding.html - Character Encoding / Line Ends
• 130_big_picture.txt - The Big Picture on Bit Patterns
• 030_binary_math.txt - Binary Integer Mathematics, unsigned, two’s complement, etc.
  • Read: Status Flags for binary integer mathematics
• 040_overflow.txt - The CARRY flag and OVERFLOW flag in binary arithmetic

2.2 From Blackboard Course DocumentsIndex

These documents have restricted distribution and cannot be put on the Course Home Page.

• 02.ppt - Data Representation
  • Read: Character encoding: ASCII, Unicode, Line Endings
  • Read: CPU Flags: slides 47-49, 53-54
  • slide 54 has an error - change second-line 0100+0010 to be 0100+0110
• 03.ppt - Boolean Algebra and Digital Logic
  • Read: Boolean Algebra
  • Ignore: Digital Logic

2.3 From the InternetIndex

• Valentine’s Day: http://www.wolframalpha.com/ - Enter this: x^2plus(y-(cuberoot(x^2)))^2=1

• http://en.wikipedia.org/wiki/ISO/IEC_8859 - a joint ISO and IEC series of standards for 8-bit character encodings

• http://czyborra.com/charsets/iso8859.html - The ISO 8859 Alphabet Soup

• http://en.wikipedia.org/wiki/UTF-8

• http://en.wikipedia.org/wiki/Bitwise_operation

• http://download.oracle.com/javase/tutorial/java/nutsandbolts/op3.html - Bitwise Operators in Java

• http://en.wikipedia.org/wiki/Introduction_to_Boolean_algebra

• http://en.wikipedia.org/wiki/Overflow_flag

• http://en.wikipedia.org/wiki/Carry_flag

2.4 From the Classroom Whiteboard/ChalkboardIndex

• Your in-class notes go here.

• Review: Famous ASCII characters: 0D 0A 20 30 41 61 7F

• Problems with all the 8-bit extended-ASCII character sets

  • can’t have Latin1 French and Latin2 Polish in the same text file: why?

• UTF-8 advantages and disadvantages

  • same as ASCII, no endian-ness, multi-byte, full Unicode, may require more bytes than Unicode

• Shifting numbers right and left

  • shifiting effect on two’s complement numbers: does it work?
  • Arithmetic Right Shift vs. Logical Right Shift
    • duplicate the top (carry) bit for Arithmetic Right Shift
  • F4240h = 1,000,000(10)
    • so F42400h = 1,000,000 * 16 = 16,000,000
    • so F424h = 1,000,000/16 = 1000000 * 0.0625 = 100 * 625 = 62,500
  • 03641100(8) = 1,000,000(10)
    • so 036411000 = 1,000,000 * 8 = 8,000,000
  • 1010(2) = 10(10)
    • so 101000 = 10 * 2 * 2 = 40

• Bitwise Operators AND, OR, XOR, NOT (in Java: & | ^ ~)

  • don’t confuse bitwise with logical: & with &&, | with ||, or ~ with !
  • if x=1, y=2 then x&y = 0 (no bits in common)
  • if x=6, y=12 then x&y = 4 (one bit in common)
  • XOR is Exclusive OR - A or B but not both
    • true OR true is true
    • true XOR true is false

• Conversions between upper/lower in ASCII using bitwise OR and AND:

  • 'A' | ' ' = 'a' (OR turns on the 20h bit, making it lower-case)
  • 'a' & ~' ' = 'A' (AND turns off the 20h bit, making it upper-case)

• Simplifying complex logic using Boolen Algebra - identities - deMorgan

  1. simplify: if ((x < y) && (y < z)) || ((x < y) && (y >=z ))
     • simplifies to just if (x < y) because:
     • let a = (x<y) and b = (y<z), therefore b' = (y>=z)
     • rewrite ab+ab' = a(b+b') = a(1) = a = (x<y)
  2. suppose a stale order means: date > 120 and amount > 0
     1. simplify: print unless stale
        • unless stale means if NOT stale
        • print if NOT (date > 120 && amt > 0)
        • using deMorgan: print if (date <= 120 || amt <= 0)
     2. simplify: print unless (stale and amt < 0)
        • print if NOT (stale && amt < 0)
        • print if ( !stale || amt >= 0)
        • print if ( date <= 120 || amt <= 0 || amt >= 0)
        • print if ( date <= 120 || TRUE )
        • therefore: always print (probably an error in the specifications!)
  3. Show that (xy + x'z + yz')' = y'(x+z')
     • (x'+y')(x+z')(y'+z) -- deMorgan
     • (x'x + x'z' + xy' + y'z')(y'+z) -- expand first two expressions
     • ( 0 + x'z' + xy' + y'z')(y'+z) -- identities
     • x'y'z' + x'z'z + xy'y' + xy'z + y'y'z' + y'z'z -- expand again
     • x'y'z' + 0 + xy' + xy'z + y'z' + 0 -- identities
     • x'y'z' + y'z' + xy' + xy'z -- regroup similar terms
     • y'z' + xy' -- absorption rule
     • y'(x+z') -- QED

• CPU Status flags: Zero, Sign, Carry, Overflow - Z S C V

  • 4-bit binary carry: 1100+1100=1000
  • 4-bit binary overflow: 0100+0100=1000
  • 4-bit binary carry and overflow: 1000+1000=0000
  • see 02.ppt
    • CPU Flags: slides 47-49, 53-54
    • slide 54 has an error - change second-line 0100+0010 to be 0100+0110

Author: 
| Ian! D. Allen  -  idallen@idallen.ca  -  Ottawa, Ontario, Canada
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