BigDecimal(double)” should not be used

BigDecimal(double)” should not be used

级别：bug

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.

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 = BigDecimal.valueOf(1.1);

To chinese：

不要使用BigDecimal(double)去构造一个BigDecimal对象，因为double类型在计算机表示方法中并不精确，因此，BigDecimal(double)构造出来的对象很可能不是预期的大小，若一定要使用double类型去构造一个BigDecimal对象，请使用BigDecimal.valueOf方法，该方法先将double转换为String，再通过String构造BigDecimal对象，通常更建议使用public BigDecimal(String val)构造方法。