Comparing doubles in Java gives strange results

I really don't understand why the following happens:

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

However, this is working as expected:

Double d = 0.0;
System.out.println(d == 0.0); // true
System.out.println(d.equals(0.0)); // true

I'm sure it's related to some way auto@R_530_2419 @Ing, but I really don't know why 0 will be used in different ways using the = = operator and When equals is called

Isn't this an implicit violation of the equality contract?

  *  It is reflexive: for any non-null reference value
  *     x,x.equals(x) should return
  *     true.

Edit:

Thank you for your quick answer I think it's a different box. The real question is: why is it boxed differently? I mean, if d = = 0d is more intuitive and expected than d.equals (0d), then it will be more intuitive, but if d = = 0 looks like an integer, it should be true than "intuitive" d.equals (0)

Solution

Just change it

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

You are comparing with integer 0

Under the cover

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

0 will be automatically boxed as an integer, and an instance of integer will be passed to the equals () method of the double class, which will be more like

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

Of course, it will return false

to update

When you use = = for comparison, it compares values, so you don't need to auto@R_530_2419 @, which operates directly on values Where equals () accepts object, so if you try to call D1 Equals (0), then 0 is not an object, so it will perform automatic boxing and package it as an integer, which is an object