为什么通过字符串的往返转换对于 double 不安全？

最近我不得不将一个 double 序列化为文本，然后再将其返回。价值似乎并不等同:

double d1 = 0.84551240822557006;
string s = d1.ToString("R");
double d2 = double.Parse(s);
bool s1 = d1 == d2;
// -> s1 is False

但是根据 标准数字格式字符串的说法，“ R”选项应该能够保证往返的安全。

往返(“ R”)格式说明符用于确保将转换为字符串的数值解析回相同的数值

为什么会这样？

10792 次浏览
小开

在我看来，这只是一个错误。你的期望完全合理。我复制了它。NET 4.5.1(x64) ，运行下面的控制台应用程序，它使用我的 DoubleConverter class.DoubleConverter.ToExactString显示了由 double表示的 一模一样值:

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);
}
}

NET 中的结果:

0.84551240822557
0.845512408225570055719799711368978023529052734375
0.84551240822556994469749724885332398116588592529296875
False

Mono3.3.0中的结果:

0.84551240822557006
0.845512408225570055719799711368978023529052734375
0.845512408225570055719799711368978023529052734375
True

如果手动指定来自 Mono 的字符串(其末尾包含“006”) ，则。NET 将把它解析回原始值。看起来问题在于 ToString("R")处理而不是解析。

正如在其他注释中指出的，这似乎是特定于在 x64CLR 下运行的。如果您编译并运行以上针对 x86的代码，没有问题:

csc /platform:x86 Test.cs DoubleConverter.cs

你得到的结果和单核细胞增多症一样。如果知道这个 bug 是否出现在 RyuJIT 下面，那将是非常有趣的——我现在自己还没有安装它。特别是，我可以想象这个 有可能是一个 JIT 错误，或者很有可能基于体系结构有完全不同的 double.ToString内部实现。

我建议你在 http://connect.microsoft.com上装个窃听器

小开
最佳答案

我找到窃听器了。

. NET 在 clr\src\vm\comnumber.cpp中执行以下操作:

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 非常简单——它只调用位于 C 运行时中的 _ecvt:

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;
}
}

结果是 _ecvt返回字符串 845512408225570。

注意到后面的零了吗? 结果显示，这一点非常重要！
当0出现时，结果实际上解析回 0.84551240822557006，这是您的 原创的数字——所以它比较相等，因此只返回15个数字。

但是，如果我将字符串从0截断到 84551240822557，那么我将返回 0.84551240822556994，也就是 没有，因此它将返回17位数。

证明: 在调试器中运行以下64位代码(其中大部分是从 Microsoft Shared Source CLI 2.0中提取的) ，并在 main末尾检查 v:

#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;
}
小开

最近，我在努力解决这个问题.通过密码指出 double.ToString (“ R”)有以下逻辑:

  1. 尝试将 double 转换为字符串，精度为15。
  2. 将字符串转换回 double 并与原始 double 进行比较。如果它们相同，则返回精度为15的转换后的字符串。
  3. 否则，将 double 转换为字符串，精度为17。

在本例中，double.ToString (“ R”)错误地选择了精度为15的结果，因此出现了错误。在 MSDN 文档中有一个官方的解决方案:

在某些情况下，使用“ R”标准数字格式字符串格式化的 Double 值，如果使用/Platform: x64或/Platform: anycpu 交换机进行编译并在64位系统上运行，则不能成功实现往返。要解决这个问题，可以使用“ G17”标准数字格式字符串格式化 Double 值。下面的示例使用“ R”格式字符串，该字符串的 Double 值不能成功往返，并且还使用“ G17”格式字符串成功往返原始值。

因此，除非这个问题得到解决，否则必须使用 double.ToString (“ G17”)进行往返。

更新 : 现在有 一个具体的问题来跟踪这个 bug。


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