What is the most portable and "right" way to do conversion from extended precision float (80-bit value, also known as "long double" in some compilers) to double (64-bit) in MSVC win32/win64?
MSVC currently (as of 2010) assumes that "long double" is "double" synonym.
I could probably write fld/fstp assembler pair in inline asm, but inline asm is not available for win64 code in MSVC. Do I need to move this assembler code to separate .asm file? Is that really so there are no good solution?
If your compiler / platform doesn't have native support for 80 bit floating point values, you have to decode the value yourself.
Assuming that the 80 bit float is stored within a byte buffer, located at a specific offset, you can do it like this:
float64 C_IOHandler::readFloat80(IColl<uint8> buffer, uint32 *ref_offset)
{
uint32 &offset = *ref_offset;
//80 bit floating point value according to the IEEE-754 specification and the Standard Apple Numeric Environment specification:
//1 bit sign, 15 bit exponent, 1 bit normalization indication, 63 bit mantissa
float64 sign;
if ((buffer[offset] & 0x80) == 0x00)
sign = 1;
else
sign = -1;
uint32 exponent = (((uint32)buffer[offset] & 0x7F) << 8) | (uint32)buffer[offset + 1];
uint64 mantissa = readUInt64BE(buffer, offset + 2);
//If the highest bit of the mantissa is set, then this is a normalized number.
float64 normalizeCorrection;
if ((mantissa & 0x8000000000000000) != 0x00)
normalizeCorrection = 1;
else
normalizeCorrection = 0;
mantissa &= 0x7FFFFFFFFFFFFFFF;
offset += 10;
//value = (-1) ^ s * (normalizeCorrection + m / 2 ^ 63) * 2 ^ (e - 16383)
return (sign * (normalizeCorrection + (float64)mantissa / ((uint64)1 << 63)) * g_Math->toPower(2, (int32)exponent - 16383));
}
This is how I did it, and it compiles fine with g++ 4.5.0. It of course isn't a very fast solution, but at least a functional one. This code should also be portable to different platforms, though I didn't try.
Just did this in x86 code...
.686P
.XMM
_TEXT SEGMENT
EXTRN __fltused:DWORD
PUBLIC _cvt80to64
PUBLIC _cvt64to80
_cvt80to64 PROC
mov eax, dword ptr [esp+4]
fld TBYTE PTR [eax]
ret 0
_cvt80to64 ENDP
_cvt64to80 PROC
mov eax, DWORD PTR [esp+12]
fld QWORD PTR [esp+4]
fstp TBYTE PTR [eax]
ret 0
_cvt64to80 ENDP
ENDIF
_TEXT ENDS
END
I've just written this one. It constructs an IEEE double number from IEEE extended precision number using bit operations. It takes the 10 byte extended precision number in little endian format:
typedef unsigned long long uint64;
double makeDoubleFromExtended(const unsigned char x[10])
{
int exponent = (((x[9] << 8) | x[8]) & 0x7FFF);
uint64 mantissa =
((uint64)x[7] << 56) | ((uint64)x[6] << 48) | ((uint64)x[5] << 40) | ((uint64)x[4] << 32) |
((uint64)x[3] << 24) | ((uint64)x[2] << 16) | ((uint64)x[1] << 8) | (uint64)x[0];
unsigned char d[8] = {0};
double result;
d[7] = x[9] & 0x80; /* Set sign. */
if ((exponent == 0x7FFF) || (exponent == 0))
{
/* Infinite, NaN or denormal */
if (exponent == 0x7FFF)
{
/* Infinite or NaN */
d[7] |= 0x7F;
d[6] = 0xF0;
}
else
{
/* Otherwise it's denormal. It cannot be represented as double. Translate as singed zero. */
memcpy(&result, d, 8);
return result;
}
}
else
{
/* Normal number. */
exponent = exponent - 0x3FFF + 0x03FF; /*< exponent for double precision. */
if (exponent <= -52) /*< Too small to represent. Translate as (signed) zero. */
{
memcpy(&result, d, 8);
return result;
}
else if (exponent < 0)
{
/* Denormal, exponent bits are already zero here. */
}
else if (exponent >= 0x7FF) /*< Too large to represent. Translate as infinite. */
{
d[7] |= 0x7F;
d[6] = 0xF0;
memset(d, 0x00, 6);
memcpy(&result, d, 8);
return result;
}
else
{
/* Representable number */
d[7] |= (exponent & 0x7F0) >> 4;
d[6] |= (exponent & 0xF) << 4;
}
}
/* Translate mantissa. */
mantissa >>= 11;
if (exponent < 0)
{
/* Denormal, further shifting is required here. */
mantissa >>= (-exponent + 1);
}
d[0] = mantissa & 0xFF;
d[1] = (mantissa >> 8) & 0xFF;
d[2] = (mantissa >> 16) & 0xFF;
d[3] = (mantissa >> 24) & 0xFF;
d[4] = (mantissa >> 32) & 0xFF;
d[5] = (mantissa >> 40) & 0xFF;
d[6] |= (mantissa >> 48) & 0x0F;
memcpy(&result, d, 8);
printf("Result: 0x%016llx", *(uint64*)(&result) );
return result;
}
Played with the given answers and ended up with this.
#include <cmath>
#include <limits>
#include <cassert>
#ifndef _M_X64
__inline __declspec(naked) double _cvt80to64(void* ) {
__asm {
// PUBLIC _cvt80to64 PROC
mov eax, dword ptr [esp+4]
fld TBYTE PTR [eax]
ret 0
// _cvt80to64 ENDP
}
}
#endif
#pragma pack(push)
#pragma pack(2)
typedef unsigned char tDouble80[10];
#pragma pack(pop)
typedef struct {
unsigned __int64 mantissa:64;
unsigned int exponent:15;
unsigned int sign:1;
} tDouble80Struct;
inline double convertDouble80(const tDouble80& val)
{
assert(10 == sizeof(tDouble80));
const tDouble80Struct* valStruct = reinterpret_cast<const tDouble80Struct*>(&val);
const unsigned int mask_exponent = (1 << 15) - 1;
const unsigned __int64 mantissa_high_highestbit = unsigned __int64(1) << 63;
const unsigned __int64 mask_mantissa = (unsigned __int64(1) << 63) - 1;
if (mask_exponent == valStruct->exponent) {
if(0 == valStruct->mantissa) {
return (0 != valStruct->sign) ? -std::numeric_limits<double>::infinity() : std::numeric_limits<double>::infinity();
}
// highest mantissa bit set means quiet NaN
return (0 != (mantissa_high_highestbit & valStruct->mantissa)) ? std::numeric_limits<double>::quiet_NaN() : std::numeric_limits<double>::signaling_NaN();
}
// 80 bit floating point value according to the IEEE-754 specification and
// the Standard Apple Numeric Environment specification:
// 1 bit sign, 15 bit exponent, 1 bit normalization indication, 63 bit mantissa
const double sign(valStruct->sign ? -1 : 1);
//If the highest bit of the mantissa is set, then this is a normalized number.
unsigned __int64 mantissa = valStruct->mantissa;
double normalizeCorrection = (mantissa & mantissa_high_highestbit) != 0 ? 1 : 0;
mantissa &= mask_mantissa;
//value = (-1) ^ s * (normalizeCorrection + m / 2 ^ 63) * 2 ^ (e - 16383)
return (sign * (normalizeCorrection + double(mantissa) / mantissa_high_highestbit) * pow(2.0, int(valStruct->exponent) - 16383));
}
来源：https://stackoverflow.com/questions/2963055/msvc-win32-convert-extended-precision-float-80-bit-to-double-64-bit