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