barbosa barbosa - 1 year ago 82
Java Question

Unpredictability of the BigDecimal(double) constructor

I started using Sonar recently in a project, and i got a PMD rule broken about using the constructor

new BigDecimal(double val)
. When i read the java documentation, i found that new BigDecimal(double val) is somewhat unpredictable and that I should use
new BigDecimal(String val)
which is predictable.

Here is what javadoc says for
public BigDecimal(double val)

Translates a double into a BigDecimal which is the exact decimal
representation of the double's binary floating-point value. The scale
of the returned BigDecimal is the smallest value such that (10scale ×
val) is an integer.


The results of this constructor can be somewhat unpredictable. One
might assume that writing
new BigDecimal(0.1)
in Java creates a
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.

The String constructor, on the other hand, is perfectly predictable:
new BigDecimal("0.1")
creates a
which is
exactly equal to 0.1, as one would expect. Therefore, it is generally
recommended that the String constructor be used in preference to this

When a double must be used as a source for a
, note that
this constructor provides an exact conversion; it does not give the
same result as converting the double to a String using the
method and then using the
constructor. To get that result, use the static

Why does this constructor really exists? Isnt
new BigDecimal(String val)
enough for that matter? When should I use the
new BigDecimal(double val)

Answer Source

Why does this constructor really exists?

It converts the actual represented value of double to a BigDecimal. The whole point of BigDecimal is to give as much precision as possible and that is what this constructor does.

If you want to take the value you would get with a small amount of rounding the Double.toString(double) uses you can use




When should I use the new BigDecimal(double val) constructor

When you want to know the value double really represents. You can apply your own rounding as required.

When you use double you should always apply a sensible rounding. But, if you did that you may find you don't need BigDecimal. ;)