Assume that I don't know if one Optional is empty, or if both are present. In the latter case, I always want to prefer a over b:
final Optional<String> a = Optional.of("1");
final Optional<String> b = Optional.empty();
if (a.isPresent() || b.isPresent()) {
// prefer a over b
Integer result = a
.map(s -> s + "0")
.map(Integer::parseInt)
.orElseGet(() -> Integer.parseInt(b.get())); // <-- warning for b.get()
System.out.println(result);
}
In my real code, Idea is warning me at this point:
'Optional.get()' without 'isPresent()' check.
Why is that? I am checking priorly if either a or b is present. Also, this code works as expected, the output is 10. If I put b = Optional.of("2") the output is still 10 because I prefer a. If I then put a = Optional.empty(), the output is 2 as expected.
Am I doing something wrong, or is Idea's linter wrong?
It's idea that is confused here, it's internal rules probably suppose the checks to isPresent in the the same Optional chaining.
This obviously lays outside the Optional checks itself due to (a.isPresent() || b.isPresent()); it can't possible tell that there's no way to call orElseGet on an empty optional b because of those checks...