[FIXED] Why can't HashMap have different NaN values as keys?

Issue

import java.util.*;

public class MyClass {
    public static void main(String args[]) {
      Map<Float, Integer> m = new HashMap<>();
      
      m.put(Float.intBitsToFloat(0x7f800001), 1);
      m.put(Float.intBitsToFloat(0x7f800002), 2);
      
      System.out.println(m.size());
    }
}

Why does the above code return 1 as the size of m? 0x7f800001 and 0x7f800002 are both NaN float values, but since NaN != NaN by definition, this should not cause a collision.

The behavior is similar to the documented handling of null keys in the Java Hashmap, but I can't find anything that indicates that NaN is handled as null by HashMap.

Solution

Float.equals(Object) is documented as:

Compares this object against the specified object. The result is true if and only if the argument is not null and is a Float object that represents a float with the same value as the float represented by this object. For this purpose, two float values are considered to be the same if and only if the method floatToIntBits(float) returns the identical int value when applied to each.

Now that sounds like different NaN values should be treated as non-equal, but floatToIntBits documentation includes (emphasis mine):

If the argument is NaN, the result is 0x7fc00000.

In all cases, the result is an integer that, when given to the intBitsToFloat(int) method, will produce a floating-point value the same as the argument to floatToIntBits (except all NaN values are collapsed to a single "canonical" NaN value).

So basically, Float.equals (and Float.hashCode, which also uses floatToIntBits(float)) treats all NaN values as equal.

Answered By - Jon Skeet

Answer Checked By - Katrina (PHPFixing Volunteer)