Comparing NaN in Kotlin

Question

So I recently started loving the language kotlin. Today, while comparing doubles, I came across the inevitable NaN.

fun main(args: Array<String>) {
    val nan = Double.NaN
    println("1: " + (nan == nan))
    println("2: " + (nan == (nan as Number)))
    println("3: " + ((nan as Number) == nan))
}

N.B: (Double is a subtype of Number)

Running the above code yields:

1: false
2: true
3: true

I understand that comparing with NaN in Java returns false, so I would expect false for all expressions.

How can this behavior be explained? What is the rationale behind it?

Answer 1

That's because (2) and (3) are compiled to boxing a primitive and then Double.equals check: on JVM, primitive double can't be compared to a boxed one.

Double.equals, in turn, checks equality by comparing doubleToLongBits(...) of the two Doubles, and for the latter there's a guarantee that

If the argument is NaN, the result is 0x7ff8000000000000L.

So, the bits returned for two NaN are equal, and the rule NaN != NaN is ignored here.

Also, as @miensol mentioned, there's another consequence of this equality check: +0 and -0 are equal according to == check and not to equals check.

Equivalent code in Java would be:

double nan = Double.NaN;
System.out.println("1: " + (nan == nan)) //false 
System.out.println("2: " + ((Double) nan).equals(((Number) nan)))
System.out.println("3: " + ((Number) nan).equals(nan));

The last two lines call Double.equals, comparing doubleToLongBits(...).