Scale() of Divide method in BigDecimal

问题

new BigDecimal("37146555.53880000").divide(new BigDecimal("1000000")).scale()

This returns 10. But according to the API, the divide method:

Returns a BigDecimal whose value is (this / divisor), and whose preferred scale is (this.scale() - divisor.scale());

So in this case, 37146555.53880000's scale is 8, and 1000000's scale is 0. So the result should have a scale of 8, not 10.

What am I missing here?

Thanks

回答1:

The actual result is 37.1465555388 whose scale must be 10 for it to be exact.

What the JavaDoc says is that the preferred scale is the difference meaning that if the result didn't actually need to be 10, then it would try to make it 8. For example if you would have divided by 2, whose scale is also 0, the result would have been 18573277.76940000 (scale 8).

EDIT: small adition - you can force the division to a certain scale by using the overloaded divide methods:

• divide(BigDecimal, RoundingMode) that will give a BigDecimal with scale of this and value rounded using the specified rounding method if the result would actually need more decimals to be exact.

• divide(BigDecimal, scale, RoundingMode) that will give a BigDecimal with specified scale, and value rounded by specified method if needed.

This might be useful if your dividing by a number you know can cause repeating decimals, like 3 (1/3 = 0.333333...) since, if that happens, the simple divide will throw an exception. Bounding it to a maximum number of decimals will help you avoid the exception but will make your computations less precise.

回答2:

These scales are the ones used by the methods which return exact arithmetic results; except that an exact divide may have to use a larger scale since the exact result may have more digits. For example, 1/32 is 0.03125.

回答3:

It says "preferred scale" not "definitely will be scale".

To be absolutely sure, I would use BigDecimal.divide(BigDecimal, int, int).

来源：https://stackoverflow.com/questions/3303503/scale-of-divide-method-in-bigdecimal