// if we are in the main screen area, or one of the extra MODE 1 contended areas:
// CPU can only access memory 1 out of every 8 T-States
// else
// CPU can only access memory 1 out of every 4 T-States
Let's assume we are executing while in ray is in screen mode 4 ("paper" area).
For example LD (HL),N is 2 bytes long and writes 1 byte to memory. So it has 3 memory accesses in total, what gives the total execution time of 24T when screen is drawn.
Note that classic Z80 specs. says 10T for this instruction. So it's a quite big difference.
Also, Edwin Blink wrote that for sprite drawing I should use the sequence of
LD (HL),nn
INC L
which is 16 T per byte - but how can it be 16T in total when each memory access takes 8T?
Or do I understand it wrongly? Please tell me... :-)
/---
Aley
