Unfortunately, JavaScript does not provide these two functions. But surely there must have been someone who needed these before me, right? Sure enough, a Google search came up with a few implementations. However, an hour later I was convinced none of them actually are fully equivalent to the C functions. They were imprecise, that is, deconstructing a float using frexp() and reconstructing it with ldexp() did not result in the original value. But that is the basic use case: for all float values, if

[mantissa, exponent] = frexp(value)then

value = ldexp(mantissa, exponent)even if the value is subnormal. None of the implementations (even the complex ones) really worked.

I had to implement it myself, and here is my implementation (also as JSFiddle):

function frexp(value) {My frexp() uses a DataView to extract the exponent bits of the IEEE-754 float representation. If those bits are 0 then it is a subnormal. In that case I normalize it by multiplying with 2

if (value === 0) return [value, 0];

var data = new DataView(new ArrayBuffer(8));

data.setFloat64(0, value);

var bits = (data.getUint32(0) >>> 20) & 0x7FF;

if (bits === 0) { // denormal

data.setFloat64(0, value * Math.pow(2, 64)); // exp + 64

bits = ((data.getUint32(0) >>> 20) & 0x7FF) - 64;

}

var exponent = bits - 1022;

var mantissa = ldexp(value, -exponent);

return [mantissa, exponent];

}

function ldexp(mantissa, exponent) {

var steps = Math.min(3, Math.ceil(Math.abs(exponent) / 1023));

var result = mantissa;

for (var i = 0; i < steps; i++)

result *= Math.pow(2, Math.floor((exponent + i) / steps));

return result;

}

^{64}, getting the bits again, and subtracting 64. After applying the bias, the exponent is ready, and used to get the mantissa by canceling out the exponent from the original value.

My ldexp() is pretty straight-forward, except it needs to be able to multiply by very large and very small numbers. The smallest positive float is 0.5

^{-1073}, and to get its mantissa we need to to multiply with 2

^{1073}. That is larger then the largest float 2

^{1023}. By multiplying in steps we can deal with that. Three steps are needed for e.g. ldexp(5e-324, 1023+1074) which otherwise would result in Infinity.

So there you have it. Hope it's useful to someone.

**Correction:**The code I originally posted here for ldexp() still had a bug, it did not test for too small exponents. I fixed it above, and updated the JSFiddle, too. Also, Nicolas Cellier noticed other rounding and underflow problems, his suggestions for ldexp() are now used above.

## 6 comments:

Hi

I trie to Unterstand your Code complety.

Ohne question. Why you so you Substrate 1022 from the exponent

Exponent = Bits -1022

Thanks for your help.

Greetings Michael

Hi Michael,

for a moment I thought you had found a bug ... but I think it's correct after all:

The exponent is stored with a bias of 1023, but then there is another implicit -1 because the mantissa is stored to be in the range of 0.5...1, that's why I only subtract 1022 not 1023. This is needed so that mantissa * 2 ^ exponent equals the original value.

Hi Bert,

Thanks.

That Sounds correct. Your first answer confused me.

I found your Blog because i'm searching for a convertion of a float 32/64Bit to float16bit. And Vice verse.

Hi.

Me again. Now i unterstand that Code completly.

For the german readers i found a very helpfull article about.

http://www.iti.fh-flensburg.de/lang/informatik/ieee-format.htm

I'm not surprised, even Microsoft can get it wrong when it's about underflow see for example http://stackoverflow.com/questions/32150888/should-ldexp-round-correctly

By the way, the code exhibited here expose a double rounding problem in case of underflow.

See Squeak image side fallback implementation - the one from Kernel-nice.900.mcz -

http://lists.squeakfoundation.org/pipermail/packages/2015-February/007538.html

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