COMP 1402

1
COMP 1402

• Winter 2008
• Tutorial 2

2
Overview of Tutorial 2

• Number representation basics
• Binary conversions
• Octal conversions
• Hexadecimal conversions
• Signed numbers (signed magnitude, ones and twos
  complement, Excess-M)
• Float conversions

3
Number representation basics

• In binary each digit (or bit) has 2 possible
  values 0 or 1. Ex (10010101)2
• In octal each digit has 8 possible values 0, 1,
  2 7. Ex (4127)8
• In hexadecimal (or hex for short) each digit has
  16 possible values 0, 1, 2 9, then A, B, C, D,
  E and F (representing 10, 11, 12, 13 ,14 and 15
  in decimal respectively). Ex (1AF6)16 or 1AF6h

4
Bases

• Arbitrary base b
• dn dn-1 dn-2... d2 d1 d0
• To calculate the value
• dnbn dn-1bn-1 ... d1b1 d0b0
• Decimals (base 10)
• How do we interpret the number 2510?
• 2103 5102 1101 0100

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Bases (2)

• Given a number dn dn-1 dn-2... d2 d1 d0
• In binary (base 2)
• value dn2n dn-12n-1 ... d121
  d020
• In octal (base 8)
• value dn8n dn-18n-1 ... d181
  d080
• In hexadecimal (base 16)
• value dn16n dn-116n-1 ... d1161
  d0160

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Fractions in bases

• Arbitrary base fractions
• 0. d-1 d-2 d-3
• value d-1b-1 d-2b-2 d-3b-3
• In decimal (base 10)
• 0.512 is equal to 510-1 110-2 210-3
• In binary (base 2)
• 0.1011 is equal to 12-1 02-2 12-3
  12-4
• 10.5 00.25 10.125 10.0625 0.6875

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Binary Most/Least Significant bits

• Most significant bit (MSB) is the leftmost bit
• Ex 10010100
• It is the bit of highest value (128 in the
  example above) but can also be used as the sign
  of the number (as well see later).
• Least significant bit (LSB) is the rightmost bit
• Ex 10010100
• Has the least value of all the bits (0 or 1).

8
Binary to decimal

• To convert binary to decimal we must add the
  digits weighed by exponents of 2 used in the
  binary number as seen previously.
• value dn2n dn-12n-1 ... d121 d020
• Ex convert (11010)2 to decimal.
• This is equal to 124 123 022 121
  020
• 16 8 0 2 0
  26
• (11010)2 (26)10

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Decimal to binary

• Converting a number from base 10 to base 2.
• Lets look again at the value of a binary number
  dn2n dn-12n-1 ... d121 d020
• We need to fill in the d0 to dn to build the
  binary number as dn dn-1 dn-2... d2 d1 d0
• Different ways to solve this
• Start with the largest power of 2 in the decimal
  number, then move down (slow)
• Algorithm using the mod approach (easier)

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Decimal to binary (2)

• Decimal to binary algorithm
• Q decimal number
• While Q is not equal to 0 do the following
• Binary digit Q mod 2
• Q Q / 2 (quotient)
• End While
• Lets try an example

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Decimal to binary (3)

• Example convert (134)10 to binary.
• 134 mod 2 0
• 67 mod 2 1
• 33 mod 2 1
• 16 mod 2 0
• 8 mod 2 0
• 4 mod 2 0
• 2 mod 2 0
• 1 mod 2 1
• We then read the numbers from bottom up 10000110
• So (134)10 (10000110)2

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Decimal fraction to binary

• Convert both sides of the period separately.
  Weve seen how to do the left side, now the
  fraction side.
• Instead of mod, we multiply by two (2). When we
  go over 1.0, subtract 1 for next round.
• Ex Convert (0.21875)10 to binary.
• 0.21875 2
  0.4375
• 0.4375 2 0.875
• 0.875 2 1.75
• 0.75 2 1.5
• 0.5 2 1.0
  (stop at 1.0)
• Then we read the numbers top to bottom 00111
• Therefore (0.21875)10 0.00111

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Octal to binary

• Since every octal digit can take 8 values we can
  convert each digit to binary using 3 bits.
• Ex convert (7213)8 to binary.
• Well convert each digit separately
• Octal 7 2 1
  3
• Binary 111 010 001 011
• Therefore (7213)8 (111010001011)2

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Binary to octal

• Group the binary number into groups of 3 bits,
  starting from the right, then convert each group
  into their octal value.
• Ex convert (11010110011101)2 to octal.
• 11 010 110 011 101
• 3 2 6 3 5
  
• Therefore (11010110011101)2 (32635)8

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Hexadecimal to binary

• Since every digit of hex has 16 possible values,
  we represent each digit using 4 binary bits.
• Ex convert (5DE9)16 to binary.
• convert each digit into four binary bits
• 5 D E
  9
• 0101 1101 1110
  1001
• Therefore (5DE9)16 (0101110111101001)2

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Binary to hexadecimal

• Group the binary number into groups of 4 bits,
  starting from the right, then convert each group
  into their hex value.
• Ex (10010010111110111)2 to hex.
• 1 0010 0101 1111 0111
• 1 2 5 F
  7
• Therefore (10010010111110111)2 (125F7)16

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Octal to decimal

• Count the digits weighed by exponents of 8, as
  seen previously
• value dn8n dn-18n-1 ... d181
  d080
• Ex convert (314)8 to decimal.
• 382 181 480
• 364 18 41
• 192 8 4
• (314)8 (204)10

