Thur, 3rd Sept 2015, 01:33 a few loose ends

My students,

A couple of loose ends I want to give you with respect to Chapter 2.

1. The floating point data type.

Recall, 32 bits, broken down as:

Bit 31 -- sign bit (0 is positive, 1 is negative)
Bits 30:23 -- exponent bits, using an excess 127 code
Bits 22:0 -- the significand bits.

(Notation: "Bits 30:23" means Bits 30,29,28,27,26,25,24,23)

So, for example, if we had the number 21 5/8: 10101.101

We normalize: 1.0101101 x 2^4

The exponent is 4, so in Bits 30:23, we store 4 + EXCESS = 4 + 127 = 131 = 10000011.

The fraction we store the significant bits BUT we do not store the 1 to the left
of the binary point because we know it is a 1 and so there is no point wasting a bit
to store it.  So, in bits 22:0, we store 01011010000000000000000.

I FORGOT to tell you not to store the most significant bit, since it is redundant.
That is, it has to be a 1, so why waste a bit of storage?

2. Some simple terminology for making a bit 0 or making a bit 1.

When we make a bit 0, we say we CLEAR the bit.  Recall device 2 was free, indicated
by bit 2 = 1.  We gave it work to do, and wanted to make bit 2 = 0 to indicate that
device 2 was now busy.  We did that by ANDing the bit vector with 111...11011,
forcing bit 2 to 0.  We say we CLEARED bit 0.

When we make a bit 1, we say we SET the bit.  Recall device 0 then finished its
work. We indicate that by making bit 0 a 1 (i.e., device 0 is free to accept more
work).  We did that by ORing the bit vector with the mask 000...0001.  We say we
SET bit 0.

3. Hexadecimal representations.

I also did not explain hexadecimal representations, which I will at the
beginning of class, before we move on to Chapter 3.

Enjoy the rest of the week, and the long weekend.  I will see you in class
next Wednesday.

Yale Patt