Yes but why? You can accurately represent every 32-bit int with a float,
but Pd clips them by using the %g format specifier to print them,
instead of %f.
In m_atom.c, at line 68, the function atom_string() converts atoms into
strings, and in the case of float atoms, uses this line to do it:
sprintf(tbuf, "%g", a->a_w.w_float);
This prints both of the floats 16777215.0 and 16777214.0 as "167772e+7".
It seems to me that
sprintf(tbuf, "%f", a->a_w.w_float);
would be better, since it prints 16777215.000000, 16777214.000000.
Martin
Roman Haefeli wrote:
I assume, this is because a 24bit integer cannot be saved with full
precision with Pd, since Pd seems to strip off some bits when saving a
floating point value (or when printing or displaying it). If the color
would be encoded as RGB 8bpp, it would look different after saving and
restoring it. So a smaller range had to be used.
When sending 'color' messages to the iemguis directly, the full 24bit
resolution can be used.
Roman
On Sat, 2010-02-13 at 18:32 -0500, Martin Peach wrote:
Ah yes, in g_all_guis.c line 281:
void iemgui_all_colfromload(t_iemgui *iemgui, int *bflcol)
{
if(bflcol[0] < 0)
{
bflcol[0] = -1 - bflcol[0];
iemgui->x_bcol = ((bflcol[0] & 0x3f000) << 6)|((bflcol[0] &
0xfc0) << 4)|
((bflcol[0] & 0x3f) << 2);
}
else
{
bflcol[0] = iemgui_modulo_color(bflcol[0]);
iemgui->x_bcol = iemgui_color_hex[bflcol[0]];
}
...so if the colour is negative it's a negated (all bits flipped) 18-bit
rgb value and if it's positive it's an indexed colour from the iemgui
palette.
111111RRRRRRGGGGGGBBBBBB
is bit-flipped to get:
000000rrrrrrggggggbbbbbb
which is shifted into this:
rrrrrr00gggggg00bbbbbb00
so the 2 LSBs of each colour are set to 0. I don't know why.
Martin
Robert Schwarz wrote:
Thanks for the quick answer.
The concept of embedding three 8 bit components in one integer was clear
to me, but I think that pd doesn't really use all 8 bits for the colors.
Or maybe there is some issue with 2-complements or something.
For example, if I want to create three bang objects, in red (#ff0000),
green (#00ff00) and blue (#0000ff), your formula gives values of:
16711680, 65280, 255 for the three colors.
But I insert them in a patch, like:
#N canvas 825 10 450 300 10;
#X obj 0 0 bng 15 200 50 0 target empty empty 0 0 0 8 16711680 0 0 ;
#X obj 0 15 bng 15 200 50 0 target empty empty 0 0 0 8 65280 0 0 ;
#X obj 0 30 bng 15 200 50 0 target empty empty 0 0 0 8 255 0 0 ;
I see the colors white, white, yellow.
Now, when I change the colors by hand, to really get red, blue and green
on the bang objects and save the file, it reads:
#N canvas 825 10 450 300 10;
#X obj 0 0 bng 15 200 50 0 target empty empty 0 0 0 8 -258049 -1 -1 ;
#X obj 0 15 bng 15 200 50 0 target empty empty 0 0 0 8 -4033 -1 -1 ;
#X obj 0 30 bng 15 200 50 0 target empty empty 0 0 0 8 -64 -1 -1 ;
So it uses negative numbers, and -64 means "full blue".
Now, when I re-open the same file and look at the properties of the blue
bang object, the color now reads: #0000fc instead of the #0000ff I
entered just before saving.
That's why I suspect some lower resolution going on. I tried to browse
this part in the sources, but all the GUI code confuses me.
For your interest, this patch results for colors of #040000, #000400 and
#000004 set by hand in the properties window:
#X obj 0 0 bng 15 200 50 0 target empty empty 0 0 0 8 -4097 -1 -1;
#X obj 0 15 bng 15 200 50 0 target empty empty 0 0 0 8 -65 -1 -1;
#X obj 0 30 bng 15 200 50 0 target empty empty 0 0 0 8 -2 -1 -1;
Setting colors to lower values, like #010000 results in getting them
rounded down to #000000.
So, the resolution is apparently 256/4 = 64 values, or 6 bits.
Indeed, if I replace the formula with:
color = (-([red]+1)/4*64*64) - (([green]+1)/4*64) - ([blue]+1)/4
I get the same values that Pure Data produces.
Hm, I might just have solved my problem.
It's still weird and some developer could check this our or change the
documentation.
Cheers, Robert
On 02/13/2010 11:08 PM, Martin Peach wrote:
That formula should read:
color = ([red] * 65536) + ([green] * 256) + ([blue])
In binary the idea is to shift the 8 'red' bits 16 to the left, then add
8 'green' bits shifted 8 bits, and finally 8 'blue' bits, so in all 24
bits are occupied.
Multiplying the blue value by -1 in the original formula has the effect
of setting the 16 bits to the left of it to 1, so you get different
shades of pure blue.
Martin
Robert Schwarz wrote:
Hi all,
I recently tried writing patches in a text editor (or from scripts) and
had problems getting the color settings right, for bang elements.
There is some documentation at
http://puredata.info/docs/developer/fileformat
with the explanation:
Color: Some graphical elements have color attributes. Per color only
one signed integer value is stored that contains the three 8-bit
color components (RGB). Formula to calculate color attribute values:
color = ( [red] * -65536) + ( [green] * -256) + ( [blue] * -1)
Where [red], [green], [blue] obviously represent the three color
components, their values range from 0 to 255. They apply to the
attributes [background color], [front color], [label color] of
various elements.
I tried that, but it didn't work. Instead of showing the whole spectrum
I just got different shades of blue. Also, when I opened one of my
handwritten patches in PureData, looked at the color settings and saved,
the resulting numbers changed. I assume that some kind of rounding is
happening, and colors are actually saved in lower resolution.
Do you have any ideas?
Also, my application is a 13x13 button matrix, each triggering different
chords via MIDI. The buttons should be color coded. Obviously, it's too
much work setting all colors individually and I might want to create
several of these patches with different colors.
Maybe there is another obvious solution I didn't see.
Any help is appreciated!
(I'm using standard pd 0.42_5 on Arch Linux, but this shouldn't make a
difference.)
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