在C＃中从double中提取尾数和指数

二进制格式不应该更改-对现有规范肯定是一个重大更改。正如Jimmy所说，它被定义为IEEE754 / IEC
60559：1989格式。（C＃3.0语言规范第1.3节； ECMA 335第8.2.2节）。DoubleConverter中的代码应该很好并且健壮。

为了将来参考，示例中代码的相关位为：

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 & (1L << 63)) != 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 negative ? "-0" : "0";    }        while((mantissa & 1) == 0)     {            mantissa >>= 1;        exponent++;    }    …}

当时的评论对我来说很有意义，但是我敢肯定我现在必须考虑一下。在第一部分之后，您将获得“原始”指数和尾数-其余代码仅有助于以一种更简单的方式对待它们。