How to convert a floating point number to its binary representation (IEEE 754) in Javascript?

Question

What's the easiest way to convert a floating point number to its binary representation in Javascript? (e.g. 1.0 -> 0x3F800000).

I have tried to do it manually, and this works to some extent (with usual numbers), but it fails for very big or very small numbers (no range checking) and for special cases (NaN, infinity, etc.):

function floatToNumber(flt)
{
    var sign = (flt < 0) ? 1 : 0;
    flt = Math.abs(flt);
    var exponent = Math.floor(Math.log(flt) / Math.LN2);
    var mantissa = flt / Math.pow(2, exponent);

    return (sign << 31) | ((exponent + 127) << 23) | ((mantissa * Math.pow(2, 23)) & 0x7FFFFF);
}

Am I reinventing the wheel?

EDIT: I've improved my version, now it handles special cases.

function assembleFloat(sign, exponent, mantissa)
{
    return (sign << 31) | (exponent << 23) | (mantissa);
}

function floatToNumber(flt)
{
    if (isNaN(flt)) // Special case: NaN
        return assembleFloat(0, 0xFF, 0x1337); // Mantissa is nonzero for NaN

    var sign = (flt < 0) ? 1 : 0;
    flt = Math.abs(flt);
    if (flt == 0.0) // Special case: +-0
        return assembleFloat(sign, 0, 0);

    var exponent = Math.floor(Math.log(flt) / Math.LN2);
    if (exponent > 127 || exponent < -126) // Special case: +-Infinity (and huge numbers)
        return assembleFloat(sign, 0xFF, 0); // Mantissa is zero for +-Infinity

    var mantissa = flt / Math.pow(2, exponent);
    return assembleFloat(sign, exponent + 127, (mantissa * Math.pow(2, 23)) & 0x7FFFFF);
}

I'm still not sure if this works 100% correctly, but it seems to work good enough. (I'm still looking for existing implementations).

Answer 1

Look at this page: http://babbage.cs.qc.edu/courses/cs341/IEEE-754references.html

Answer 2

Here's a function that works on everything I've tested it on, except it doesn't distinguish -0.0 and +0.0.

It's based on code from http://jsfromhell.com/classes/binary-parser, but it's specialized for 32-bit floats and returns an integer instead of a string. I also modified it to make it faster and (slightly) more readable.

// Based on code from Jonas Raoni Soares Silva
// http://jsfromhell.com/classes/binary-parser
function encodeFloat(number) {
    var n = +number,
        status = (n !== n) || n == -Infinity || n == +Infinity ? n : 0,
        exp = 0,
        len = 281, // 2 * 127 + 1 + 23 + 3,
        bin = new Array(len),
        signal = (n = status !== 0 ? 0 : n) < 0,
        n = Math.abs(n),
        intPart = Math.floor(n),
        floatPart = n - intPart,
        i, lastBit, rounded, j, exponent;

    if (status !== 0) {
        if (n !== n) {
            return 0x7fc00000;
        }
        if (n === Infinity) {
            return 0x7f800000;
        }
        if (n === -Infinity) {
            return 0xff800000
        }
    }

    i = len;
    while (i) {
        bin[--i] = 0;
    }

    i = 129;
    while (intPart && i) {
        bin[--i] = intPart % 2;
        intPart = Math.floor(intPart / 2);
    }

    i = 128;
    while (floatPart > 0 && i) {
        (bin[++i] = ((floatPart *= 2) >= 1) - 0) && --floatPart;
    }

    i = -1;
    while (++i < len && !bin[i]);

    if (bin[(lastBit = 22 + (i = (exp = 128 - i) >= -126 && exp <= 127 ? i + 1 : 128 - (exp = -127))) + 1]) {
        if (!(rounded = bin[lastBit])) {
            j = lastBit + 2;
            while (!rounded && j < len) {
                rounded = bin[j++];
            }
        }

        j = lastBit + 1;
        while (rounded && --j >= 0) {
            (bin[j] = !bin[j] - 0) && (rounded = 0);
        }
    }
    i = i - 2 < 0 ? -1 : i - 3;
    while(++i < len && !bin[i]);
    (exp = 128 - i) >= -126 && exp <= 127 ? ++i : exp < -126 && (i = 255, exp = -127);
    (intPart || status !== 0) && (exp = 128, i = 129, status == -Infinity ? signal = 1 : (status !== status) && (bin[i] = 1));

    n = Math.abs(exp + 127);
    exponent = 0;
    j = 0;
    while (j < 8) {
        exponent += (n % 2) << j;
        n >>= 1;
        j++;
    }

    var mantissa = 0;
    n = i + 23;
    for (; i < n; i++) {
        mantissa = (mantissa << 1) + bin[i];
    }
    return ((signal ? 0x80000000 : 0) + (exponent << 23) + mantissa) | 0;
}

Answer 3

See http://stackoverflow.com/questions/3077718/converting-a-decimal-value-to-a-32bit-floating-point-hexadecimal/3117567#3117567, the second code snippet.