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 = … Read more
BigDecimal value = new BigDecimal(Float.toString(123.4f)); From the javadocs, the string constructor is generally the preferred way to convert a float into a BigDecimal, as it doesn’t suffer from the unpredictability of the BigDecimal(double) constructor. Quote from the docs: Note: For values other float and double NaN and ±Infinity, this constructor is compatible with the values … Read more
You can’t use the equals() method to compare BigDecimals, like this assertion does. That is because this equals function will compare the scale. If the scale is different, equals() will return false, even if they are the same number mathematically. You can however use compareTo() to do what you want: As @assylias points out, you … Read more
You can use setScale() e.g. double d = … BigDecimal db = new BigDecimal(d).setScale(12, BigDecimal.ROUND_HALF_UP);
You can specify locale for DecimalFormat this way: DecimalFormatSymbols symbols = new DecimalFormatSymbols(Locale.US); DecimalFormat format = new DecimalFormat(“##.########”, symbols);
You have two BigDecimal numbers both of which require over a 1000 decimal places. Trying to set the scale to only have 113 decimal places means you will lose precision and therefore you need to round. You can use the setScale methods that take a RoundingMode to prevent the exception but not the rounding.
Try this // Create a DecimalFormat that fits your requirements DecimalFormatSymbols symbols = new DecimalFormatSymbols(); symbols.setGroupingSeparator(‘,’); symbols.setDecimalSeparator(‘.’); String pattern = “#,##0.0#”; DecimalFormat decimalFormat = new DecimalFormat(pattern, symbols); decimalFormat.setParseBigDecimal(true); // parse the string BigDecimal bigDecimal = (BigDecimal) decimalFormat.parse(“10,692,467,440,017.120”); System.out.println(bigDecimal); If you are building an application with I18N support you should use DecimalFormatSymbols(Locale) Also keep in mind … Read more
This is fine, remember that using the constructor of BigDecimal to declare a value can be dangerous when it’s not of type String. Consider the below… BigDecimal valDouble = new BigDecimal(0.35); System.out.println(valDouble); This will not print 0.35, it will infact be… 0.34999999999999997779553950749686919152736663818359375 I’d say your solution is probably the safest because of that.
You haven’t specified a scale for the result. Please try this 2019 Edit: Updated answer for JDK 13. Cause hopefully you’ve migrated off of JDK 1.5 by now. import java.math.BigDecimal; import java.math.RoundingMode; public class Main { public static void main(String[] args) { BigDecimal a = new BigDecimal(“1”); BigDecimal b = new BigDecimal(“3”); BigDecimal c = … Read more
I think this describes solution to your problem: Java Traps: Big Decimal and the problem with double here From the original blog which appears to be down now. Java Traps: double Many traps lay before the apprentice programmer as he walks the path of software development. This article illustrates, through a series of practical examples, … Read more