How to get a nice round on a real?

if this is scientific data then the usage of double is correct and the
result is also "correct"

if this is money data then the usage of double is wrong and you should be
using BigDecimal instead and you will get 1.4 then

this is the nature of floating point

(as a 99% sure workaround you can print with a 3 decimals)

Hello!

Thanks for your reply! It really helped me out!

package mathtester;

import java.math.*;

public class MathTester
{
    public MathTester()
    {
        double d = 1.4;
        double multiplier1 = 0.000001;
        double multiplier2 = 0.0001;
        d = d * multiplier1;
        d = d / multiplier2;
        System.out.println(d);

        int decimalPlace = 2;

        BigDecimal bigDecimal = new BigDecimal(d);
        bigDecimal = bigDecimal.setScale(decimalPlace ,BigDecimal.ROUND_HALF_UP);
        d = bigDecimal.doubleValue();
        System.out.println(d);


    }

    public static void main(String[] args)
    {
        new MathTester();
    }
}

Please give a svar so I can reward yiu mate!

Best regards
Fredrik

I was actually thinking about:

        BigDecimal xd = new BigDecimal("1.4");
        BigDecimal xmultiplier1 = new BigDecimal("0.000001");
        BigDecimal xmultiplier2 = new BigDecimal("0.0001");
        xd = xd.multiply(xmultiplier1);
        xd = xd.divide(xmultiplier2);
        System.out.println(xd);

It is worth noting that DECIMAL/NUMERIC is part of all databases and that C# actually
has decimal as a simple data type

svar

Hello!

I'm just able to use 1.4 so I guess I need to use:

        BigDecimal bigDecimal = new BigDecimal(d);
        bigDecimal = bigDecimal.setScale(decimalPlace ,BigDecimal.ROUND_HALF_UP);
        d = bigDecimal.doubleValue();
Best regards
Fredrik

21:26:03 should work in almost any Java version

Hello!

Perhaps I could ask you.

Would we have the same problem if we used float instead of double?

We have tried with som long float numbers but all those calculations seems to work fine.

Perhaps we just were lucky?

Best regards
Fredrik

float and double share the same characteristics

they just have different precision

you can see the same effect for floats, but probably for other numbers

When i hit this issue i usually get around it in a "noobish" kind of way, but it still works.
Ex:
If i get the double 14.50006 and only want 2 digits ill just multiply with 100 and parse to integer and then back to double:

double d = 14.50006*100;
int i = Integer.parseInt(d);
d = i/100;

Just wanted to mention this little shortcut

I assume that you mean:

        double d = 14.50006*100;
        int i = (int)d;
        d = i/100.0;

because what posted does really not work.

That is an OK solution if you want to round well within the precision of floating
point.

It does not really solve the precision problem.

And as a side note:

x = ((int)(x * 100))/100.0;

always round down - there is a relative good probability that

x = ((int)(x * 100 + 0.5))/100.0;

which rounds to nearest is better