如何将Rational转换为"pretty"细绳?

本文介绍了如何将Rational转换为"pretty"细绳?的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！

问题描述

我想以小数位数显示一些Rational值.即，我宁愿显示0.75，也不显示3 % 4.我希望此函数的类型为Int -> Rational -> String.第一个Int用于指定最大小数位数，因为Rational扩展名可能是无终止的.

I want to display some Rational values in their decimal expansion. That is, instead of displaying 3 % 4, I would rather display 0.75. I'd like this function to be of type Int -> Rational -> String. The first Int is to specify the maximum number of decimal places, since Rational expansions may be non-terminating.

嘘而针对Data.Ratio的黑线没有帮帮我.在哪里可以找到此功能?

Hoogle and the haddocks for Data.Ratio didn't help me. Where can I find this function?

推荐答案

这是不使用浮点数的任意精度解决方案:

Here is an arbitrary precision solution that doesn't use floats:

import Data.Ratio

display :: Int -> Rational -> String
display len rat = (if num < 0 then "-" else "") ++ (shows d ("." ++ take len (go next)))
    where
        (d, next) = abs num `quotRem` den
        num = numerator rat
        den = denominator rat

        go 0 = ""
        go x = let (d, next) = (10 * x) `quotRem` den
               in shows d (go next)

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rational