问题
Recently I have had to serialize a double into text, and then get it back. The value seems to not be equivalent:
double d1 = 0.84551240822557006;
string s = d1.ToString("R");
double d2 = double.Parse(s);
bool s1 = d1 == d2;
// -> s1 is False
But according to MSDN: Standard Numeric Format Strings, the "R" option is supposed to guarantee round-trip safety.
The round-trip ("R") format specifier is used to ensure that a numeric value that is converted to a string will be parsed back into the same numeric value
Why did this happen?
回答1:
I found the bug.
.NET does the following in clr\src\vm\comnumber.cpp:
DoubleToNumber(value, DOUBLE_PRECISION, &number);
if (number.scale == (int) SCALE_NAN) {
gc.refRetVal = gc.numfmt->sNaN;
goto lExit;
}
if (number.scale == SCALE_INF) {
gc.refRetVal = (number.sign? gc.numfmt->sNegativeInfinity: gc.numfmt->sPositiveInfinity);
goto lExit;
}
NumberToDouble(&number, &dTest);
if (dTest == value) {
gc.refRetVal = NumberToString(&number, 'G', DOUBLE_PRECISION, gc.numfmt);
goto lExit;
}
DoubleToNumber(value, 17, &number);
DoubleToNumber is pretty simple -- it just calls _ecvt, which is in the C runtime:
void DoubleToNumber(double value, int precision, NUMBER* number)
{
WRAPPER_CONTRACT
_ASSERTE(number != NULL);
number->precision = precision;
if (((FPDOUBLE*)&value)->exp == 0x7FF) {
number->scale = (((FPDOUBLE*)&value)->mantLo || ((FPDOUBLE*)&value)->mantHi) ? SCALE_NAN: SCALE_INF;
number->sign = ((FPDOUBLE*)&value)->sign;
number->digits[0] = 0;
}
else {
char* src = _ecvt(value, precision, &number->scale, &number->sign);
wchar* dst = number->digits;
if (*src != '0') {
while (*src) *dst++ = *src++;
}
*dst = 0;
}
}
It turns out that _ecvt returns the string 845512408225570.
Notice the trailing zero? It turns out that makes all the difference!
When the zero is present, the result actually parses back to 0.84551240822557006, which is your original number -- so it compares equal, and hence only 15 digits are returned.
However, if I truncate the string at that zero to 84551240822557, then I get back 0.84551240822556994, which is not your original number, and hence it would return 17 digits.
Proof: run the following 64-bit code (most of which I extracted from the Microsoft Shared Source CLI 2.0) in your debugger and examine v at the end of main:
#include <stdlib.h>
#include <string.h>
#include <math.h>
#define min(a, b) (((a) < (b)) ? (a) : (b))
struct NUMBER {
int precision;
int scale;
int sign;
wchar_t digits[20 + 1];
NUMBER() : precision(0), scale(0), sign(0) {}
};
#define I64(x) x##LL
static const unsigned long long rgval64Power10[] = {
// powers of 10
/*1*/ I64(0xa000000000000000),
/*2*/ I64(0xc800000000000000),
/*3*/ I64(0xfa00000000000000),
/*4*/ I64(0x9c40000000000000),
/*5*/ I64(0xc350000000000000),
/*6*/ I64(0xf424000000000000),
/*7*/ I64(0x9896800000000000),
/*8*/ I64(0xbebc200000000000),
/*9*/ I64(0xee6b280000000000),
/*10*/ I64(0x9502f90000000000),
/*11*/ I64(0xba43b74000000000),
/*12*/ I64(0xe8d4a51000000000),
/*13*/ I64(0x9184e72a00000000),
/*14*/ I64(0xb5e620f480000000),
/*15*/ I64(0xe35fa931a0000000),
// powers of 0.1
