比较BigDecimal值和double值是否等于两个小数位

Question

I'm working with an API that returns a double value. I need to compare that value to a BigDecimal value for equality to two decimal places.

Should I convert BigDecimal to double and then compare

Math.abs(myBigDecimal.doubleValue() - apiDouble) >= .01

Or are there issues with this approach? Perhaps I should convert the double into a BigDecimal and and then compare?

Answer 1

No. Although IEEE double has guard bits, it cannot exactly represent a decimal value. Convert the double to a BigDecimal, using rounding, and then use compareTo(). Note that equals() takes into account decimal positions, so compareTo() is safer. You may need to round the BigDecimal as well.

Answer 2

Shift desired decimal positions multiplying both doubles by 100, use Math.ceil or Math.floor, as you prefer, to get rid by the rest of decimal positions:

Math.ceil(myBigDecimal.doubleValue()*100) == Math.ceil(apiDouble*100)

this is equivalent to truncate without rounding. If you are worring about rounding, multiply by 1000, cast to long divide by 10 and compare.