CS 232: Computer Architecture II

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CS 232 Computer Architecture II

• Prof. Laxmikant (Sanjay) Kale
• Floating point arithmetic

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Floating Point (a brief look)

• We need a way to represent
• numbers with fractions, e.g., 3.1416
• very small numbers, e.g., .000000001
• very large numbers, e.g., 3.15576 ? 109
• Representation
• sign, exponent, significand (1)sign
  ???significand ???2exponent
• more bits for significand gives more accuracy
• more bits for exponent increases range
• IEEE 754 floating point standard
• single precision 8 bit exponent, 23 bit
  significand
• double precision 11 bit exponent, 52 bit
  significand

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Floating point representation

• The idea is to normalize all numbers, so the
  significand has exactly one digit to the left of
  the decimal point.
• 12345 1.2345 104
• .0000012345 1.2345 10-6
• Do this in binary 1.01110 x 2(1011)
• IEEE FP representation
• (/-) 1.0101010101010101010101 2 ( 10101010)
• This is single precision
• Double precision 64 bits in all.
• Where does one need accuracy of that level?

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Floating point numbers

• Representation issues
• sign bit, exponent, significand
• Question how to represent each field
• Question which order to lay them out in a word?
• Factor should be easy to do comparisons (for
  sorting)
• For arithmetic, we will have special hardware
  anyway
• Choice
• Sign magnitude representation
• Sign bit, followed by exponent, then significand
  (why?)
• exponent represented with a bias add 127
  (1023 for double precision)
• significand assume implicit 1. (so 00001 means
  1.00001)

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Floating point representation

• So
• (/-) x (1 significand) x 2 (exponent - bias)
  is the value of a floating point number
• Example 0 00001000 01010000000000000000000
• Example convert -.41 to single precision form

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IEEE 754 floating-point standard

• Leading 1 bit of significand is implicit
• Exponent is biased to make sorting easier
• all 0s is smallest exponent all 1s is largest
• bias of 127 for single precision and 1023 for
  double precision
• summary (1)sign ?????significand)
  ???2exponent bias
• Example
• decimal -.75 -3/4 -3/22
• binary -.11 -1.1 x 2-1
• floating point exponent 126 01111110
• IEEE single precision 10111111010000000000000000
  000000

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Floating point addition

• The problem is the exponents of numbers being
  added may be different
• 2.0 101 3.0 10(-1)
• 2.0 101 .03 10 1 Now we can add them
• 2.03 10 1
• But we are not necessarily done!
• E.g. 9.74 100 3.3 10(-1)
• 10.07 100 is not correct form!
• Shift again to get the correct form 1.037 101

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You can get different results

• A B C A (BC) (AB) C
• Right?
• Can you see a problem?
• When do you lose bits?

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Floating point multiplication

• Add exponents, but subtract bias
• Then multiply significands
• Then normalize

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Guard and Round bits

• Not all numbers are representable
• To minimize rounding error
• As the machine performs steps of the addition or
  multiplication algorithm, intermediate results
  are stored with 3 additional bits
• Consider first just 2 of them called round and
  guard
• Adding those 2 bits allows us to round correctly
• 1 enough for addition, but 2 needed for
  multiplication
• What to do when the result 2.50? Sticky bit. And
  rounding modes.

3.45 100 2.11 10-2 Assume 3 digits can be
used. 3.45 0.0283 ----------- 3.4783
---gt 3.48