/*1*/ I64(0xcccccccccccccccd),
/*2*/ I64(0xa3d70a3d70a3d70b),
/*3*/ I64(0x83126e978d4fdf3c),
/*4*/ I64(0xd1b71758e219652e),
/*5*/ I64(0xa7c5ac471b478425),
/*6*/ I64(0x8637bd05af6c69b7),
/*7*/ I64(0xd6bf94d5e57a42be),
/*8*/ I64(0xabcc77118461ceff),
/*9*/ I64(0x89705f4136b4a599),
/*10*/ I64(0xdbe6fecebdedd5c2),
/*11*/ I64(0xafebff0bcb24ab02),
/*12*/ I64(0x8cbccc096f5088cf),
/*13*/ I64(0xe12e13424bb40e18),
/*14*/ I64(0xb424dc35095cd813),
/*15*/ I64(0x901d7cf73ab0acdc),
};
static const signed char rgexp64Power10[] = {
// exponents for both powers of 10 and 0.1
/*1*/ 4,
/*2*/ 7,
/*3*/ 10,
/*4*/ 14,
/*5*/ 17,
/*6*/ 20,
/*7*/ 24,
/*8*/ 27,
/*9*/ 30,
/*10*/ 34,
/*11*/ 37,
/*12*/ 40,
/*13*/ 44,
/*14*/ 47,
/*15*/ 50,
};
static const unsigned long long rgval64Power10By16[] = {
// powers of 10^16
/*1*/ I64(0x8e1bc9bf04000000),
/*2*/ I64(0x9dc5ada82b70b59e),
/*3*/ I64(0xaf298d050e4395d6),
/*4*/ I64(0xc2781f49ffcfa6d4),
/*5*/ I64(0xd7e77a8f87daf7fa),
/*6*/ I64(0xefb3ab16c59b14a0),
/*7*/ I64(0x850fadc09923329c),
/*8*/ I64(0x93ba47c980e98cde),
/*9*/ I64(0xa402b9c5a8d3a6e6),
/*10*/ I64(0xb616a12b7fe617a8),
/*11*/ I64(0xca28a291859bbf90),
/*12*/ I64(0xe070f78d39275566),
/*13*/ I64(0xf92e0c3537826140),
/*14*/ I64(0x8a5296ffe33cc92c),
/*15*/ I64(0x9991a6f3d6bf1762),
/*16*/ I64(0xaa7eebfb9df9de8a),
/*17*/ I64(0xbd49d14aa79dbc7e),
/*18*/ I64(0xd226fc195c6a2f88),
/*19*/ I64(0xe950df20247c83f8),
/*20*/ I64(0x81842f29f2cce373),
/*21*/ I64(0x8fcac257558ee4e2),
// powers of 0.1^16
/*1*/ I64(0xe69594bec44de160),
/*2*/ I64(0xcfb11ead453994c3),
/*3*/ I64(0xbb127c53b17ec165),
/*4*/ I64(0xa87fea27a539e9b3),
/*5*/ I64(0x97c560ba6b0919b5),
/*6*/ I64(0x88b402f7fd7553ab),
/*7*/ I64(0xf64335bcf065d3a0),
/*8*/ I64(0xddd0467c64bce4c4),
/*9*/ I64(0xc7caba6e7c5382ed),
/*10*/ I64(0xb3f4e093db73a0b7),
/*11*/ I64(0xa21727db38cb0053),
/*12*/ I64(0x91ff83775423cc29),
/*13*/ I64(0x8380dea93da4bc82),
/*14*/ I64(0xece53cec4a314f00),
/*15*/ I64(0xd5605fcdcf32e217),
/*16*/ I64(0xc0314325637a1978),
/*17*/ I64(0xad1c8eab5ee43ba2),
/*18*/ I64(0x9becce62836ac5b0),
/*19*/ I64(0x8c71dcd9ba0b495c),
/*20*/ I64(0xfd00b89747823938),
/*21*/ I64(0xe3e27a444d8d991a),
};
static const signed short rgexp64Power10By16[] = {
// exponents for both powers of 10^16 and 0.1^16
/*1*/ 54,
/*2*/ 107,
/*3*/ 160,
/*4*/ 213,
/*5*/ 266,
/*6*/ 319,
/*7*/ 373,
/*8*/ 426,
/*9*/ 479,
/*10*/ 532,
/*11*/ 585,
/*12*/ 638,
/*13*/ 691,
/*14*/ 745,
/*15*/ 798,
/*16*/ 851,
/*17*/ 904,
/*18*/ 957,
/*19*/ 1010,
/*20*/ 1064,
/*21*/ 1117,
};
static unsigned DigitsToInt(wchar_t* p, int count)
{
wchar_t* end = p + count;
unsigned res = *p - '0';
for ( p = p + 1; p < end; p++) {
res = 10 * res + *p - '0';
}
return res;
}
#define Mul32x32To64(a, b) ((unsigned long long)((unsigned long)(a)) * (unsigned long long)((unsigned long)(b)))
static unsigned long long Mul64Lossy(unsigned long long a, unsigned long long b, int* pexp)
{
// it's ok to losse some precision here - Mul64 will be called
// at most twice during the conversion, so the error won't propagate
// to any of the 53 significant bits of the result
unsigned long long val = Mul32x32To64(a >> 32, b >> 32) +
