Home Java javaTutorial Why Does Java's Double Produce Inaccurate Rounding, and How Can BigDecimal Solve It?

Why Does Java's Double Produce Inaccurate Rounding, and How Can BigDecimal Solve It?

Dec 16, 2024 pm 09:25 PM

Why Does Java's Double Produce Inaccurate Rounding, and How Can BigDecimal Solve It?

How to Overcome Java Rounding Double Anomalies

In Java, floating point calculations can sometimes result in unexpected rounding issues. To illustrate, let's examine the following code:

double tempCommission = targetPremium.doubleValue() * rate.doubleValue() / 100d;
double netToCompany = targetPremium.doubleValue() - tempCommission;
double dCommission = request.getPremium().doubleValue() - netToCompany;
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The expected value for dCommission is 877.85, but the actual result is slightly inaccurate at 877.8499999999999. To resolve this issue, we need to employ more precise calculations.

The solution lies in utilizing java.math.BigDecimal. This class provides control over floating point arithmetic precision. Here's an updated version of the code using BigDecimal:

import java.math.BigDecimal;

BigDecimal premium = BigDecimal.valueOf("1586.6");
BigDecimal netToCompany = BigDecimal.valueOf("708.75");
BigDecimal commission = premium.subtract(netToCompany);
System.out.println(commission + " = " + premium + " - " + netToCompany);
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With BigDecimal, the result accurately reflects the expected value:

877.85 = 1586.6 - 708.75
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