I think it's still possible to use BigDecimal for the calculation if the scale value is set higher. Then round off the final value to 2 or 3 decimal places.

Joyce [ September 22, 2004: Message edited by: Joyce Lee ]

Keep in mind that you have yourself specified the number of significant digits to carry. BigDecimal has another divide method that does not require a scale:

public BigDecimal divide(BigDecimal val, int roundingMode)Returns a BigDecimal whose value is (this / val), and whose scale is this.scale(). If rounding must be performed to generate a result with the given scale, the specified rounding mode is applied.

I have yet to see this method fail in standard calculations with a known-to-be-precise answer.

HOWEVER...

*You can always cover yourself with a NumberFormat that only allows 2 decimal places. *BigDecimal also provides many rounding modes to choose from, *You can always modify operator order to clarify what the result should be *Your operands define the inital scale of the result

Play around with the following code some. Note that I don't specify a scale.

If you pass "415" without the ".0" you'll get 5 as a result -- one of the operands had a zero scale.

Anyway, good luck and have fun.

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