Question

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.

Answer 1

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());