You can try BigDecimal for this purpose
Double toBeTruncated = new Double("3.5789055");
Double truncatedDouble = BigDecimal.valueOf(toBeTruncated)
.setScale(3, RoundingMode.HALF_UP)
.doubleValue();
Answer from Neel on Stack OverflowVideos
Java Double Precision
Java Programming Help: How to set Precision for double value | Codersarts
Java Double Precision : java
[Java] Precision related question on double value vs BigDecimal object
You can try BigDecimal for this purpose
Double toBeTruncated = new Double("3.5789055");
Double truncatedDouble = BigDecimal.valueOf(toBeTruncated)
.setScale(3, RoundingMode.HALF_UP)
.doubleValue();
Answer from Neel on Stack OverflowYou can try BigDecimal for this purpose
Double toBeTruncated = new Double("3.5789055");
Double truncatedDouble = BigDecimal.valueOf(toBeTruncated)
.setScale(3, RoundingMode.HALF_UP)
.doubleValue();
You can't set the precision of a double (or Double) to a specified number of decimal digits, because floating-point values don't have decimal digits. They have binary digits.
You will have to convert into a decimal radix, either via BigDecimal or DecimalFormat, depending on what you want to do with the value later.
See also my answer to this question for a refutation of the inevitable *100/100 answers.
I came across a piece of code in a legacy Java 8 application at work which adds two doubles and gives out a double. I observed that the resulting doubles for various inputs had variable number of digits after the decimal point. Some were very precise with 12 digits after the decimal point and some had merely a digit after the decimal point.
Iโm curious to know what factors affect certain doubles to be so very precise and certain doubles not as much.
Examples:
double one = 3880.95; double two = 380.9; Result: 4261.849999999999
double one = 1293.65; double two = 1293.6; Result: 2587.25