Hi all,

i need to access the vram in a 256 byte aligned fashion. So my addresses are always on 256 boundary.

Once i've selected a 256 aligned block address to access i need to write about 50 bytes on this block, randomly.

actually i have this code (well known) (supposing hl=pointing to vram)

ld a,l
out(0x99),a
ld a,h
or 64
out (0x99),a

; write the byte
out (0x98),a

My guess is: because in my routine i never change the H part of HL can i set once the full 14 bit address then only the
low 8 bits of vram ptr as shown below?

; first set the full 14 bit address

ld a,l
out (0x99),a
ld a,h
or 64
out (0x99),a

then, after the first setup do only the setting of address of the least significant byte

ld c,0x99
; set only lsb
out (c),l
; send data
out (0x98),a

does this works on the real machine? (in bluemsx does)
does this works on the v9938 and tms?

As far as I know, it will not work: the first byte written to 0x99 is put into an internal VDP buffer and only when the second byte is written it will know what to do with the first byte: it could be an address selection (bit7=0) or a register write (bit7=1).

Quote:

As far as I know, it will not work: the first byte written to 0x99 is put into an internal VDP buffer and only when the second byte is written it will know what to do with the first byte: it could be an address selection (bit7=0) or a register write (bit7=1).

so, this tech works only on sega master system VDP that is derived from original...

you cannot set the vram address with only one out on port 99h,
but, if you do not need to change h, store the bit 6 in h itself, try
something like this

ld c,0x99
set 6,h
out (c),l
out (c),h
out (0x98),a

the first time and

out (c),l
out (c),h
out (0x98),a

the other times, the cost is only one extra out (c),h

I'll test it in the real thing, but it doesn't seem to be feasible.