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Now we will calculatethe vertical numerical valuesof the columnsof dots, andenterthemunderneaththegrid.Eachverticalcolumnis firstdividedinto three groupsof eight dots. Each group of eight dots is representedby one byte, which consistsof eightbits.
Thisiswherethenumbersdowntheleft sideof thegrid come in. Noticethat thereis a number for each row of dots and that each number is twicethe numberbelowit. By makingthesenumberspowersof two we can take any combinationof dots in a verticalcolumnand assignthem a uniquevalue.
Assigning the Index Table data
Unlikedefiningin the Standardmode,youmustassigntheIndexTablewith the IBM mode. This Index Table is preparedfor the informationof each
Each characterrequires9 indextable data.
Thefirst andthe secondbytes(ml andTn2)indicatesthepositionof the first dotpatterninthe memory.ml is thehighorderbyte,andrn2is theloworder byte.
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.
Thisbyteindicatesthe characterwidthto be printed,andinformationof the repetitiondots for blockgraphics“characters.
If the characterisa normalcharacter,add 192to the characterwidth.
If the characterisa 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 telephonesymbolisanormalpica character,so the characterwidthis 35, and this byte shouldbe 227.
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