Because of floating point imprecision, you're unlikely to get the value you expect from the BigDecimal(double)
constructor.
From the JavaDocs:
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.
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.
double d = 1.1; BigDecimal bd1 = new BigDecimal(d); // Noncompliant; see comment above BigDecimal bd2 = new BigDecimal(1.1); // Noncompliant; same result
double d = 1.1; BigDecimal bd1 = BigDecimal.valueOf(d); BigDecimal bd2 = new BigDecimal("1.1"); // using String constructor will result in precise value