IEEE 754 浮点表示
如何从十进制转换为 IEEE 745 浮点单精度？ 我可以使用 0.5、0.75 等小数字 我的问题是我不知道如何处理较小的数字。 例如，
12.1325 * 10^-13
How do I convert from decimal to IEEE 745 Floating point single precision ?
I can work with small numbers like 0.5, 0.75, etc
My problem is that I've no idea what to do with smaller numbers.
For example,
12.1325 * 10^-13
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我假设困难在于二进制转换而不是 IEEE 浮点编码（因为你说你知道如何转换 0.5 和 0.75）。对于不太简单的示例，例如 12.1325 * 10-13，我会使用我的 二进制转换器）。到 70 位，0.00000000000121325 是
0.0000000000000000000000000000000000000001010101010111111111100000001111...
（手动）四舍五入到 24 个有效位，即
0.00000000000000000000000000000000000000010101010101111111111
从那里开始，可以很简单地将其编码为浮点数。
另一种方法是分析（不过，您需要一个好的任意精度计算器的帮助，例如 PARI/GP）。首先将 12.1325 * 10-13 重写为 121325/1017，并尝试找到最接近的具有 24 位分子和二次分母幂的数字：
121325/10 17 = x/2n
x = (2n * 121325)/1017
您会发现 n=63 给出了您想要的结果：x = 11190256.123714056749056 = 11190256（四舍五入到最接近的值时）整数）。 （11190256 = 101010101011111111110000 二进制，您会看到它与上面的答案匹配。）标准化浮点时，您将从指数中减去 23，得到： 1.0101010101111111111 * 2-40
（对于后一种方法见我的文章使用大整数将小数正确转换为浮点.)
I'll assume the difficulty is the binary conversion and not the IEEE float encoding (since you say you know how to convert 0.5 and 0.75). For less straightforward examples, like 12.1325 * 10-13, I would use my binary converter). To 70 places, 0.00000000000121325 is
0.0000000000000000000000000000000000000001010101010111111111100000001111...
Rounded (by hand) to 24 significant bits, that's
0.00000000000000000000000000000000000000010101010101111111111
From there, it's straitforward to encode this as a float.
Another way to do this is analytically (you'll need the help of a good arbitrary-precision calculator though, like PARI/GP). Start by rewriting 12.1325 * 10-13 as 121325/1017, and try to find the closest number with a 24-bit numerator and power of two denominator:
121325/1017 = x/2n
x = (2n * 121325)/1017
You'll discover that n=63 gives you what you're after: x = 11190256.123714056749056 = 11190256 (when rounded to the nearest integer). (11190256 = 101010101011111111110000 in binary, and you'll see that matches the answer above.) When you normalize the float, you'll subtract 23 from the exponent, giving: 1.0101010101111111111 * 2-40
(For the latter method, see my article Correct Decimal To Floating-Point Using Big Integers.)