转换双人/浮点数到字符串 [英] Convert Double/Float to string

问题描述

我需要一个浮点数转换为等效的字符串十进制（或其他碱）。转换首先，需要在格式做 XE + 0 ，其中 X 是浮点数。

I need to convert a floating point number to an equivalent string in decimal (or other base). Conversion at first, needs to be done in the format xE+0 where x is the floating point number.

这个想法我已经是浮点数首先截断到一个临时整数，则该整数转换成字符串，然后再考虑小数部分，用 10 而小数部分不会成为 0 。小数部分被转移到小数点的左边后，再次应用整数字符串功能和分数部分转换为字符串。有没有更好的办法，这将是比这个更快？这种方法将促使任何种类的副作用？

The idea i have is to first truncate the floating point number into a temporary integer and then convert that integer into string, and then consider the fractional part, multiply it with 10 while the fractional part does not become 0. After the fractional part is transferred into the left side of the decimal point, apply the integer to string function again and convert the fraction part to string. Is there a better way, which will be faster than this? Will this method induce any kind of side effects?

要转换浮点数为指数重新presentation我该怎么同上，然后调整电源？或者直接在位​​掩码IEEE754浮点重新presentation，每个部分转换成字符串。

To convert the floating point number into exponential representation shall i do the same as above and then adjust the power? Or directly bitmask the IEEE754 floating point representation and convert each part into string.

的注：没有其他功能可以使用的，因为我有机会获得绝对没有库函数。这code进入玩具内核。的

推荐答案

唯一确切的解决方案是为基数转换执行arbitrary- precision小数运算，因为准确的值可以非常长 - 80位长双，高达约10000位小数。幸运的是它的只有高达约700位左右的IEEE 双击。

The only exact solution is to perform arbitrary-precision decimal arithmetic for the base conversion, since the exact value can be very long - for 80-bit long double, up to about 10000 decimal places. Fortunately it's "only" up to about 700 places or so for IEEE double.

，而不是与个别十进制数字的工作，这是有益的，而不是工作基-1-十亿（最高功率的10，适合在一32位整数），然后这些基-1-十亿数字转换为9十进制数字每个在你计算结束。

Rather than working with individual decimal digits, it's helpful to instead work base-1-billion (the highest power of 10 that fits in a 32-bit integer) and then convert these "base-1-billion digits" to 9 decimal digits each at the end of your computation.

我有一个非常密集的（而难以阅读），但有效的实施在这里，在<击> LGPL MIT许可证：

I have a very dense (rather hard to read) but efficient implementation here, under LGPL MIT license:

的http：// git.musl-libc.org/cgit/musl/blob/src/stdio/vfprintf.c?h=v1.1.6

如果你去掉所有的十六进制浮点支持，无限/楠支持，％G / ％F / ％E 变化的支持，倒圆（这将永远不会被需要，如果你只想要确切的答案），你可能不需要其他的东西，剩下的code是相当简单

If you strip out all the hex float support, infinity/nan support, %g/%f/%e variation support, rounding (which will never be needed if you only want exact answers), and other things you might not need, the remaining code is rather simple.

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