Comparing doubles in Java gives odd results

Question

I really can\'get my head around why the following happens:

Double d = 0.0;
System.out.println(d == 0); // is true
System.out.println(d.equals(0)); // is false ?         


        

Answer 1

When you perform

d == 0

this is upcast to

d == 0.0

however there are no upcasting rules for autoboxing and even if there were equals(Object) gives no hits that you want a Double instead of an Integer.

Answer 2

just change it to

System.out.println(d.equals(0d)); // is false ?! now true

You were comparing double with Integer 0

Under the cover

System.out.println(d.equals(0)); // is false ?!

0 will be autoboxed to Integer and an instance of Integer will be passed to equals() method of Double class, where it will compare like

@Override
    public boolean equals(Object object) {
        return (object == this)
                || (object instanceof Double)
                && (doubleToLongBits(this.value) == doubleToLongBits(((Double) object).value));
    }

which is going to return false of course.

Update

when you do comparison using == it compares values so there is no need to autobox , it directly operates on value. Where equals() accepts Object so if you try to invoke d1.equals(0) , 0 is not Object so it will perform autoboxing and it will pack it to Integer which is an Object.

Answer 3

It's probably worth noting that you should compare floating point numbers like this:

|x - y| < ε, ε very small

Answer 4

d.equals(0) : 0 is an int. The Double.equals() code will return true only for Double objects.

Answer 5

Number objects only equal to numbers with the same value if they are of the same type. That is:

new Double(0).equals(new Integer(0));
new BigInteger("0").equals(new BigDecimal("0"));

and similar combinations are all false.

In your case, the literal 0 is boxed into an Integer object.