Transcript CS 232: Computer Architecture II

CS 232:
Computer Architecture II
Prof. Laxmikant (Sanjay) Kale
Floating point arithmetic
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
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 * 10^4
– .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?
Floating point numbers
• Representation issues:
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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)
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
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:
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decimal: -.75 = -3/4 = -3/22
binary: -.11 = -1.1 x 2-1
floating point: exponent = 126 = 01111110
IEEE single precision:
10111111010000000000000000000000
Floating point addition
• The problem is: the exponents of numbers being added
may be different
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2.0 * 10^1 + 3.0 * 10^(-1)
2.0 * 10^1 + .03 * 10^ 1 : Now we can add them
2.03 * 10 ^1
But we are not necessarily done!
E.g. 9.74 * 10^0 + 3.3 * 10^(-1)
10.07 * 10^0 is not correct form!
Shift again to get the correct form: 1.037 * 10^1
You can get different results
• A + B + C = A + (B+C) = (A+B) + C
– Right?
• Can you see a problem?
• When do you lose bits?
Floating point multiplication
• Add exponents, but subtract bias
• Then multiply significands
• Then normalize