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I
Now we
will
calculatethe
vertical numerical
valuesof the columnsof dots,
andenterthemunderneaththegrid.Eachverticalcolumnis firstdivided
into
three groupsof eight dots. Each group of eight dots is representedby
one
byte, which consistsof eightbits.
This
is
wherethenumbersdowntheleft sideof thegrid
come in.
Noticethat
there is a
number for each
row
of dots and
that
each number is
twice
the
numberbelow
it.
By makingthesenumberspowersof
two we can
take any
combinationof dots in
a
verticalcolumnand assignthem
a
unique
value.
Assigning the Index Table data
Unlikedefining
in the
Standardmode,youmustassigntheIndexTablewith
the IBM mode. This Index Table is preparedfor the informationof each
charactersattributedata,such
as
charactertype(Normal24-dothigh,
or
30-
dot high block graphics), the dot pattern data
in the
memory, and the
compressionmask bit data.
Each characterrequires9 indextable data.
Thefirst andthe secondbytes(ml andTn2) i ndicatesthepositionof the first
dotpattern
in
the memory.ml is thehighorderbyte,andrn2is theloworder
byte.
Thethirdbyte,m-?, i ndicatesthecharacter
type
andthedotpatterndatawidth
in the memory.
If the ch~acter
isnormal,
simply enter the width of dot pattern
inthe
memory.If the characteris
a
block graphic, add 128
to the
width of dot
pattern in the memory.
Ourtelephonesymbolis
a
normalcharacterandthe widthof the dotpattern
in the memory shouldbe 25, so this
value is
25.
The fourth byte, m4, indicatesthe printingattribute.
This
byteindicatesthe characterwidth
to
be printed,andinformationof the
repetitiondots for blockgraphics"characters.
If the character
isa normal
character,add 192
to the
characterwidth.
If the character
isa
blockcharacter,and it shouldbe printed
as a
line draw
character,add 64
to
the characterwidth.If the block characteris
not a line
draw character,this byte shouldbe the same
as the
characterwidth.
Our telephonesymbol
isa normal
pica character,so the characterwidthis
35, and this byte shouldbe 227.
93
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