point in Java
Is there a Java library anywhere that can perform computations on IEEE 754 half-precision numbers or convert them to and from double-precision?
Either of these approaches would be suitable:
Edit : conversion needs to be 100% accurate - there are lots of NaNs, infinities and subnormals in the input files.
Related question but for JavaScript: Decompressing Half Precision Floats in Javascript
You can Use Float.intBitsToFloat() and Float.floatToIntBits() to convert them to and from primitive float values. If you can live with truncated precision (as opposed to rounding) the conversion should be possible to implement with just a few bit shifts.
I have now put a little more effort into it and it turned out not quite as simple as I expected at the beginning. This version is now tested and verified in every aspect I could imagine and I'm very confident that it produces the exact results for all possible input values. It supports exact rounding and subnormal conversion in either direction.
// ignores the higher 16 bits
public static float toFloat( int hbits )
{
int mant = hbits & 0x03ff; // 10 bits mantissa
int exp = hbits & 0x7c00; // 5 bits exponent
if( exp == 0x7c00 ) // NaN/Inf
exp = 0x3fc00; // -> NaN/Inf
else if( exp != 0 ) // normalized value
{
exp += 0x1c000; // exp - 15 + 127
if( mant == 0 && exp > 0x1c400 ) // smooth transition
return Float.intBitsToFloat( ( hbits & 0x8000 ) << 16
| exp << 13 | 0x3ff );
}
else if( mant != 0 ) // && exp==0 -> subnormal
{
exp = 0x1c400; // make it normal
do {
mant <<= 1; // mantissa * 2
exp -= 0x400; // decrease exp by 1
} while( ( mant & 0x400 ) == 0 ); // while not normal
mant &= 0x3ff; // discard subnormal bit
} // else +/-0 -> +/-0
return Float.intBitsToFloat( // combine all parts
( hbits & 0x8000 ) << 16 // sign << ( 31 - 15 )
| ( exp | mant ) << 13 ); // value << ( 23 - 10 )
}
// returns all higher 16 bits as 0 for all results
public static int fromFloat( float fval )
{
int fbits = Float.floatToIntBits( fval );
int sign = fbits >>> 16 & 0x8000; // sign only
int val = ( fbits & 0x7fffffff ) + 0x1000; // rounded value
if( val >= 0x47800000 ) // might be or become NaN/Inf
{ // avoid Inf due to rounding
if( ( fbits & 0x7fffffff ) >= 0x47800000 )
{ // is or must become NaN/Inf
if( val < 0x7f800000 ) // was value but too large
return sign | 0x7c00; // make it +/-Inf
return sign | 0x7c00 | // remains +/-Inf or NaN
( fbits & 0x007fffff ) >>> 13; // keep NaN (and Inf) bits
}
return sign | 0x7bff; // unrounded not quite Inf
}
if( val >= 0x38800000 ) // remains normalized value
return sign | val - 0x38000000 >>> 13; // exp - 127 + 15
if( val < 0x33000000 ) // too small for subnormal
return sign; // becomes +/-0
val = ( fbits & 0x7fffffff ) >>> 23; // tmp exp for subnormal calc
return sign | ( ( fbits & 0x7fffff | 0x800000 ) // add subnormal bit
+ ( 0x800000 >>> val - 102 ) // round depending on cut off
>>> 126 - val ); // div by 2^(1-(exp-127+15)) and >> 13 | exp=0
}
I implemented two small extensions compared to the book because the general precision for 16 bit floats is rather low which could make the inherent anomalies of floating point formats visually perceivable compared to larger floating point types where they are usually not noticed due to the ample precision.
The first one are these two lines in the toFloat() function:
if( mant == 0 && exp > 0x1c400 ) // smooth transition
return Float.intBitsToFloat( ( hbits & 0x8000 ) << 16 | exp << 13 | 0x3ff );
Floating point numbers in the normal range of the type size adopt the exponent and thus the precision to the magnitude of the value. But this is not a smooth adoption, it happens in steps: switching to the next higher exponent results in half the precision. The precision now remains the same for all values of the mantissa until the next jump to the next higher exponent. The extension code above makes these transitions smoother by returning a value that is in the geographical center of the covered 32 bit float range for this particular half float value. Every normal half float value maps to exactly 8192 32 bit float values. The returned value is supposed to be exactly in the middle of these values. But at the transition of the half float exponent the lower 4096 values have twice the precision as the upper 4096 values and thus cover a number space that is only half as large as on the other side. All these 8192 32 bit float values map to the same half float value, so converting a half float to 32 bit and back results in the same half float value regardless of which of the 8192 intermediate 32 bit values was chosen. The extension now results in something like a smoother half step by a factor of sqrt(2) at the transition as shown at the right picture below while the left picture is supposed to visualize the sharp step by a factor of two without anti aliasing. You can safely remove these two lines from the code to get the standard behavior.
