DecimalFormat the answer?

Question

Please forgive my ignorance, I'm a beginner.

Can anyone tell me why I am getting my error code in the if/else statement below? I think my result for the ln multiplication is is not exactly the same as the standard multiplication (there is a .00000000006 of a difference or something like that). Is there a work around for this? I have tried to use DecimalFormat but to no avail.

If I add:

DecimalFormat fmt = new DecimalFormat ("0.###");

to the tester and

    if (fmt.format(z) != fmt.format(result)){

to the if statement I receive my own same error statement. What is wrong!?

Thank you very much.

Joe

import java.util.Scanner; 
import java.lang.Math;
import java.text.DecimalFormat;


public class Logs {

  //integer x field & encapsulation
  private static double x;

  // x encapsulation
  public double getX(){
    return x;
  }
  public void setX (double ex){
    if (ex >= 0){
      x = ex;
    }
    else{
    System.out.println("You may not choose a negative number, 'x' = 0");
    }
  }

  //integer y field 
  private static double y;
  // y encapsulation
  public double getY(){
    return y;
  }

  public void setY (double why){
    if (why >= 0){
      y = why;
    }
    else{
    System.out.println("You may not choose a negative number, 'y' = 0");
    }
  }

  //tester
  public static void main (String [] args){
    Logs sample =new Logs();
    Scanner var = new Scanner(System.in);
    DecimalFormat fmt = new DecimalFormat ("0.###");
    sample.setX (var.nextDouble());
    sample.setY (var.nextDouble());

    x = sample.getX();
    y = sample.getY();
    double z = (x*y);
    double logX = Math.log(y);
    double logY = Math.log(x); 
    double logZ = (logX +logY); 
    double result = (Math.exp(logZ));



    if (z != result){
      System.out.printf("%s", "Incorrect answer: be a better coder");
    }
    else {
      System.out.printf("%s %.3d %s %.3d %s %.3d", 
                        "The product of the values you chose is: ", + z,
                        "\nln(x) + ln(y) is: ", + logZ,
                        "\nCompute to e: ", + result);

    }

  }
}

see http://stackoverflow.com/questions/14749647/why-the-result-of-0-3-3-is-0-89999999999999-in-scala-language for example :) — twillouer, Feb 07 '13 at 17:28
You're comparing strings with `==`. See http://tinyurl.com/so-java-string-equality — Jon Skeet, Feb 07 '13 at 17:29

score 1 · Accepted Answer · answered Feb 07 '13 at 17:26

1

You're comparing string references rather than their values, and you want String.equals() e.g. z.equals(result)

However, I think what you're trying to do is compare two decimal numbers to a certain precision. It's more intuitive to calculate the difference and determine if that's within an acceptable error bound e.g.

if (Math.abs(z - result) < 0.01) {
   // ok
}

See Math.abs(double a) for more details

answered Feb 07 '13 at 17:26

Brian Agnew

268,207
37
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I tried this and received an error: [line: 54] double cannot be dereferenced – Joe Austin Feb 07 '13 at 17:57

score 0 · Answer 2 · answered Feb 07 '13 at 17:26

0

I'd suggest to try

if (!fmt.format(z).equals(fmt.format(result))) {

answered Feb 07 '13 at 17:26

marty

4,005
22
19

score 0 · Answer 3 · edited May 23 '17 at 11:48

0

I think it's because of double. For best practice use BigDecimal instead. Please, take a look at "Why do I see a double variable initialized to some value like 21.4 as 21.399999618530273?".

edited May 23 '17 at 11:48

Community

1
1

answered Feb 07 '13 at 17:26

Igor K

470
6
10

Do you mean BigDecimal instead of BigInteger? – Aaron Kurtzhals Feb 07 '13 at 17:38
Sure, BigDecimal. Thanks for correction. – Igor K Feb 07 '13 at 17:41

score 0 · Answer 4 · answered Feb 07 '13 at 17:32

0

Use BigDecimal instead of double
Use StrictMath instead of Math

answered Feb 07 '13 at 17:32

Aaron Kurtzhals

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3
17
21

DecimalFormat the answer?

4 Answers4