Assigning a new value to Float objects

Question

Hi,
With BigDecimals, you can assign a value without explicitly creating a new object like this:
BigDecimal b = new BigDecimal(5.23f);
b = BigDecimal.valueOf(625, 2);
As far as I can see, the only ways of doing this with Float is:
Float f = new Float(5.23f);
f = new Float(6,25f);
//OR:
f = Float.valueOf("6.25");
Are there any other ways? Which is most efficient?
Cheers,
Rewbs.

Answer 1

Hi,
With BigDecimals, you can assign a value without
explicitly creating a new object like this:
BigDecimal b = new BigDecimal(5.23f);
b = BigDecimal.valueOf(625, 2);BigDecimal.valueOf:
"Translates a long value into a BigDecimal with a scale of zero. This 'static factory method' is provided in preference to a (long) constructor because it allows for reuse of frequently used BigDecimals."
"frequently used BigDecimals" seems to mean all values form -16 up to 16. For all other values a new BigDecimal object is created anyhow.
>
As far as I can see, the only ways of doing this with
Float is:
Float f = new Float(5.23f);
f = new Float(6,25f);
//OR:
f = Float.valueOf("6.25");
Are there any other ways? Which is most efficient?
No there are no other ways, and the first is definitely the most efficient one. The constructor does no more than creating a new object and assigning the value to it's value field. The other method involves parsing the string, check whether it is a valid representation of a float, ... , and then create a new Float object.