(Mul32x32To64(a >> 32, b) >> 32) +
(Mul32x32To64(a, b >> 32) >> 32);
// normalize
if ((val & I64(0x8000000000000000)) == 0) { val <<= 1; *pexp -= 1; }
return val;
}
void NumberToDouble(NUMBER* number, double* value)
{
unsigned long long val;
int exp;
wchar_t* src = number->digits;
int remaining;
int total;
int count;
int scale;
int absscale;
int index;
total = (int)wcslen(src);
remaining = total;
// skip the leading zeros
while (*src == '0') {
remaining--;
src++;
}
if (remaining == 0) {
*value = 0;
goto done;
}
count = min(remaining, 9);
remaining -= count;
val = DigitsToInt(src, count);
if (remaining > 0) {
count = min(remaining, 9);
remaining -= count;
// get the denormalized power of 10
unsigned long mult = (unsigned long)(rgval64Power10[count-1] >> (64 - rgexp64Power10[count-1]));
val = Mul32x32To64(val, mult) + DigitsToInt(src+9, count);
}
scale = number->scale - (total - remaining);
absscale = abs(scale);
if (absscale >= 22 * 16) {
// overflow / underflow
*(unsigned long long*)value = (scale > 0) ? I64(0x7FF0000000000000) : 0;
goto done;
}
exp = 64;
// normalize the mantisa
if ((val & I64(0xFFFFFFFF00000000)) == 0) { val <<= 32; exp -= 32; }
if ((val & I64(0xFFFF000000000000)) == 0) { val <<= 16; exp -= 16; }
if ((val & I64(0xFF00000000000000)) == 0) { val <<= 8; exp -= 8; }
if ((val & I64(0xF000000000000000)) == 0) { val <<= 4; exp -= 4; }
if ((val & I64(0xC000000000000000)) == 0) { val <<= 2; exp -= 2; }
if ((val & I64(0x8000000000000000)) == 0) { val <<= 1; exp -= 1; }
index = absscale & 15;
if (index) {
int multexp = rgexp64Power10[index-1];
// the exponents are shared between the inverted and regular table
exp += (scale < 0) ? (-multexp + 1) : multexp;
unsigned long long multval = rgval64Power10[index + ((scale < 0) ? 15 : 0) - 1];
val = Mul64Lossy(val, multval, &exp);
}
index = absscale >> 4;
if (index) {
int multexp = rgexp64Power10By16[index-1];
// the exponents are shared between the inverted and regular table
exp += (scale < 0) ? (-multexp + 1) : multexp;
unsigned long long multval = rgval64Power10By16[index + ((scale < 0) ? 21 : 0) - 1];
val = Mul64Lossy(val, multval, &exp);
}
// round & scale down
if ((unsigned long)val & (1 << 10))
{
// IEEE round to even
unsigned long long tmp = val + ((1 << 10) - 1) + (((unsigned long)val >> 11) & 1);
if (tmp < val) {
// overflow
tmp = (tmp >> 1) | I64(0x8000000000000000);
exp += 1;
}
val = tmp;
}
val >>= 11;
exp += 0x3FE;
if (exp <= 0) {
if (exp <= -52) {
// underflow
val = 0;
}
else {
// denormalized
val >>= (-exp+1);
}
}
else
if (exp >= 0x7FF) {
// overflow
val = I64(0x7FF0000000000000);
}
else {
val = ((unsigned long long)exp << 52) + (val & I64(0x000FFFFFFFFFFFFF));
}
*(unsigned long long*)value = val;
done:
if (number->sign) *(unsigned long long*)value |= I64(0x8000000000000000);
}
int main()
{
NUMBER number;
number.precision = 15;
double v = 0.84551240822557006;
char *src = _ecvt(v, number.precision, &number.scale, &number.sign);
int truncate = 0; // change to 1 if you want to truncate
if (truncate)
{
while (*src && src[strlen(src) - 1] == '0')
{
src[strlen(src) - 1] = 0;
}
}
wchar_t* dst = number.digits;
