How to round 0.745 to 0.75 using BigDecimal.ROUND_HALF_UP?

问题

I tried the following,

   double doubleVal = 1.745;
   double doubleVal1 = 0.745;
   BigDecimal bdTest = new BigDecimal(  doubleVal);
   BigDecimal bdTest1 = new BigDecimal(  doubleVal1 );
   bdTest = bdTest.setScale(2, BigDecimal.ROUND_HALF_UP);
   bdTest1 = bdTest1.setScale(2, BigDecimal.ROUND_HALF_UP);
   System.out.println("bdTest:"+bdTest); //1.75
   System.out.println("bdTest1:"+bdTest1);//0.74    problemmmm ????????????

but got weird results. Why?

回答1:

Never construct BigDecimals from floats or doubles. Construct them from ints or strings. floats and doubles loose precision.

This code works as expected (I just changed the type from double to String):

public static void main(String[] args) {
  String doubleVal = "1.745";
  String doubleVal1 = "0.745";
  BigDecimal bdTest = new BigDecimal(  doubleVal);
  BigDecimal bdTest1 = new BigDecimal(  doubleVal1 );
  bdTest = bdTest.setScale(2, BigDecimal.ROUND_HALF_UP);
  bdTest1 = bdTest1.setScale(2, BigDecimal.ROUND_HALF_UP);
  System.out.println("bdTest:"+bdTest); //1.75
  System.out.println("bdTest1:"+bdTest1);//0.75, no problem
}

回答2:

double doubleVal = 1.745;
double doubleVal1 = 0.745;
System.out.println(new BigDecimal(doubleVal));
System.out.println(new BigDecimal(doubleVal1));

outputs:

1.74500000000000010658141036401502788066864013671875
0.74499999999999999555910790149937383830547332763671875

Which shows the real value of the two doubles and explains the result you get. As pointed out by others, don't use the double constructor (apart from the specific case where you want to see the actual value of a double).

More about double precision:

here
there

回答3:

Use BigDecimal.valueOf(double d) instead of new BigDecimal(double d). The last one has precision errors by float and double.

回答4:

This will maybe give you a hint on what went wrong.

import java.math.BigDecimal;

public class Main {
    public static void main(String[] args) {
        BigDecimal bdTest = new BigDecimal(0.745);
        BigDecimal bdTest1 = new BigDecimal("0.745");
        bdTest = bdTest.setScale(2, BigDecimal.ROUND_HALF_UP);
        bdTest1 = bdTest1.setScale(2, BigDecimal.ROUND_HALF_UP);
        System.out.println("bdTest:" + bdTest);
        System.out.println("bdTest1:" + bdTest1); 
    }
}

The problem is, that your input (a double x=0.745;) can not represent 0.745 exactly. It actually saves a value slightly lower. For BigDecimals, this is already below 0.745, so it rounds down...

Try not to use the BigDecimal(double/float) constructors.

回答5:

For your interest, to do the same with double

double doubleVal = 1.745;
double doubleVal2 = 0.745;
doubleVal = Math.round(doubleVal * 100 + 0.005) / 100.0;
doubleVal2 = Math.round(doubleVal2 * 100 + 0.005) / 100.0;
System.out.println("bdTest: " + doubleVal); //1.75
System.out.println("bdTest1: " + doubleVal2);//0.75

or just

double doubleVal = 1.745;
double doubleVal2 = 0.745;
System.out.printf("bdTest: %.2f%n",  doubleVal);
System.out.printf("bdTest1: %.2f%n",  doubleVal2);

both print

bdTest: 1.75
bdTest1: 0.75

I prefer to keep code as simple as possible. ;)

As @mshutov notes, you need to add a little more to ensure that a half value always rounds up. This is because numbers like 265.335 are a little less than they appear.

来源：https://stackoverflow.com/questions/12460482/how-to-round-0-745-to-0-75-using-bigdecimal-round-half-up

标签

java

bigdecimal

rounding