提取尾数和指数的双用C＃ [英] extracting mantissa and exponent from double in c#

问题描述

有没有简单的方法来获得在C＃中（一般或.NET）的尾数和指数从双？

Is there any straightforward way to get the mantissa and exponent from a double in c# (or .NET in general)?

我发现这个例子使用谷歌，但我不知道这将是多么强大的是。莫非二进制重新presentation的双重变化框架将来的某个版本，等等？

I found this example using google, but I'm not sure how robust it would be. Could the binary representation for a double change in some future version of the framework, etc?

我发现另一种方法是使用System.Decimal，而不是双并使用<一个href="http://msdn.microsoft.com/en-us/library/system.decimal.getbits(VS.80).aspx">Decimal.GetBits()方法来提取它们。

The other alternative I found was to use System.Decimal instead of double and use the Decimal.GetBits() method to extract them.

有什么建议？

推荐答案

二进制格式不应该改变 - 这肯定会是一个重大更改现有的规范。它的定义是在IEEE754 / IEC 60559：1989的格式，如吉米说。 （C＃3.0语言规范第1.3节; ECMA 335第8.2.2节）。在code在DoubleConverter应罚款和强大。

The binary format shouldn't change - it would certainly be a breaking change to existing specifications. It's defined to be in IEEE754 / IEC 60559:1989 format, as Jimmy said. (C# 3.0 language spec section 1.3; ECMA 335 section 8.2.2). The code in DoubleConverter should be fine and robust.

有关为了将来参考，的code中的例子中的相关位是：

For the sake of future reference, the relevant bit of the code in the example is:

public static string ToExactString (double d)
{
    …

    // Translate the double into sign, exponent and mantissa.
    long bits = BitConverter.DoubleToInt64Bits(d);
    // Note that the shift is sign-extended, hence the test against -1 not 1
    bool negative = (bits < 0);
    int exponent = (int) ((bits >> 52) & 0x7ffL);
    long mantissa = bits & 0xfffffffffffffL;

    // Subnormal numbers; exponent is effectively one higher,
    // but there's no extra normalisation bit in the mantissa
    if (exponent==0)
    {
        exponent++;
    }
    // Normal numbers; leave exponent as it is but add extra
    // bit to the front of the mantissa
    else
    {
        mantissa = mantissa | (1L<<52);
    }

    // Bias the exponent. It's actually biased by 1023, but we're
    // treating the mantissa as m.0 rather than 0.m, so we need
    // to subtract another 52 from it.
    exponent -= 1075;

    if (mantissa == 0) 
    {
        return "0";
    }

    /* Normalize */
    while((mantissa & 1) == 0) 
    {    /*  i.e., Mantissa is even */
        mantissa >>= 1;
        exponent++;
    }

    …
}

的意见是有道理的，我的时间，但我敢肯定，我现在已经想了关于他们的一段时间。的第一部分之后，你已经得到了原始指数和尾数 - 的code只是帮助治疗他们在一个更简单的方式，其余

The comments made sense to me at the time, but I'm sure I'd have to think for a while about them now. After the very first part you've got the "raw" exponent and mantissa - the rest of the code just helps to treat them in a simpler fashion.

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