Roman Thilenius

i understand and accept that and why expr %($f1) cannot work

maybe its just too simple for me, but i dont find a working

i understand and accept that and why expr %($f1) cannot work
but please how else could i write it?

maybe its just too simple for me, but i dont find a working
expression on my own.

expr-f1

Since fmod() is the standard C math function for floating-point modulo, one might expect [expr fmod($f1, $f2)] to work.

One could file a feature request for expr to support fmod() (or, better still, a % float-operator).

In the meantime, there's Ch's suggestion.

And, of course, the % object, which started supporting floats in an early Max 4 release.

thanks Ch, well hopefully you dont think i was just to lazy to

do it myself but yesterday i wasnt able to make it myself.

i have never made a float % abstraction myself (back in OS9 

when % did not support floats) because i simply never needed 

it to do floats, that is why i did not knew it offhand.

but in this case i need stuff to work inside expr as this 

actually it is suprising that [expr] does not support "fmod" 

or "%", which is probably a less excotic function than natural 

well, there will be a reason for it, we discussed the nature 

thanks Ch, well hopefully you dont think i was just to lazy to
do it myself but yesterday i wasnt able to make it myself.

i have never made a float % abstraction myself (back in OS9 
when % did not support floats) because i simply never needed 
it to do floats, that is why i did not knew it offhand.

but in this case i need stuff to work inside expr as this 
saves patchcords and improves timing.

actually it is suprising that [expr] does not support "fmod" 
or "%", which is probably a less excotic function than natural 
log (which noone ever uses) isnt it?

well, there will be a reason for it, we discussed the nature 
of calling expr´s function earlier.

Roman Thilenius wrote on Mon, 27 July 2009 21:48

which is probably a less excotic function than natural 

Claiming "no one ever uses" any given function is just begging, pleading to be contradicted, isn't it?

Frankly, I use natural log as often as any of the trig/log functions.

But more to the point is the rather idiosyncratic choices made by expr of which functions from the C math library to support and which to ignore. fmod() would definitely be useful.

Ideally, expr should support all the functions in ISO/IEC 9899:1999 that simply return a value. (ie, functions like frexp(double, int *) with a secondary return value as a pointer argument won't fit into the expr paradigm).

Does Cycling 74 need an 'official' bug report to do anything about this?

Roman Thilenius wrote on Mon, 27 July 2009 21:48which is probably a less excotic function than natural 
log (which noone ever uses) isnt it?

Frankly, I use natural log as often as any of the trig/log functions. 

Does Cycling 74 need an 'official' bug report to do anything about this?


thats already the second new thing i learned about expr/if

this week. the whole thing seems to be more inconsistent than

thats already the second new thing i learned about expr/if
this week. the whole thing seems to be more inconsistent than
in your most wet dream.

[expr $f1 % $f2] outputs the same result as [expr $i1 % $i2],

the only difference is that if $f2 is between 0. and 1., then max crashes (the divide by 0 isn't detected).

by the way, a nice thing about expr that is stated in the reference :

"Additional functions can be added by means of external code resources placed in Max's startup folder."

[expr $f1 % $f2] outputs the same result as [expr $i1 % $i2],
which is not what i expect.
the only difference is that if $f2 is between 0. and 1., then max crashes (the divide by 0 isn't detected).

by the way, a nice thing about expr that is stated in the reference :
"Additional functions can be added by means of external code resources placed in Max's startup folder."
Have you ever tried that?

but the % sign alone was a surprise for me - i mainly asked

for a formula because i thought % would not work at all.

somehow i tend to think that everything which is fully 

supported by the object is noted in the helpfile.

(like i said in the other thread the only additional "undocumented" 

feature i knew about was comparison operators.)

and i assume i would need to write c classes for that ...

but i dont have time to discuss that now, as i am going to 

implement a coffeemachine in my startup folder asap.

Ch wrote on Fri, 31 July 2009 08:49
What do you mean Roman?

but the % sign alone was a surprise for me - i mainly asked
for a formula because i thought % would not work at all.

somehow i tend to think that everything which is fully 
supported by the object is noted in the helpfile.

