十进制和浮动的精确值为双倍？ [英] Precise value of double in decimal and floating?

问题描述

DBL_MAX 是

  179769313486231570814527423731704356798070567525844996598917476803157260 
 780028538760589558632766878171540458953514382464234321326889464182768467 
 546703537516986049910576551282076245490090389328944075868508455133942304 
 583236903222948165808559332123348274797826204144723168738177180919299881 
 250404026184124858368.000000 
  

但如果我这样做：

  double a = 1234567890123456789.0; 
 printf（％f，a）; 
  

1234567890123456768.000000

这里的精度是17位数。

  double a = 0.1234567890123456789; 
 printf（％。20f，a）; 
  

0.1234567890123456773

这里浮点后的精度是17位数：

  double a = 1234567890.1234567890123456789 ; 
 printf（％。20f，a）; 
  

这将生成：

1234567890.12345671653747558594

现在，浮点后的十进制+ 7的精度将为10位， / p>

这是否意味着我只有17位才能获得双精度的精确值？如果是，为什么DBL_MAX的数字超过300位数字？

解决方案

典型的 double 具有15-16个十进制数字的意义 - 与下一个 double 相比。

所有有限的 double 具有如OP的 DBL_MAX = 1797693134862315708 ... 的确切值。浮点数是对数分布的。所以通常有两种不同的 double 在0.5到1.0之间，因为在2到4之间。

意味着OP的第二大数字只匹配 DBL_MAX 到前300位数字的前16位。见下文。

---

这是否意味着我只有17位才能获得双精度值？如果是，使用DBL_MAX的所有数字的数字是多于300位数字？

double ，因为 binary64 可以正确地表示 / em>约2 ** 64个不同的数字。 0.1234567890123456789 不是其中之一。最近的 double 是 0.1234567890123456773 ... 注意它匹配至少15 （DBL_DIG） 数位。

通常这意味着 double 确切 $ c>与下一个 double 的相对差异不重要。

1e300 或 1e-300 是范围之一，而不是精确度。

---

  int main（void）{
 printf（％。* e\\\
 DBL_DECIMAL_DIG + 2，DBL_MAX）; 
 printf（％。* e\\\
，DBL_DECIMAL_DIG + 2，nextafter（DBL_MAX，0））; 
 printf（％。* e\\\
，DBL_DECIMAL_DIG + 2，nextafter（0，1））; 
} 
  

输出

  1.7976931348623157081e + 308 
 1.7976931348623155086e + 308 
 4.9406564584124654418e-324 
  

考虑使用 printf（％。* e，DBL_DECIMAL_DIG-1，a）; 而不是 printf（ ％.20f，a）; 可以更清楚地查看十进制符号意义一个 double 。如果您准备好十六进制符号，请尝试 printf（％a，a）;

DBL_MAX is

179769313486231570814527423731704356798070567525844996598917476803157260
780028538760589558632766878171540458953514382464234321326889464182768467
546703537516986049910576551282076245490090389328944075868508455133942304
583236903222948165808559332123348274797826204144723168738177180919299881
250404026184124858368.000000

But if I do:

double a=1234567890123456789.0;
printf("%f",a);

1234567890123456768.000000

Here the precision is 17 digits.

double a=0.1234567890123456789;
printf("%.20f",a);

0.1234567890123456773

Here also the precision after the floating point is 17 digits:

double a=1234567890.1234567890123456789;
printf("%.20f",a);

That will generate:

1234567890.12345671653747558594

Now, the precision will be 10 digits for decimal + 7 after the floating point which makes 17.

Does that mean that I have only 17 digits to get a precise value for double? If yes, why is the number of digits of DBL_MAX over 300 digits?

解决方案

Typical double has 15-16 decimal digits of significance - as compared to the next double.

All finite double have exact values like with OP's DBL_MAX = 1797693134862315708.... Floating point numbers are distributed logarithmically. So typically there are as many different double between 0.5 and 1.0 as there are between 2 and 4.

This distribution means OP's 2nd largest number only matches DBL_MAX to first 16 places out of the 300+ digits. See below.

---

Does it mean that I have only 17 digits to get a precise value for double ? If yes what is the use of all that number of digits for DBL_MAX more than 300 digit?

double, as binary64 can represent exactly about 2**64 different numbers. 0.1234567890123456789 is not one of them. The closest double is 0.1234567890123456773... Notice it matches to at least 15 (DBL_DIG) digits.

Usually this means the exact value of a double is not as important as its relative difference to the next double.

The value of 1e300 or 1e-300 is one of range, not precision.

---

int main(void) {
  printf("%.*e\n", DBL_DECIMAL_DIG + 2, DBL_MAX);
  printf("%.*e\n", DBL_DECIMAL_DIG + 2, nextafter(DBL_MAX,0));
  printf("%.*e\n", DBL_DECIMAL_DIG + 2, nextafter(0, 1));
}

Output

1.7976931348623157081e+308
1.7976931348623155086e+308
4.9406564584124654418e-324

Consider using printf("%.*e", DBL_DECIMAL_DIG-1, a); rather than printf("%.20f",a); to get a more clear view of the decimal notation significance of a double. If you are ready for hexadecimal notation, try printf("%a",a);

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