Question

I am converting four bytes to float and I m getting NaN as a result, but I want the value 0.0. What am I doing wrong?

This is my code:

public class abc 
{
    public static void main(String[] args) 
    {
        int[] arry = { 255, 255, 255, 255 };
        int num = ((arry[0] << 24) & 0xFF000000) | ((arry[1] << 16) & 0xFF0000)
            | ((arry[2] << 8) & 0xFF00) | (arry[3] & 0xFF);
        float f = Float.intBitsToFloat(num);

        f= (float) ((f < 0 ? Math.ceil(f * 10) : Math.floor(f * 10)) / 10);

        System.out.println(f);
    }
}

Answer 1

Your main problem is that 0xFFFFFFFF is indeed a NaN.

A float with a value of 0 is... 0.

Changing the array to

int[] arry = { 0x00, 0x00, 0x00, 0x00 };

Will change the resulting value to a 0.0f float.

Answer 2

Well, your bit pattern happens to actually be NaN:

IEEE 754 NaNs are represented with the exponential field filled with ones and some non-zero number in the significand. A bit-wise example of a IEEE floating-point standard single precision NaN: x111 1111 1axx xxxx xxxx xxxx xxxx xxxx where x means don t care. If a = 1, it is a quiet NaN, otherwise it is a signalling NaN.

Since all your bits are 1, it obviously fits above criteria and is a quiet NaN (although Java iirc doesn t even support quiet or signalling NaNs).

Answer 3

NaN = Not a Number, iaw, the bit pattern 0xFFFFFFFF does not represent a ~~float value~~ number. (Joachim, you re right .. again :-) )

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