1. CS 232: Computer Architecture II
Prof. Laxmikant (Sanjay) Kale
Floating point arithmetic

2. Floating Point (a brief look)
We need a way to represent
numbers with fractions, e.g.,
very small numbers, e.g.,
very large numbers, e.g.,  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

3. 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 = * 10^4
= * 10^-6
Do this in binary: x 2^(1011)
IEEE FP representation
(+/-) * 2 ^ ( )
This is single precision
Double precision: 64 bits in all.
Where does one need accuracy of that level?

4. 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 means )

5. Floating point representation
So:
(+/-) x (1 + significand) x 2 ^ (exponent - bias) is the value of a floating point number
Example:
Example: convert -.41 to single precision form

6. 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: = -3/4 = -3/22
binary: = -1.1 x 2-1
floating point: exponent = 126 =
IEEE single precision:

7. Floating point addition
The problem is: the exponents of numbers being added may be different
2.0 * 10^ * 10^(-1)
2.0 * 10^ * 10^ 1 : Now we can add them
2.03 * 10 ^1
But we are not necessarily done!
E.g * 10^ * 10^(-1)
10.07 * 10^0 is not correct form!
Shift again to get the correct form: * 10^1

8. 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?

9. Floating point multiplication
Add exponents, but subtract bias
Then multiply significands
Then normalize

10. 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 * 10^ * 10^-2
Assume 3 digits can be used.
3.45
> 3.48