ieee-754 - 将科学计数法中的十进制数转换为 IEEE 754

Question

我已经阅读了一些文本和线程，展示了如何从小数转换为 IEEE 754，但我仍然对如何在不扩展小数的情况下转换数字（以科学记数法表示）感到困惑

我特别使用的数字是9.07 * 10^23，但任何数字都可以；对于我的特定示例，我将弄清楚如何做到这一点。

Answer 1

我假设您希望结果是最接近十进制数的浮点数，并且您使用的是双精度浮点数。

对于大多数数字，有一种方法可以相对快速地完成。简而言之，这是它的工作原理。

您需要将数字拆分为具有精确表示为浮点数的乘积或数字的一部分。可精确表示的 10 的最大幂是 10^22。因此，要获得浮点形式的 9.07e+23，我们可以这样写：

9.07e+23 = 907 * 10^21

根据 IEEE-754 标准，保证单个浮点运算被正确舍入，因此上述乘积计算为 2 个双精度浮点数的乘积，将给出正确舍入的结果。

如果您要在转换函数中使用它，您可能会将 10 的幂存储在一个数组中。

请注意，您不能将此方法用于 9.07e-23。这个数字等于 907 / 10^23，因此分母太大而无法精确表示。在这种情况下，以及其他处理非常大或非常小的数字时，您必须使用某种形式的高精度算术。

有关更多详细信息和示例，请参阅快速路径十进制到浮点转换。

Answer 2

Converting a number from a decimal string to binary IEEE is fairly straight-forward if you know how to do IEEE floating-point addition and multiplication. (or if you're using any basic programming language like C/C++)

There's a lot of different approaches to this, but the easiest is to evaluate 9.07 * 10^23 directly.

First, start with 9.07:

9.07 = 9 + 0 * 10^-1 + 7 * 10^-2

Now evaluate 10^23. This can be done by starting with 10 and using any powering algorithm.

Then multiply the results together.

Here's a simple implementation in C/C++:

double mantissa = 9;
mantissa += 0 / 10.;
mantissa += 7 / 100.;

double exp = 1;
for (int i = 0; i < 23; i++){
    exp *= 10;
}

double result = mantissa * exp;

Now, going backwards (IEEE -> to decimal) is a lot harder.

Again, there's also a lot of different approaches. Here's the easiest one I can think of it.

I'll use 1.0011101b * 2^40 as the example. (the mantissa is in binary)

First, convert the mantissa to decimal: (this should be easy, since there's no exponent)

1.0011101b * 2^40 = 1.22656 * 2^40

Now, "scale" the number such that the binary exponent vanishes. This is done by multiplying by an appropriate power of 10 to "get rid" of the binary exponent.

1.22656 * 2^40 = 1.22656 * (2^40 * 10^-12) * 10^12
               = 1.22656 * (1.09951) * 10^12
               = 1.34861 * 10^12

So the answer is:

1.0011101b * 2^40 = 1.34861 * 10^12

In this example, 10^12 was needed to "scale away" the 2^40. Determining the power of 10 that is needed is simply equal to:

power of 10 = (power of 2) * log(2)/log(10)