Does the Java compiler remove multiplications by 1, when talking about BigDecimal?
I'm standing in something similar to this:
BigDecimal bd = getBigDecimal();//get arbitrary bigDecimal, could be anyone.
bd = bd.multiply(new BigDecimal(getOneTwoOrThree());
Where the getOneTwoOrThree method is declared as:
/*
* Always returns Integers 1, 2 or 3.
*/
Integer getOneTwoOrThree(){
//some code
}
So, basically.
if getOneTwoOrThree() returns 1, will the compiler perform the multiplication? Or will it nop the instruction?
This is somewhat of an existential doubt, but I guess that I'm at some level early - optimizing.
No. BigDecimal is a library class (it's not even in java.lang), so the compiler treats it as any other class.
BigDecimal could special-case this internally, but apparently doesn't.
(Edit: I should add that it's possible that the JIT compiler could work some magic, but I would have to do some tests to be sure.)
I would only suggest that you change your code to use BigDecimal.valueOf(), because 1, 2, and 3 are some of the special cases which are cached internally by BigDecimal.
bd = bd.multiply(BigDecimal.valueOf(getOneTwoOrThree());