Thanks :)
On 16/12/2017 19:36, Peter Graf via Ql-Users wrote:
The least significant of the 6 green bits.
Am 16.12.2017 um 19:12 schrieb pjwitte via Ql-Users:
Sorry for yanking your chain again so soon but, on going the other
way, ie from mode 32 to 33, what is the best value for W? g6, 0, 1..?
The least significant of the 6 green bits.
Am 16.12.2017 um 19:12 schrieb pjwitte via Ql-Users:
> Sorry for yanking your chain again so soon but, on going the other
> way, ie from mode 32 to 33, what is the best value for W? g6, 0, 1..?
>
> Per
>
> On 16/12/2017 18:13, pjwitte via Ql-Users wrot
Sorry for yanking your chain again so soon but, on going the other
way, ie from mode 32 to 33, what is the best value for W? g6, 0, 1..?
Per
On 16/12/2017 18:13, pjwitte via Ql-Users wrote:
Aah! Perfect! Thanks Marcel, youre a star!
So in fact I interpreted the input wrong. It should have bee
Aah! Perfect! Thanks Marcel, youre a star!
So in fact I interpreted the input wrong. It should have been:
GGGggRRR rrBBBbbW <- input
and
ggWBBBbb RRRrrGGG -> output
Seems so obvious now ;)
Per
On 16/12/2017 15:30, Marcel Kilgus via Ql-Users wrote:
320 c$ = c$(4 to 5) & c$(16) & c$(11 to 15)
pjwitte via Ql-Users wrote:
> With
>
> 290 REMark GGGBBBbb RRRrrgg0
> 300 c$ = c$(3 TO 5) & c$(11 TO 15) & c$(6 TO 10) & c$(1 TO 2) & '0'
>
> I got the best results. While, with 300 REMarked out and 320 un-REMarked:
>
> 310 ggWBBBbb RRRrrGGG -> RRRrrGGG ggWBBBbb in big-endian
> 320 c$ = c$(1 TO
Thanks both for your input!
However, Wolfgang, Im having trouble with your suggestion. Perhaps Ive
interpreted wrongly?
Heres how I got on:
100 REMark Convert screens mode 33 to 32
110 REMark POC, pjw, December 16th 2017
120 :
140 fnm$ = 'ram2_dmp1024x768_scr'
150 :
160 t = DATE
170 ERT ScrCv3
pjwitte via Ql-Users wrote:
> I havent tested your suggestion yet, Wolfgang, but what I found so far
> was that gggbrgg0 appears (to my eye) to look cleaner than
> gggbrggW. Is that so wrong? ;)
It is right, because gggbrggW has the W at the wrong bit. It
must be the leas
Oops! That should be:
GGGBBBbb RRRrrgg0 = mode 32 translated
P
On 16/12/2017 11:43, pjwitte via Ql-Users wrote:
I havent tested your suggestion yet, Wolfgang, but what I found so far
was that gggbrgg0 appears (to my eye) to look cleaner than
gggbrggW. Is that so wrong? ;)
BTW,
Hi Per,
The PC switches the bytes around.
So
gggb rgg0
actually means
rgg0 gggb
In other words, you're switching the third highest byte for green on or off.
If you sure that's what you want, then that's fine.
Wolfgang
On 16/12/2017 11:43, pjwitte via Ql-Users wrote:
I have
But the same as my proposal :)
Wolf via Ql-Users wrote:
> No, not the same as %gggbrggW, as suggested in the original post.
>
> Wolfgang
>
> On 16/12/2017 10:18, Peter Graf via Ql-Users wrote:
>> Wolfgang Lenerz via Ql-Users wrote:
>>> I'd do it this way
>>>
>>> %ggWbrggg
>>
>> W
I havent tested your suggestion yet, Wolfgang, but what I found so far
was that gggbrgg0 appears (to my eye) to look cleaner than
gggbrggW. Is that so wrong? ;)
BTW, when converting the translation to assembler, I found the
following representation helpful:
GGGggRRR rrBBBbbW
No, not the same as %gggbrggW, as suggested in the original post
to which I was replying.
Wolfgang
On 16/12/2017 10:18, Peter Graf via Ql-Users wrote:
Wolfgang Lenerz via Ql-Users wrote:
I'd do it this way
%ggWbrggg
Which is the same :)
_
No, not the same as %gggbrggW, as suggested in the original post.
Wolfgang
On 16/12/2017 10:18, Peter Graf via Ql-Users wrote:
Wolfgang Lenerz via Ql-Users wrote:
I'd do it this way
%ggWbrggg
Which is the same :)
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QL-Users M
Wolfgang Lenerz via Ql-Users wrote:
> I'd do it this way
>
> %ggWbrggg
Which is the same :)
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