Java浮动到双 - 上限和下限？

Question

As most here will know, double -> float incurs a loss in precision. This means, multiple double values may be mapped to the same float value. But how do I go the other way? Given a normal (I'm not caring about the extreme cases) float, how do I find the upper and lower value of double precision that are still mapped to the same float?

Or, to speak in code:

function boolean testInterval(float lowF, float highF, double queryD) {
    float queryF = (float) queryD;
    return (lowF <= queryF) && (queryF <= highF);
}

and

function boolean testInterval(float lowF, float highF, double queryD) {
    double lowD = (double) lowF;
    double highD = (double) highF;
    return (lowD <= queryD) && (queryD <= highD);
}

do not always give the same result. I'm looking for two functions float-> double to make the second function return the same result at the first.

This could work, but it looks like a hack and not the proper solution to me.

function boolean testIntervalHack(float lowF, float highF, double queryD) {
    double lowD = (double) lowF - Float.MIN_VALUE;
    double highD = (double) highF + Float.MIN_VALUE;
    return (lowD <= queryD) && (queryD <= highD);
}

Answer 1

Your testIntervalHack doesn't work, the range of double values mapping to the same float varies. For example, with x = 2^24-1 , every double between x-0.5 and x+0.5 will be mapped to the same value (the float value of x ), but x +/- Float.MIN_VALUE == x .

I'm not aware of any convenient API methods, so the best I can offer is

1. convert to double
2. convert the double to the bit representation via doubleTo(Raw)LongBits
3. add or subtract one of 2 ²⁸ or 2 ²⁸ -1, depending on whether you want the upper or lower bound and the 2 ²⁹ -bit is 0 or 1 (because of round-to-even)
4. convert that long to double via longBitsToDouble

Well, that's for finite values in float range. For NaN s, you can stop after step 1., for infinities, it's a bit more delicate, since double values larger than or equal to 2 ¹²⁸ -2 ¹⁰³ are converted to (float)Infinity , which is quite a bit away from the bit representation of (double)Infinity .