C ++中的double类型数字 [英] double type digits in C++

问题描述

IEE754（64位）浮点数应该正确表示15位有效数字，尽管内部表示有17位数据。有没有办法强制第16和第17位数字为零？

Ref：
http://msdn.microsoft.com/en-us/library/system.double（VS.80）。 aspx ：
。
。

请记住，浮点数只能接近十进制数，而浮点数的精度决定了这个数的近似值十进制数字。默认情况下，Double值包含十五个精度的十进制数字，但内部最多保留17位数字。浮点数的精度有几个结果：
。
。

示例：
d1 = 97842111437.390091

d2 = 97842111437.390076

d1和d2小数点后16位和小数点17位的数字不应该是重要的。寻找方法迫使他们归零。即
d1 = 97842111437.390000
d2 = 97842111437.390000

解决方案

反例：将两个最接近的浮点数转换为有理数

  1.11111111111118 
  

（有十进制数字15）是

  1.1111111111111799942818834097124636173248291015625 
 1.1111111111111802163264883347437717020511627197265625 
  

换句话说，没有浮点数以 1.1111111111111800 。

The IEE754 (64 bits) floating point is supposed to correctly represent 15 significant digit although the internal representation has 17 ditigs. Is there a way to force the 16th and 17th digits to zero ??

Ref: http://msdn.microsoft.com/en-us/library/system.double(VS.80).aspx : . .

Remember that a floating-point number can only approximate a decimal number, and that the precision of a floating-point number determines how accurately that number approximates a decimal number. By default, a Double value contains 15 decimal digits of precision, although a maximum of 17 digits is maintained internally. The precision of a floating-point number has several consequences: . .

Example nos: d1 = 97842111437.390091
d2 = 97842111437.390076
d1 and d2 differ in 16th and 17th decimal places that are not supposed to be significant. Looking for ways to force them to zero. ie d1 = 97842111437.390000 d2 = 97842111437.390000

解决方案

No. Counter-example: the two closest floating-point numbers to a rational

1.11111111111118

(which has 15 decimal digits) are

1.1111111111111799942818834097124636173248291015625
1.1111111111111802163264883347437717020511627197265625

In other words, there is not floating-point number that starts with 1.1111111111111800.

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