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Hexadecimal to decimal

• Count the digits weighed by exponents of 16, as
  seen previously
• value dn16n dn-116n-1 ... d1161
  d0160
• Ex convert (4A1C)16 to decimal.
• 4163 10162 1161 12160
• 44096 10256 116
  121
• 16384 2560 16 12
• (4A1C)16 18972

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Part I of the exercisesthen correction
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Signed Numbers

• Signed Magnitude
• Ones Complement
• Twos Complement
• Excess-M (Bias)

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Signed Magnitude

• MSB is 1 if the number is negative (-), 0 if the
  number is positive ().
• The rest of the number is converted to binary
  like weve seen before.
• What happens with numbers smaller than -127?
• overflow
• Example (-14)10
• 14 in binary is 00001110
• -14 in signed magnitude is 10001110

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Ones complement

• MSB is the sign bit
• 1 is negative
• 0 is positive
• Rules to change sign
• Flip all the bits (change 0s to 1s and 1s to
  0s)
• Two zeroes 00000000 and 11111111
• Harder to know the value of a negative number
  (have to be complemented first)

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Ones Complement (decimal to binary)

• First, convert the decimal to binary
• If the number is negative, flip every bit (0
  become 1, 1 becomes 0). Positive numbers are
  converted to binary without change.
• Example (-44)10 in ones complement
• 44 in binary is 00101100
• -44 in ones complement is 11010011
• Example (38)10 in ones complement
• 38 in binary is
  00100110
• 38 in ones complement is 00100110
  (same)

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Ones Complement (binary to decimal)

• If the MSB is 1, we must flip every bit before
  converting to decimal.
• Ex convert (10011100)2 from ones complement to
  decimal.
• Flip the bits since M.S.B. is a 1 01100011
• Then find the decimal value
• 643221 99 and we add the sign (-99)10
• If the number is positive (MSB is 0) we convert
  to decimal without any other changes.

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Twos complement

• MSB is the sign bit
• 1 is negative
• 0 is positive
• Rules to change sign
• Take ones complement, then add 1.
• Only one zero 00000000
• Have to take the twos complement of negatives in
  order to know the value.

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Twos Complement (decimal to binary)

• Convert the decimal to binary, then to ones
  complement, then add 1.
• Ex convert (-57)10 to twos complement.
• 57 in binary is 00111001
• in ones complement 11000110
• add 1
  1
• in twos complement 11000111

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Twos Complement (binary to decimal)

• If the MSB is 1, take ones complement (flip
  every bit) then add 1.
• Ex convert (11001100)2 from twos complement to
  decimal.
• Take ones complement 00110011
• add 1
  1
• 
  00110100
• and convert to decimal 32164 52 so the
  answer is (-52)10

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Excess-M (Bias)

• 8-bit version has 256 values (-127 to 128).
• Used to store the exponent in the float rep.
• For floating point, well use Excess-127.
• Ex Convert 32 to Excess-127
• 32 127 159 or 10011111
• To convert Excess-127 back to decimal convert to
  decimal then subtract 127.
• Ex Convert 11001010 in Excess-127 to decimal
• 11001110 206
• 206 127 (79)10

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Addition in Ones complement

• First convert both decimals in ones complement
  then add. If there is a carry, add it to the
  right.
• Ex add (-15)10 and (69)10 in ones complement.
• -15 in ones complement 11110000
• 69 in ones complement 01000101
• add
  100110101
• add carry to right
  1
• answer
  00110110 or 54

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Addition in Twos complement

• First convert both decimals in twos complement,
  then add. If there is a carry, discard it.
• Ex add (-22)10 and (19)10 in twos complement
• -22 in twos complement 11101010
• 19 in twos complement 00010011
• add
  11111101
• Since MSB is 1, take twos complement to get
  the answer 00000011 or
  -3

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Float representation

• Used to represent fractional numbers
• Overall sign
• Mantissa (fraction part)
• Exponent
• Base
• In C 32-bit float (4 bytes)
• Sign Exponent Mantissa

1 bit
8 bits
23 bits
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Float to decimal

• Convert 1 00000011 11000000000000000000000 from
  float (1-bit sign, 8-bit exponent and 23-bit
  mantissa) to decimal.
• Sign - since the bit sign is 1
• Exponent 00000011 to decimal 3, then
  subtract 127 to convert from Excess-127 to true
  value 3-127 -124.
• Mantissa 11 followed by zeroes.
• We supply the leading 1 (always).
• So the number is -(1.11)2 2-124 -1.75
  2-124

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Decimal to float

• Convert (3.6875)10 to float (1-bit sign, 8-bit
  exponent and 23-bit mantissa).
• Sign 0 (since the number is positive)
• To figure out the mantissa and the exponent, we
  must first convert the decimal to binary.
• 3 to binary 11
• .6875 to binary 0.6875 2 1.375
• 0.375 2
  0.75
• 0.75 2
  1.5
• 0.5 2
  1.0
• So 3.6875 in binary is 11.1011
  (continued next slide)

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Decimal to float (2)

• Floats must always be in the format 1.x so we
  must move the period to have 1.x in our binary
  number
• 11.1011 (we have to shift the period one
  place to the left).
• So 11.1011 1.11011 21
• We now have the mantissa 11011(followed by
  zeroes)
• And the exponent is 1, converted to
  Excess-127 128. In binary 128 10000000.
• Answer 0 10000000 11011000000000000000000
• Convert back to decimal to verify the answer.

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Part II of the exercises then correction