covered number space on either side of the returned value:
6.0E-8 ####### ##########
4.5E-8 | #
3.0E-8 ######### ########
The second extension is in the fromFloat() function:
{ // avoid Inf due to rounding
if( ( fbits & 0x7fffffff ) >= 0x47800000 )
...
return sign | 0x7bff; // unrounded not quite Inf
}
This extension slightly extends the number range of the half float format by saving some 32 bit values form getting promoted to Infinity. The affected values are those that would have been smaller than Infinity without rounding and would become Infinity only due to the rounding. You can safely remove the lines shown above if you don't want this extension.
I tried to optimize the path for normal values in the fromFloat() function as much as possible which made it a bit less readable due to the use of precomputed and unshifted constants. I didn't put as much effort into 'toFloat()' since it would not exceed the performance of a lookup table anyway. So if speed really matters could use the toFloat() function only to fill a static lookup table with 0x10000 elements and than use this table for the actual conversion. This is about 3 times faster with a current x64 server VM and about 5 times faster with the x86 client VM.
I put the code hereby into public domain.
The code by x4u encodes the value 1 correctly as 0x3c00 (ref: https://en.wikipedia.org/wiki/Half-precision_floating-point_format). But the decoder with smoothness improvements decodes that into 1.000122. The wikipedia entry says that integer values 0..2048 can be represented exactly. Not nice...
Removing the "| 0x3ff" from the toFloat code ensures that toFloat(fromFloat(k)) == k for integer k in the range -2048..2048, probably at the cost of a bit less smoothness.
Before I saw the solution posted here, I had whipped up something simple:
public static float toFloat(int nHalf)
{
int S = (nHalf >>> 15) & 0x1;
int E = (nHalf >>> 10) & 0x1F;
int T = (nHalf ) & 0x3FF;
E = E == 0x1F
? 0xFF // it's 2^w-1; it's all 1's, so keep it all 1's for the 32-bit float
: E - 15 + 127; // adjust the exponent from the 16-bit bias to the 32-bit bias
// sign S is now bit 31
// exp E is from bit 30 to bit 23
// scale T by 13 binary digits (it grew from 10 to 23 bits)
return Float.intBitsToFloat(S << 31 | E << 23 | T << 13);
}
I do like the approach in the other posted solution, though. For reference:
// notes from the IEEE-754 specification:
// left to right bits of a binary floating point number:
// size bit ids name description
// ---------- ------------ ---- ---------------------------
// 1 bit S sign
// w bits E[0]..E[w-1] E biased exponent
// t=p-1 bits d[1]..d[p-1] T trailing significant field
// The range of the encoding’s biased exponent E shall include:
// ― every integer between 1 and 2^w − 2, inclusive, to encode normal numbers
// ― the reserved value 0 to encode ±0 and subnormal numbers
// ― the reserved value 2w − 1 to encode +/-infinity and NaN
// The representation r of the floating-point datum, and value v of the floating-point datum
// represented, are inferred from the constituent fields as follows:
// a) If E == 2^w−1 and T != 0, then r is qNaN or sNaN and v is NaN regardless of S
// b) If E == 2^w−1 and T == 0, then r=v=(−1)^S * (+infinity)
// c) If 1 <= E <= 2^w−2, then r is (S, (E−bias), (1 + 2^(1−p) * T))
// the value of the corresponding floating-point number is
// v = (−1)^S * 2^(E−bias) * (1 + 2^(1−p) * T)
// thus normal numbers have an implicit leading significand bit of 1
// d) If E == 0 and T != 0, then r is (S, emin, (0 + 2^(1−p) * T))
// the value of the corresponding floating-point number is
// v = (−1)^S * 2^emin * (0 + 2^(1−p) * T)
// thus subnormal numbers have an implicit leading significand bit of 0
// e) If E == 0 and T ==0, then r is (S, emin, 0) and v = (−1)^S * (+0)
// parameter bin16 bin32
// -------------------------------------------- ----- -----
// k, storage width in bits 16 32
// p, precision in bits 11 24
// emax, maxiumum exponent e 15 127
// bias, E-e 15 127
// sign bit 1 1
// w, exponent field width in bits 5 8
// t, trailing significant field width in bits 10 23
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