Why does the Math.round(float f) method in Android check for NaN?

Question

I was looking at the Math sources in the Android Framework and here is what the round() method looks like :

public static int round(float f) {
    // check for NaN
    if (f != f) {
        return 0;
    }
    return (int) floor(f + 0.5f);
}

I'm trying to wrap my head around what the first check does without success. How could f be different from itself ? I also can't think of any test case that would make this any different... Any idea ?

See [JLS 15.21.1](http://docs.oracle.com/javase/specs/jls/se7/html/jls-15.html#jls-15.21.1). That's just how it's defined. — ajb, Feb 18 '14 at 01:52

user207421 · Accepted Answer · 2014-02-18T02:05:44.077

0

How could f be different from itself?

If f is a NaN it will test as different from all floats including another NaN. That's the definition. From JLS #15.21.1:

"If either operand is NaN, then the result of == is false but the result of != is true. Indeed, the test x!=x is true if and only if the value of x is NaN."

I also can't think of any test case that would make this any different... Any idea?

Err, Float.NaN.

edited Feb 18 '14 at 02:05

answered Feb 18 '14 at 01:50

user207421

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I find this a bit confusing but I guess it makes sense. Thanks EJP, will mark the answer as accepted when possible – ben Feb 18 '14 at 01:58
@downvoter You're arguing with the JLS here. If you have an over-riding citation please provide it. – user207421 Jul 03 '14 at 06:50

score 0 · Answer 2 · edited May 23 '17 at 11:49

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NaN is the only variable for which x != x holds true. Take a look at this answer for more details.

edited May 23 '17 at 11:49

Community

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answered Feb 18 '14 at 01:51

Chthonic Project

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Thanks! This thread definitely made it easier for me to understand the reason why it's how it works – ben Feb 18 '14 at 01:59

Why does the Math.round(float f) method in Android check for NaN?

2 Answers2