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.
Noncompliant Code Example
double d = 1.1;
BigDecimal bd1 = new BigDecimal(d); // Noncompliant; see comment above
BigDecimal bd2 = new BigDecimal(1.1); // Noncompliant; same result
Compliant Solution
double d = 1.1;
BigDecimal bd1 = BigDecimal.valueOf(d);
BigDecimal bd2 = new BigDecimal("1.1"); // using String constructor will result in precise value
See
- CERT, NUM10-J. - Do not construct BigDecimal objects from floating-point literals
