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precision lost

Posted on 2004-09-27
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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.

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Question by:jtcy
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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?
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LVL 11

Expert Comment

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

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

Accepted Solution

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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);
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LVL 35

Expert Comment

by:girionis
ID: 12160159
...although you should be able to cast a double to int.
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Expert Comment

by:TimYates
ID: 12160235
I think he wasn't using the cast...
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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
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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);
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LVL 86

Expert Comment

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

Expert Comment

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

Expert Comment

by:zzynx
ID: 12160417
athours = authors
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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();

?
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LVL 35

Expert Comment

by:girionis
ID: 12163772
:)
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