What is the purpose of the OptionalInt.isPresent field?

Question

Looking at the source code for java.util.OptionalInt, an optional int is made of an int value and a boolean isPresent. The only way of obtaining an empty optional int is through the OptionalInt.empty() method which returns OptionalInt.EMPTY, the common instance for all empty optional ints.

If this is the case, then why is the isPresent() method implemented as return isPresent rather than this == EMPTY and reducing memory usage by getting rid of the isPresent field?

Answer 1

It's an implementation choice and the only people who can give a clear answer are the people who wrote the implementation.

But very likely it's a choice that puts readability, clarity and maintainability of code over memory micro optimizations.

It really doesn't make much sense to be worried about the space a boolean takes up in an object that wraps an int. If that space is relevant, an OptionalInt shouldn't be used in the first place, (or Java for that matter,) considering the object header needs at least 8 bytes (on 32 bit JVMs, more on 64 bit) already.

Java isn't for writing memory constrained applications, it's for writing easily maintainable code. And implementing isPresent() as getter for isPresent is easier to read, less error prone when refactoring and fits with established Java coding practices.

On a side note: Since Java objects are 8 byte aligned it probably doesn't even make the class smaller when you remove isPresent. As Eugene pointed out in a comment, it actually does increase the size, as the field isPresent lies right on the boundary and then 7 more bytes are added for padding.