(like i said in the other thread the only additional "undocumented" 
feature i knew about was comparison operators.)

Quote:
"Additional functions can be added by means of external code resources placed in Max's startup folder."
Have you ever tried that?

and i assume i would need to write c classes for that ... 

but i dont have time to discuss that now, as i am going to 
implement a coffeemachine in my startup folder asap.

My apologies (as I fear I've asked about this before) but I'm currently trying to track down the solution for a true looping euclidean modulo for floats in [expr]:

-10. -9. -8. -7. -6. -5. -4. -3. -2. -1. 0. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. = 

0. 1. 2. 3. 4. 0. 1. 2. 3. 4. 0. 1. 2. 3. 4. 0. 1. 2. 3. 4. 0. 1. 2. 3. 4.

My apologies (as I fear I've asked about this before) but I'm currently trying to track down the solution for a true looping euclidean modulo for floats in [expr]:

ie. % 5. =
-10. -9. -8. -7. -6. -5. -4. -3. -2. -1. 0. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. =   
0. 1. 2. 3. 4. 0. 1. 2. 3. 4. 0. 1. 2. 3. 4. 0. 1. 2. 3. 4. 0. 1. 2. 3. 4. 

you can not loop at all within the object. but you can do it around it.

So say if I needed to use [vexpr @scalarmode 1], to act on lists, I'd still need to [iter]?

not on lists, but for your euler thing you need the result of the first round for the second calculation, isnt it?

there are many options to perform a "for loop", including counter & gate or using poly~ in serial mode.

really funny to see my own nonsese question above after all the years... and now i am sitting on the other side of the desk. :P


not on lists, but for your euler thing you need the result of the first round for the second calculation, isnt it?

there are many options to perform a "for loop", including counter & gate or using poly~ in serial mode.

really funny to see my own nonsese question above after all the years... and now i am sitting on the other side of the desk. :P


Yeah, right now I'm convinced this is the simplest option:

If anyone knows of a more efficient method, I'd love to see it. :D

Here is another version, a little convoluted because of the kind of float modulo:

@Jean -- ...This is incredible. I believe this is exactly what I was looking for. Thank you. So. Damn. Much.

Ok. Final question. Is there a way of recreating the [round] function --

@Jean -- ...This is incredible. I believe this is exactly what I was looking for. Thank you. So. Damn. Much.

Ok. Final question. Is there a way of recreating the [round] function -- ie. [round 0.5] -- within [vexpr]?

Thank you so much, again. 
- Aaron

There is a round function, but as per C, it rounds to the nearest integer, but we can make it work to the nearest 0.5 too;

Ah ok, I see. Thanks for the response. :)

Is there a way to output a value out a second outlet? (Ie. output a value calculated halfway through, as well as output the overall expression out the first outlet as usual?) 

I realize you could just break into 2 separate [vexpr] objects, but yeah. :)

Ok, actual last question:

Is there a way to output a value out a second outlet? (Ie. output a value calculated halfway through, as well as output the overall expression out the first outlet as usual?) 

I realize you could just break into 2 separate [vexpr] objects, but yeah. :)

No, I don't think there is. It would also be nice is we could do this

and get a tuple of both results at the (single) output, but as you say, it can also be done with two vexpr.

Indeed, 2 separate [vexpr] are just great. Use [t l l] if you want a result before the other.

Ok, another question. (Apologies. I keep encountering unforeseen snags, lol.)

I'm guessing it wouldn't be possible to consolidate the following into a single expression object:

Ok, duly noted. :) -- Thanks, guys. 

Ok, another question. (Apologies. I keep encountering unforeseen snags, lol.)

I'm guessing it wouldn't be possible to consolidate the following  into a single expression object:

sometimes it can be required to delete most of a patch.

grafik.png


sometimes it can be required to delete most of a patch.


@Aaron yes, you trigger different calculations from two flonums. It's possibile to do all the maths in one (v)expr, but you have to store value from "B" outside of it.