Java - Common Gotchas [closed]

Answer 1

Floats

I don't know many times I've seen

floata == floatb

where the "correct" test should be

Math.abs(floata - floatb) < 0.001

I really wish BigDecimal with a literal syntax was the default decimal type...

Answer 2

Try reading Java Puzzlers which is full of scary stuff, even if much of it is not stuff you bump into every day. But it will destroy much of your confidence in the language.

Answer 3

Going first, here's one I caught today. It had to do with Long/long confusion.

public void foo(Object obj) {
    if (grass.isGreen()) {
        Long id = grass.getId();
        foo(id);
    }
}
private void foo(long id) {
    Lawn lawn = bar.getLawn(id);
    if (lawn == null) {
        throw new IllegalStateException("grass should be associated with a lawn");
    }   
}

Obviously, the names have been changed to protect the innocent :)

Answer 4

Another one I'd like to point out is the (too prevalent) drive to make APIs generic. Using well-designed generic code is fine. Designing your own is complicated. Very complicated!

Just look at the sorting/filtering functionality in the new Swing JTable. It's a complete nightmare. It's obvious that you are likely to want to chain filters in real life but I have found it impossible to do so without just using the raw typed version of the classes provided.

Answer 5

Comparing equality of objects using == instead of .equals() -- which behaves completely differently for primitives.

This gotcha ensures newcomers are befuddled when "foo" == "foo" but new String("foo") != new String("foo").

Answer 6

When you create a duplicate or slice of a ByteBuffer, it does not inherit the value of the order property from the parent buffer, so code like this will not do what you expect:

ByteBuffer buffer1 = ByteBuffer.allocate(8);
buffer1.order(ByteOrder.LITTLE_ENDIAN);
buffer1.putInt(2, 1234);

ByteBuffer buffer2 = buffer1.duplicate();
System.out.println(buffer2.getInt(2));
// Output is "-771489792", not "1234" as expected