Java Casting/Conversion Problem

The Corinthian

I'm trying to format into a more readable fashion a float by turning it into an int. However, it does not appear to be doing anything at all. Here's the code:

public static String formatResult(float iRaw) {
  int iTemp = (int)iRaw;
  String sTemp = new String();
  sTemp = Convert.toString(iTemp);
  return sTemp;
}

I'm using some odd objects/methods and also can't use either double or long data types as it's for a 16-bit platform (PDA). Any suggestions?

No doubt I'm doing something stupid

Enygma

Take a look at the classes in the java.text.* package for formatting numbers.

If you just want to convert a float to a String try this:
[PHP]
Float f = new Float(yourFloatValue);
String floatStr = f.toString();
[/PHP]

The Corinthian

Originally posted by Enygma
Take a look at the classes in the java.text.* package for formatting numbers.

Can't, I'm limited to the classes that come with Waba, which is a cut down library.

Originally posted by Enygma
If you just want to convert a float to a String try this:

I'm already doing that alright, but I'm looking to format the number in a more readable fashion (i.e. 123.45 rather than 1.2345000e02).

Enygma

OK then, the easiest and most portable way out is to mulitply the float by 100 and cast it into an int, which removes anything after two decimal places. Now divide that Integer by 100 and cast the result into a float for printing. (Note no rounding or anything here, I'll leave that up to you, there's an earlier thread on this anyway.)
[PHP]
float f =123.345;
f = f * 100;
int intVal = (int)f;
float formattedFloat= (float)intVal / 100;
System.out.println("formattedFloat is: [" + formattedFloat + "]");
[/PHP]

Hope this helps.

Trojan

Perhaps you want to do a check on size just before you multiply by 100 there, or is that unnessary in Java?

Al.

satchmo

Well strictly speaking you should, but the largest positive float possible is 3.40282347e+38, and the smallest is 1.40239846e-45F, so unless you're working with inordinately large numbers you don't need to.

Trojan

Originally posted by Jazz
Well strictly speaking you should, but the largest positive float possible is 3.40282347e+38, and the smallest is 1.40239846e-45F, so unless you're working with inordinately large numbers you don't need to.

Always check for boundary conditions, they're always the gotcha

Al.

Evil Phil

I felt lucky in google and came up with this. The Rounding Class isn't mine:

iimport java.lang.*;
import java.math.*;

public class ****{

 public static void main(String[] args)
	{
		float i;
		i = 289.988f;
		System.out.println(formatResult(i));
	}

 public static String formatResult(float iRaw)
	{
	      	double d;
                int i;
		Float f= new Float(iRaw);
                d = f.doubleValue();
                i = 4;

		String s = new String(Rounding.toString(d, i));
	      	return s;
	  }

}

class Rounding
  {
    public static String toString (double d, int place)
    {
      if (place <= 0)
        return ""+(int)(d+((d > 0)? 0.5 : -0.5));
      String s = "";
      if (d < 0)
        {
  	s += "-";
  	d = -d;
        }
      d += 0.5*Math.pow(10,-place);
      if (d > 1)
        {
  	int i = (int)d;
  	s += i;
  	d -= i;
        }
      else
        s += "0";
      if (d > 0)
        {
  	d += 1.0;
  	String f = ""+(int)(d*Math.pow(10,place));
  	s += "."+f.substring(1);
        }
      return s;
    }
}
    

It returnes 289.9880. I hope this works with the reduced JVM.

The Corinthian

Originally posted by Enygma
OK then, the easiest and most portable way out is to mulitply the float by 100 and cast it into an int, which removes anything after two decimal places. Now divide that Integer by 100 and cast the result into a float for printing.

Unfortunately, that resulted in numbers being formatted again as the original float - so back to square one. The solution I found was a filthy dirty hack that fomatted the number once is was a string:

[PHP]
public static String formatResult(float iRaw) {
String sTemp = new String();

iRaw = iRaw * 100;
int iTemp = (int)iRaw;
sTemp = Convert.toString(iTemp);
if (iTemp > 0) {
sTemp = sTemp.substring(0, sTemp.length() - 2) + "."
+ sTemp.substring(sTemp.length() - 2, sTemp.length());
if (iTemp < 100) sTemp = "0" + sTemp;
} else {
sTemp = "0.00";
}
return sTemp;
}
[/PHP]

Not Pretty, but it appears works (debugging will come later :rolleyes: ).

Originally posted by Trojan
Always check for boundary conditions, they're always the gotcha

That's done prior to both the formatting and the core calculation in a seperate method.

Thanx all for the pointers.

The Corinthian

Originally posted by Evil Phil
I felt lucky in google and came up with this.

Thankx m8, but as from my previous post I solved it already. Anyhow, I could (just glancing at the code) see problems with it right away.

For example, it makes use of doubles, which I can't use as it's running on a 16-bit machine and they require 32-bit. Also, while I did find a custom Math class, it does not come with the default Waba libraries.

Anyhow thanx again.

Evil Phil

Oh yeah, I've just re-read your initial post (doh!).

bonkey

I havent tried running it, but your initial code actually looks solid.

I would be inclined to guess that this is serious breakage in the maths lib you are using...or in the Java implementation, as it should be technically impossible for a primitive int to contain decimal spaces....the basic storage mechanism in memory alone should prevent it.

So, allowing for that, I would say that your workaround is probably the "right" solution.

jc

dogs

Heya,

Sorry to nit-pick, but am I missing something here...

[PHP]
String sTemp = new String();
sTemp = Convert.toString(iTemp);
[/PHP]

...why do you create a new String() object and then
immediatley dereference it ? Convert.toString() is returning
a reference to a String anyway...

am i just dumb ?

The Corinthian

Originally posted by dogs
Sorry to nit-pick, but am I missing something here...

Yes you're nit-picking. Would you prefer:

[PHP]
String sTemp = Convert.toString(iTemp);
[/PHP]