你如何打印（无的printf）的IEEE754多少？ [英] How do you print out an IEEE754 number (without printf)?

问题描述

有关这个问题的目的，我做的不的必须使用的printf 设施（我不能告诉你为什么，可惜的能力，但我们只是现在假设我知道我在做什么）。

For the purposes of this question, I do not have the ability to use printf facilities (I can't tell you why, unfortunately, but let's just assume for now that I know what I'm doing).

有关的IEEE754单precision号，您有以下位：

For an IEEE754 single precision number, you have the following bits:

SEEE EEEE EFFF FFFF FFFF FFFF FFFF FFFF

其中，取值为符号，电子是指数和˚F是小部分。

where S is the sign, E is the exponent and F is the fraction.

打印标志是相对容易的所有情况下，为的是抓住所有的特殊情况下，如 NaN的（电子== 0xFF的，女！= 0 ），天道酬勤（电子== 0xFF的，女== 0 ）和 0 （电子== 0，女== 0 ，视为特殊只是因为指数的偏见是不是在使用情况）。

Printing the sign is relatively easy for all cases, as is catching all the special cases like NaN (E == 0xff, F != 0), Inf (E == 0xff, F == 0) and 0 (E == 0, F == 0, considered special just because the exponent bias isn't used in that case).

我有两个问题。

首先是如何最好地把denormalised号码（如电子== 0，女！= 0 ）转换成标准化的号码（如 1＆LT = E＆LT; = 0xFE的）？我怀疑这将是必要的简化答案的一个问题（但我可能是错的可以随意来教育我）。

The first is how best to turn denormalised numbers (where E == 0, F != 0) into normalised numbers (where 1 <= E <= 0xfe)? I suspect this will be necessary to simplify the answer to the next question (but I could be wrong so feel free to educate me).

第二个问题是如何打印出的归一化数字。我希望能够打印出来有两种方式，像指数 -3.74195E3 和非指数如 3741.95 。虽然，只是看着那两个并排的一面，它应该是相当容易，只需移动小数点扭转前者为后者。因此，让我们把心思的指数形式。

The second question is how to print out the normalised numbers. I want to be able to print them out in two ways, exponential like -3.74195E3 and non-exponential like 3741.95. Although, just looking at those two side-by-side, it should be fairly easy to turn the former into the latter by just moving the decimal point around. So let's just concentrate on the exponential form.

我有我以前早就打印出来PI，你使用的日益减少公式之一，并保持对可能性的上限和下限，输出数字时，这两个限制答应了，移算法的模糊的记忆计算由10（所以当上限和下限分别为 3.2364 和 3.1234 ，您可以输出的一个因素 3 ，并在计算调整的）。

I have a vague recollection of an algorithm I used long ago for printing out PI where you used one of the ever-reducing formulae and kept an upper and lower limit on the possibilities, outputting a digit when both limits agreed, and shifting the calculation by a factor of 10 (so when the upper and lower limits were 3.2364 and 3.1234, you could output the 3 and adjust for that in the calculation).

但它是一个的长的，因为我这样做，所以我甚至不知道这是在这里采取合适的进场时间。如此看来，因为每个位的值的一半，通过小数部分移动时（在previous位的 1/2 ， 1 / 4 ， 1/8 等）。

But it's been a long time since I did that so I don't even know if that's a suitable approach to take here. It seems so since the value of each bit is half that of the previous bit when moving through the fractional part (1/2, 1/4, 1/8 and so on).

我真的preFER的不的有去通过的printf 源$ C ​​$ C跋涉，除非绝对必要的，因此，如果任何人可以帮助出这一点，我会感激不尽。

I would really prefer not to have to go trudging through printf source code unless absolutely necessary so, if anyone can help out with this, I'll be eternally grateful.

推荐答案

如果你想获得对每次转换确切的结果，你将不得不使用arbitrary- precision算法，在做的printf（）实现。如果你想要得到的结果是接近，也许只有在它们的最小显著位（S）不同，那么一个很简单的双precision基于算法就足够了：对于整数部分，反复除以十追加余数形成十进制字符串（反向）;为小数部分，反复十繁殖并减去关闭的整数部分以形成十进制字符串

If you want to get exact results for every conversion, you'll have to use arbitrary-precision arithmetic, as done in printf() implementations. If you want to get results that are "close," perhaps differing only in their least significant digit(s), then a very simple double-precision based algorithm will suffice: for the integer part, repeatedly divide by ten and append the remainders to form the decimal string (in reverse); for the fractional part, repeatedly multiply by ten and subtract off the integer parts to form the decimal string.

我最近写了一篇关于这种方法的文章：的http://www.exploringbinary.com/quick-and-dirty-floating-point-to-decimal-conversion/ 。它不打印科学记数法，但是这应该是微不足道的补充。该算法打印正规数（我印的那些准确的走了出来，但你必须做更全面的测试）。

I recently wrote an article about this method: http://www.exploringbinary.com/quick-and-dirty-floating-point-to-decimal-conversion/ . It does not print scientific notation, but that should be trivial to add. The algorithm prints subnormal numbers (the ones I printed came out accurately, but you'd have to do more thorough testing).

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