Quotient and Modulo Mathematical Operations Request

raykholo

Hi David,

Would it be possible to have the mathematical operations of "Quotient" (divide without remainder) and Modulo (just the remainder)? I did not see any reference to these in the Meta Commands documentation.

I want to build a filament changer (MMU2 style to feed different filament types/ colors) on top of a tool changer (grab different nozzle sizes) so that any combination of filament/ color and nozzle size is possible.

In an example of 3 tools (say a 0.25, 0.4, and 0.8 volcano nozzle sizes) and 5 filament colors, we have 15 combinations to resolve. I would use the existing T# functionality to handle tool changing, and would write a similar construct in Macros to handle the filament color changer.

I would use a Macro to resolve the 15 possible combinations into the existing T# commands to select tool, and Macros would handle the filament switching commands.

As you can see in the spreadsheet image here, we would need a Quotient and Modulo operation to do this. Quotient is Tool #, Modulo is Filament Color #.

Thanks!

dc42

This could be provided, but I think it would be simpler to provide a round-to-integer operator. Then you could get the quotient by using round(a/b) and the remainder from a-b*round(a/b).

A Former User

Maybe i missunderstood that, but

a  b div mod  round(a/b) a-b*round(a/b)
1  5   0   1           0              1
0  5   0   0           0              0
2  5   0   2           0              2
3  5   0   3           1             -2
4  5   0   4           1             -1
5  5   1   0           1              0
6  5   1   1           1              1
7  5   1   2           1              2
8  5   1   3           2             -2
9  5   1   4           2             -1
10 5   2   0           2              0
11 5   2   1           2              1
12 5   2   2           2              2
13 5   2   3           3             -2
14 5   2   4           3             -1

division without remainder would be equivalent of floor(a/b) discarding, round(a/b) would round up?
(public, don't mess it up too badly https://docs.google.com/spreadsheets/d/1jaGuxY3Faa-QBwInEJwWeWvN_0XcVKpzSUmcAWgXuP8/edit#gid=0)

dc42

I was assuming that round would behave like floor. In fact I should call it floor, not round.

A Former User

to me mod() would be much better readability than a-b*floor(a/b), but I guess not everyone will be familiar with div() and mod() either. floor gets the job done at least

dc42

I could introduce the % operator to mean mod (often the same as remainder), but integer divide would need another symbol. I deliberately made the / operator mean floating point division, so that users needn't worry about whether the operands are floats or ints.

raykholo

I agree that having to use a-b*floor(a/b) for remainder hurts readability, especially when inside complex expressions. Please please please have a dedicated mod operation. Either syntax of mod(a,b) or a%b is completely fine.

For the quotient, rounding is surely another option and floor(a,b) would be fine for our purposes.

Also, please note that I used syntax of function (a,b) and not function (a/b) above. I have usually seen it as comma separated parameters and I am concerned that using / would confuse the system into thinking it is division inside the function.

Thank you very much for the consideration.

dc42

I have implemented floor(x) and mod(x, y) in the latest internal builds of RRF3 at https://www.dropbox.com/sh/3azy1njy3ayjsbp/AACquxr2m00eV568RZg5QG5wa?dl=0.