Int32 divide probleme

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I try to divide 2 integers and get the remainder using the Int32 library from Parallax site.

Below my code:

import stamp.math.Int32;
public class DivisionTest {
· static Int32 num1 = new Int32();
· static Int32 num2 = new Int32();
· public static void main() {
······ num1.set(14512);
······ num2.set(5);
······ System.out.println("Before divide:");
······ System.out.println("num1: " + num1.toString());
······ System.out.println("num2: " + num2.toString());
······ num1.divide(num2);
······ num1.remainder(num2);
······ System.out.println("After divide:");
······ System.out.println("num1: " + num1.toString());
······ System.out.println("num2: " + num2.toString());
· }
}

Here's the output:

Before divide:
num1: 14512
num2: 5
After divide:
num1: 2902
num2: 2

The remainder·never seem to be exact. Is it a error in the library or there something that I don't catch?

Thanks in advance!!

Peter Verkaik

The results look ok to me
14512/5 = 2902.4
int(14512/5) = 2902
remain(14512/5) = 14512 - (2902*5) = 14512 - 14510 = 2

Edit:
num1.divide(5) = 14512/5 = 2902 (is now num1)
num1.remainder(5) = 2902 % 5 = 2

In this case both remainders are 2.
Is the calculation the one you wanted
or did you actually wanted to calculate 14512 % 5?

regards peter



Post Edited (Peter Verkaik) : 11/6/2006 8:33:33 PM GMT

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I thougth that the remainder function return the fractionnal part of the division like the .4 of 14512/5 = 2902.4

I'm french so I think I just misunderstand the words;

So the remainder function actualy return the modulo. Is it right?

Thank you very much Peter

Peter Verkaik

Yes, the remainder is the modulo.
if z = x % y
then·fraction = z/y
You can calculate the fraction (for ints)
using UnsignedIntMath.frac(z,y) or UnsignedIntMath.ufrac(z,y)
You could rewrite those functions for Int32
(and also for·the other functions in UnsignedIntMath class)

regards peter

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Cool that perfect! You seem to be a lot more proficient en math then me. Luck you!!

Thank you very much for your help!