[PD] vbitmap

Mathieu Bouchard matju at sympatico.ca
Wed Dec 8 06:58:11 CET 2004

On Tue, 7 Dec 2004, Frank Barknecht wrote:
> Christopher Charles hat gesagt: // Christopher Charles wrote:
> > the other thing i was thinking about is the inlet behaviour: for now, it 
> > doesn't clip nor mod the value it sends to it's outlet(that's crap, for 
> > when you send a bigger value to the inlet, there is no graphical way to 
> > affect the bits which are out of range). i could either let it clip to 
> > 2^number-1, so in case a bigger number is sent, all bits would stay on. 
> > the other option (the one i'd prefer) is performing a modulo 2^number on 
> > the inlet. any other options?
> I'd vote for clipping, as that is consistent with the IEMgui objects,
> and if someone wants modulo, it's still possible to add a [mod] in
> front.

the other iemgui objects don't work with bitfields. Float may be the
_type_ in use, but the _interpretation_ of the numbers is different. With
bitfields you are expected to use ^ and & in place of + and *, for
example, according to Ring Theory ;-)

a slider clipping does min(b,max(a,x)) to keep x within [a,b].

the most used definitions of min and max for bitfields are:

x<y means x&y==x
x>y means x&y==y

(side note: but it's not guaranteed that you have either less than,
greater than, or equal. (see Partial Order Theory). then you don't
have true min/max's, instead you have their cousins inf/sup, but let's 
still write it min/max)

usual bit definitions of min/max:

min(x,y) = x&y
max(x,y) = x|y

the notion of bitfield intervals is pretty weird, so I will explain it
only if really wanted, but let's just say we use [0,(1<<n)-1] as you say,
but with the definition of min/max I have given. Then the clipping to that
interval is (x|0)&((1<<n)-1), or just x&((1<<n)-1), which behaves exactly
like x mod 1<<n.

So I think Chris is right (as in, his solution is more consistent).

Mathieu Bouchard -=- Montréal QC Canada -=- http://artengine.ca/matju

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