CS 208 w20 lecture 6 outline

1 Poll

What is the base 10 equivalent of the 6-bit two's complement fixed-point number 0b101.110?

\(V = (-1)^s \times M \times 2^E\)

s encodes \(s\)
exp encodes \(E\) in biased form
- normally, \(E\) = exp \(- Bias\) where the \(k\) bits of exp are treated as an unsigned integer and \(Bias = 2^{k-1} - 1\)
frac is the binary fraction \(0.f_{n-1}\cdots f_1f_0\), and the significand is \(M = 1 + f\)

Take the same six bits from before, 0b1 011 10, what value do they represent under this scheme?
- s is 1, so value is negative
- exp is 3, so \(E = 3 - 3 = 0\)
- frac is 0.5, so \(M = 1 + 0.5 = 1.5\)
- V = -1.5 × 2⁰ = -1.5

we want to represent very small and very large numbers, so we need the exponent to be signed
- this suggests encoding exp as a two's complement integer
we want floating-point operations to be fast in hardware
- easier to compare floats if more 1s in exp means bigger number
clever trick: store exp as unsigned with implicit bias
- in fact, by putting exp in between s and frac, the same hardware can do two's complement comparisons and floating-point comparisons

When exp is 0, the representation switches from normalized to denormalized form
\(E = 1 - Bias\)
\(M = f\)

IEEE standard specifies four rounding modes
- typically round-to-even, helps avoid statistical bias in practice by distributing rounding between rounding up and rounding down
In general, perform exact computation and then round to something representable with available bits
- can underflow if closest representable value is 0
- can overflow if \(E\) is too big to fit in exp (result is \(\pm\infty\))
- rounding breaks associtivity