Solved

precision lost

Posted on 2004-09-27
14
250 Views
Last Modified: 2012-05-05
Suppose I have two double values that I wanna compare them for equality. How can I lose the precision so that it returns true once all numbers before the dot are equal (ignoring the decimal)? I cant conver them to int becuz it reports double value cannot be dereferenced error.

0
Comment
Question by:jtcy
[X]
Welcome to Experts Exchange

Add your voice to the tech community where 5M+ people just like you are talking about what matters.

  • Help others & share knowledge
  • Earn cash & points
  • Learn & ask questions
  • 6
  • 3
  • 3
  • +2
14 Comments
 
LVL 37

Expert Comment

by:zzynx
ID: 12160123
>> I cant conver them to int becuz it reports double value cannot be dereferenced error.
???

Can you give you code snippet?
0
 
LVL 11

Expert Comment

by:cjjclifford
ID: 12160133
java.lang.Math.round()

long a = Math.round( 10.0001 );
0
 
LVL 35

Accepted Solution

by:
girionis earned 20 total points
ID: 12160145
Try to convert it to float and then cast it to int:

double d = 1.2345;
float f = new Double(d).floatValue();
System.out.println((int) f);
0
Salesforce Has Never Been Easier

Improve and reinforce salesforce training & adoption using WalkMe's digital adoption platform. Start saving on costly employee training by creating fast intuitive Walk-Thrus for Salesforce. Claim your Free Account Now

 
LVL 35

Expert Comment

by:girionis
ID: 12160159
...although you should be able to cast a double to int.
0
 
LVL 35

Expert Comment

by:TimYates
ID: 12160235
I think he wasn't using the cast...
0
 
LVL 37

Expert Comment

by:zzynx
ID: 12160289
       double d1 = 1.2345;
        double d2 = 1.5432;
        Double dOne = new Double(d1);
        Double dTwo = new Double(d2);
        int i1 = (int)dOne.doubleValue();
        int i2 = (int)dTwo.doubleValue();
        System.out.println("i1=" + i1 + " - i2=" + i2);

Nicely prints:
  i1=1 - i2=1
0
 
LVL 37

Expert Comment

by:zzynx
ID: 12160309
And this too:

        double d1 = 1.2345;
        double d2 = 1.5432;
        int i1 = (int)d1;
        int i2 = (int)d2;
        System.out.println("i1=" + i1 + " - i2=" + i2);
0
 
LVL 86

Expert Comment

by:CEHJ
ID: 12160356
if ((int)d1 == (int)d2) {
    // do it
}
0
 
LVL 11

Expert Comment

by:cjjclifford
ID: 12160357
Sorry, meant to say Math.floor(), as Math.round() may round up as well as down, but Math.floor() will return the lower, even when Math.round() goes up (e.g. for 1.51 Math.round() returns 2, Math.floor() returns 1).

So...

double a = 1.2;
double b = 1.51;
System.out.println( "a = " + Math.floor( a ) );
System.out.println( "b = " + Math.floor( b ) );

will print:

a = 1
b = 1

cleaner than wrapping, and casting...

btw, zzynx, why do you need the Double object, as the call to doubleValue() simply returns the "double" primitive - you should have just cast "d1" and "d2" directly....
0
 
LVL 11

Expert Comment

by:cjjclifford
ID: 12160370
oops, sorry, zzynx, forgot to refresh before submit...
0
 
LVL 37

Expert Comment

by:zzynx
ID: 12160409
>> zzynx, why do you need the Double object
When question athours talk about "a double" you're never sure if it is a Double object or a double.
That's why I gave it for both situations.
0
 
LVL 37

Expert Comment

by:zzynx
ID: 12160417
athours = authors
0
 
LVL 37

Expert Comment

by:zzynx
ID: 12160480
jtcy, can you explain why you choose that comment instead of the more "intuitive"/"logical"

       int i = (int)dOne.doubleValue();

?
0
 
LVL 35

Expert Comment

by:girionis
ID: 12163772
:)
0

Featured Post

Online Training Solution

Drastically shorten your training time with WalkMe's advanced online training solution that Guides your trainees to action. Forget about retraining and skyrocket knowledge retention rates.

Question has a verified solution.

If you are experiencing a similar issue, please ask a related question

After being asked a question last year, I went into one of my moods where I did some research and code just for the fun and learning of it all.  Subsequently, from this journey, I put together this article on "Range Searching Using Visual Basic.NET …
Java had always been an easily readable and understandable language.  Some relatively recent changes in the language seem to be changing this pretty fast, and anyone that had not seen any Java code for the last 5 years will possibly have issues unde…
Viewers learn how to read error messages and identify possible mistakes that could cause hours of frustration. Coding is as much about debugging your code as it is about writing it. Define Error Message: Line Numbers: Type of Error: Break Down…
Viewers will learn about the different types of variables in Java and how to declare them. Decide the type of variable desired: Put the keyword corresponding to the type of variable in front of the variable name: Use the equal sign to assign a v…

717 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.

Join & Ask a Question