Because of floating point imprecision, you're unlikely to get the value you expect from the ``++BigDecimal(double)++`` constructor.
From http://docs.oracle.com/javase/7/docs/api/java/math/BigDecimal.html#BigDecimal(double)[the JavaDocs]:
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The results of this constructor can be somewhat unpredictable. One might assume that writing new BigDecimal(0.1) in Java creates a BigDecimal which is exactly equal to 0.1 (an unscaled value of 1, with a scale of 1), but it is actually equal to 0.1000000000000000055511151231257827021181583404541015625. This is because 0.1 cannot be represented exactly as a double (or, for that matter, as a binary fraction of any finite length). Thus, the value that is being passed in to the constructor is not exactly equal to 0.1, appearances notwithstanding.
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Instead, you should use ``++BigDecimal.valueOf++``, which uses a string under the covers to eliminate floating point rounding errors, or the constructor that takes a ``++String++`` argument.
\[~ann.campbell.2] we might want to refer or quote the JavaDoc of BigDecimal double constructor here : \http://docs.oracle.com/javase/7/docs/api/java/math/BigDecimal.html#BigDecimal(double)
=== on 10 Oct 2014, 15:13:53 Freddy Mallet wrote:
Perfect !
=== on 16 Jan 2015, 09:32:59 Sébastien Gioria wrote:
Coudl be tag "security" as it's part of the CERT Secure Coding for Java (\https://www.securecoding.cert.org/confluence/display/java/NUM10-J.+Do+not+construct+BigDecimal+objects+from+floating-point+literals)