if (*src != '0') {
while (*src) *dst++ = *src++;
}
*dst++ = 0;
NumberToDouble(&number, &v);
return 0;
}
回答2:
It seems to me that this is simply a bug. Your expectations are entirely reasonable. I've reproduced it using .NET 4.5.1 (x64), running the following console app which uses my DoubleConverter class.DoubleConverter.ToExactString shows the exact value represented by a double:
using System;
class Test
{
static void Main()
{
double d1 = 0.84551240822557006;
string s = d1.ToString("r");
double d2 = double.Parse(s);
Console.WriteLine(s);
Console.WriteLine(DoubleConverter.ToExactString(d1));
Console.WriteLine(DoubleConverter.ToExactString(d2));
Console.WriteLine(d1 == d2);
}
}
Results in .NET:
0.84551240822557
0.845512408225570055719799711368978023529052734375
0.84551240822556994469749724885332398116588592529296875
False
Results in Mono 3.3.0:
0.84551240822557006
0.845512408225570055719799711368978023529052734375
0.845512408225570055719799711368978023529052734375
True
If you manually specify the string from Mono (which contains the "006" on the end), .NET will parse that back to the original value. To it looks like the problem is in the ToString("R") handling rather than the parsing.
As noted in other comments, it looks like this is specific to running under the x64 CLR. If you compile and run the above code targeting x86, it's fine:
csc /platform:x86 Test.cs DoubleConverter.cs
... you get the same results as with Mono. It would be interesting to know whether the bug shows up under RyuJIT - I don't have that installed at the moment myself. In particular, I can imagine this possibly being a JIT bug, or it's quite possible that there are whole different implementations of the internals of double.ToString based on architecture.
I suggest you file a bug at http://connect.microsoft.com
回答3:
Recently, I'm trying to resolve this issue. As pointed out through the code , the double.ToString("R") has following logic:
- Try to convert the double to string in precision of 15.
- Convert the string back to double and compare to the original double. If they are the same, we return the converted string whose precision is 15.
- Otherwise, convert the double to string in precision of 17.
In this case, double.ToString("R") wrongly chose the result in precision of 15 so the bug happens. There's an official workaround in the MSDN doc:
In some cases, Double values formatted with the "R" standard numeric format string do not successfully round-trip if compiled using the /platform:x64 or /platform:anycpu switches and run on 64-bit systems. To work around this problem, you can format Double values by using the "G17" standard numeric format string. The following example uses the "R" format string with a Double value that does not round-trip successfully, and also uses the "G17" format string to successfully round-trip the original value.
So unless this issue being resolved, you have to use double.ToString("G17") for round-tripping.
Update: Now there's a specific issue to track this bug.
来源：https://stackoverflow.com/questions/24299692/why-is-a-round-trip-conversion-via-a-string-not-safe-for-a-double