Function Dragon4_PrintFloat_Intel_extended

numpy/core/src/multiarray/dragon4.c:2506–2591

* Intel's 80-bit IEEE extended precision floating-point format * * "long doubles" with this format are stored as 96 or 128 bits, but * are equivalent to the 80 bit type with some zero padding on the high bits. * This method expects the user to pass in the value using a 128-bit * FloatVal128, so can support 80, 96, or 128 bit storage formats, * and is endian-independent. * * sign: 1 bi

 * mantissa: 63 bits, first u64
 */
static npy_uint32
Dragon4_PrintFloat_Intel_extended(
    Dragon4_Scratch *scratch, FloatVal128 value, Dragon4_Options *opt)
{
    char *buffer = scratch->repr;
    const npy_uint32 bufferSize = sizeof(scratch->repr);
    BigInt *bigints = scratch->bigints;

    npy_uint32 floatExponent, floatSign;
    npy_uint64 floatMantissa;

    npy_uint64 mantissa;
    npy_int32 exponent;
    npy_uint32 mantissaBit;
    npy_bool hasUnequalMargins;
    char signbit = '\0';

    /* deconstruct the floating point value (we ignore the intbit) */
    floatMantissa = value.lo & bitmask_u64(63);
    floatExponent = value.hi & bitmask_u32(15);
    floatSign = (value.hi >> 15) & 0x1;

    /* output the sign */
    if (floatSign != 0) {
        signbit = '-';
    }
    else if (opt->sign) {
        signbit = '+';
    }

    /* if this is a special value */
    if (floatExponent == bitmask_u32(15)) {
        /*
         * Note: Technically there are other special extended values defined if
         * the intbit is 0, like Pseudo-Infinity, Pseudo-Nan, Quiet-NaN. We
         * ignore all of these since they are not generated on modern
         * processors. We treat Quiet-Nan as simply Nan.
         */
        return PrintInfNan(buffer, bufferSize, floatMantissa, 16, signbit);
    }
    /* else this is a number */

    /* factor the value into its parts */
    if (floatExponent != 0) {
        /*
         * normal
         * The floating point equation is:
         *  value = (1 + mantissa/2^63) * 2 ^ (exponent-16383)
         * We convert the integer equation by factoring a 2^63 out of the
         * exponent
         *  value = (1 + mantissa/2^63) * 2^63 * 2 ^ (exponent-16383-63)
         *  value = (2^63 + mantissa) * 2 ^ (exponent-16383-63)
         * Because of the implied 1 in front of the mantissa we have 64 bits of
         * precision.
         *   m = (2^63 + mantissa)
         *   e = (exponent-16383+1-64)
         */
        mantissa            = (1ull << 63) | floatMantissa;