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.


== Noncompliant Code Example

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double d = 1.1;

BigDecimal bd1 = new BigDecimal(d); // Noncompliant; see comment above
BigDecimal bd2 = new BigDecimal(1.1); // Noncompliant; same result
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== Compliant Solution

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double d = 1.1;

BigDecimal bd1 = BigDecimal.valueOf(d);
BigDecimal bd2 = new BigDecimal("1.1"); // using String constructor will result in precise value
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== See

* https://wiki.sei.cmu.edu/confluence/x/kzdGBQ[CERT, NUM10-J.] - Do not construct BigDecimal objects from floating-point literals