Rounding in java shows different result [duplicate]

Question

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?

Answer 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
}

Answer 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