我如何在C ++中用N个小数字符串化分数

问题描述

我想以可变精度将C ++中的一部分无符号整数字符串化.因此，使用2的precision将1/3打印为0.33.我知道float和std::ostream::precision可以用于快速而肮脏的解决方案:

I want to stringify a fraction of unsigned integers in C++ with variable precision. So 1/3 would be printed as 0.33 using a precision of 2. I know that float and std::ostream::precision could be used for a quick and dirty solution:

std::string stringifyFraction(unsigned numerator,
                              unsigned denominator,
                              unsigned precision)
{
    std::stringstream output;
    output.precision(precision);
    output << static_cast<float>(numerator) / denominator;
    return output.str();
}

但是，这还不够好，因为float的精度有限，并且实际上不能精确地表示十进制数.我还有什么其他选择?如果我想输入100位左右，或者出现重复的小数，即使double也会失败.

However, this is not good enough because float has limited precision and can't actually represent decimal numbers accurately. What other options do I have? Even a double would fail if I wanted 100 digits or so, or in case of a recurring fraction.

推荐答案

始终可以执行长除法以逐位数字化字符串.注意结果由整数部分和小数部分组成.我们可以通过使用/运算符简单地除法并调用std::to_string来获得不可或缺的部分.对于其余部分，我们需要以下功能:

It's always possible to just perform long division to stringify digit-by-digit. Note that the result consists of an integral part and a fractional part. We can get the integral part by simply dividing using the / operator and calling std::to_string. For the rest, we need the following function:

#include <string>

std::string stringifyFraction(const unsigned num,
                              const unsigned den,
                              const unsigned precision)
{
    constexpr unsigned base = 10;

    // prevent division by zero if necessary
    if (den == 0) {
        return "inf";
    }

    // integral part can be computed using regular division
    std::string result = std::to_string(num / den);

    // perform first step of long division
    // also cancel early if there is no fractional part
    unsigned tmp = num % den;
    if (tmp == 0 || precision == 0) {
        return result;
    }

    // reserve characters to avoid unnecessary re-allocation
    result.reserve(result.size() + precision + 1);

    // fractional part can be computed using long divison
    result += '.';
    for (size_t i = 0; i < precision; ++i) {
        tmp *= base;
        char nextDigit = '0' + static_cast<char>(tmp / den);
        result.push_back(nextDigit);
        tmp %= den;
    }

    return result;
}

只需将base用作模板参数，就可以轻松地将其扩展为与其他库一起使用，但是您就不能再使用std::to_string了.

You could easily extend this to work with other bases as well, by just making base a template parameter, but then you couldn't just use std::to_string anymore.

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