Best way of finding the bigdecimal is greater than 1 ?

Snapshot of the code :
1)

2)

I would say a variation of the first one. But don't create a new BigDecimal representation of 1 each time. Use the static BigDecimal ONE from the BigDecimal class (added in Java 1.5). Also, don't use the > 1.00 since that is confusing. The compareTo method returns one of three distinct values. Use them.

Mark Vedder wrote:The compareTo method returns one of three distinct values. Use them.

Actually... the API documentation for that method says:

The suggested idiom for performing these comparisons is: (x.compareTo(y) <op> 0), where <op> is one of the six comparison operators.

I would agree that's the most understandable way to write the code.

Excellent Mark. Thanks

Paul Clapham wrote:
Actually... the API documentation for that method says:

The suggested idiom for performing these comparisons is: (x.compareTo(y) <op> 0), where <op> is one of the six comparison operators.

I would agree that's the most understandable way to write the code.

Good point Paul. Thanks. I think it was the comparison to 1.00 that was looking weird to me. Consistently comparing to 0 would indeed be clearer.

OK using 0, will allow the use of 6 operators. I was wondering, how to use 6 operators with 1.

x.compareTo(y) returns a positive number if x > y.