williams-et-al-1983-apple-ii-computer-graphics

72 APPLE II COMPUTER GRAPHICSSolutionIf you count eight lines per box, the fourteenth row of boxes ends withthe 112th line of dots (8* 14= 112), so the 116th line is the fourth line inthe fifteenth box. Therefore, by using the memory map in Figure 8-3,the row address is 9000, and the position in the box address is 3072.Dots 57, 58, and 59 of that line are the second, third, and fourth dots inthe eighth box, which has column address 7 ($7), so the final address is:9({l({l({lJ(J72712({l79$2328$(JC({l({l$ 7$2F2FThe display fo r that byte would look like:-XXX ---where - represents a dot which is off, and X a dot which is on.Reversing the order and supplying the "missing" left zero wouldgive you the bit pattern: 00001 110, which is equal to 14 ($0E).Now that you have determi ned the address (12079 or $2F2F) andthe value (14 or $0E), type:HGRPOKE 12({l79, 14orHGRCALL -1512F2F : EThis is not so bad after you get used to it, is it? Perhaps onemore exampleExample 3Turn on the dots 27 and 30 in line 16 of the Hi-Res screen.SolutionLine 16 is the last byte in the second row of boxes, so the row addressis 8320 ($2©8©) and the position address is 7168 ($1 C00). Dots 27 and 3©are in separate boxes-the fo urth and fifth columns-so they will havedifferent column addresses, and we will have to POKE each byte separately.