Using the % modulo in math in script
by J Sears · in Torque Game Builder · 04/18/2007 (11:18 am) · 5 replies
Basically I have this code set up
%result = $a / $b;
%leftOver = $a % $b;
%resultTwo = %result - %leftOver;
now I have echos after each of these and to help explain what I'm doing (it's part of my work in progress AI for playing cards) %a is the round, each round more cards get dealt out, %b is the number of players so it stays constant for a whole game is just defined at the start of a new game. To help in my calculating of how much the AI should bid on his hand I want the average amount per player.
So when a and b are both 3 the result is 1, leftover is 0 and resultTwo is 1. That part makes sense to me and is how I expected it to work
when a is 4 and b is 3, result is 1.3333333 which is fine but leftOver is 1 making resultTwo .333333
I though % was supposed to result in the remainder so leftOver would be .33333 so that resultTwo would be 1.
Now I believe (my c/c++ coding is limited) that in C/C++ you can force variables to be INT, I'm not sure if that works in script or not but going to give it a shot. But even if I get it to work trying forcing of INT I'm still confused as to why leftOver is 1.
%result = $a / $b;
%leftOver = $a % $b;
%resultTwo = %result - %leftOver;
now I have echos after each of these and to help explain what I'm doing (it's part of my work in progress AI for playing cards) %a is the round, each round more cards get dealt out, %b is the number of players so it stays constant for a whole game is just defined at the start of a new game. To help in my calculating of how much the AI should bid on his hand I want the average amount per player.
So when a and b are both 3 the result is 1, leftover is 0 and resultTwo is 1. That part makes sense to me and is how I expected it to work
when a is 4 and b is 3, result is 1.3333333 which is fine but leftOver is 1 making resultTwo .333333
I though % was supposed to result in the remainder so leftOver would be .33333 so that resultTwo would be 1.
Now I believe (my c/c++ coding is limited) that in C/C++ you can force variables to be INT, I'm not sure if that works in script or not but going to give it a shot. But even if I get it to work trying forcing of INT I'm still confused as to why leftOver is 1.
#2
then is there a command for getting the decimal part, I am trying to end up with whole numbers.
04/18/2007 (12:32 pm)
I was certainly thinking about % the wrong way then. Since the remainder ends up as the decimal portion of the number I figured it just grabbed the decimal.then is there a command for getting the decimal part, I am trying to end up with whole numbers.
#3
%result = $a / $b;
%leftOver = $a % $b;
%decimal = %leftover / $b
%resultTwo = %result - %decimal
and that does what I want in a round about way.
04/18/2007 (12:41 pm)
Well I just did it this way, thanks for the correction on how % worked gary%result = $a / $b;
%leftOver = $a % $b;
%decimal = %leftover / $b
%resultTwo = %result - %decimal
and that does what I want in a round about way.
#4
If you want to do it at hand, I guess the easiest way is:
%leftOver = %a % %b;
%result = (%a - %leftOver) / %b;
04/19/2007 (12:06 pm)
If you want to do something like 4/3 = 1 instead of 1.333 you can do mFloor(4/3), which gives you the highest integer thats below or equal the number you passed as a parameter.If you want to do it at hand, I guess the easiest way is:
%leftOver = %a % %b;
%result = (%a - %leftOver) / %b;
Torque Owner Gary Preston