Christian Christian - 2 years ago 88
Java Question

Android: java rounding error. Can't understand why?

Can anybody explain why on Earth these "same" expressions returns different values?

(new BigDecimal(String.valueOf(131.7d))).multiply(new BigDecimal(String.valueOf(0.95d))).doubleValue() = 125.115

(new BigDecimal( 131.7d )).multiply(new BigDecimal( 0.95d )).doubleValue() = 125.11499999999998

What BigDecimal is doing different between them?

Answer Source

If you read the API documentation, you will find taht String.valueOf(dobule) uses Double.toString(double) to format the value. It's perhaps not obvious, but Double.toString(double) rounds the value, before formatting it as a string:

How many digits must be printed for the fractional part of m or a? There must be at least one digit to represent the fractional part, and beyond that as many, but only as many, more digits as are needed to uniquely distinguish the argument value from adjacent values of type double. That is, suppose that x is the exact mathematical value represented by the decimal representation produced by this method for a finite nonzero argument d. Then d must be the double value nearest to x; or if two double values are equally close to x, then d must be one of them and the least significant bit of the significand of d must be 0.

The result of this is that String.valueOf(131.7d) will return the string "131.7" even if the exact value of the argument is 131.69999999999998863131622783839702606201171875. The reason for this is that decimal fractions cannot always be represented exactly using binary fractions (as used with floats and doubles).

So, new new BigDecimal(String.valueOf(131.7)) will create a BigDecimal with the exact value 131.7. new BigDecimal(131.7) will create a BigDecimal with the exact value 131.69999999999998863131622783839702606201171875.

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