将表示二进制数字的字符串转换为基本10字符串haskell [英] Convert a string representing a binary number to a base 10 string haskell

问题描述

我有字符串1001，我想要字符串9。

数字库有（相当笨拙的）showIntAtBase，但我没有能够找到相反的结果。

解决方案

这里或多或少是你从Prelude中寻找的东西。
从数字：

（注意：readInt是showIntAtBase的双重，readDec是showInt的双重，不一致的命名是历史事故。）

  import Data.Char（digitToInt）
 import Data.Maybe（listToMaybe）
 import Numeric（readInt） 
 
 readBin :: Integral a =>字符串 - >也许
 readBin = fmap fst。 listToMaybe。 readInt 2（`elem`01）digitToInt 
  -  readBin1001==只是9 
  

I have the string "1001" and I want the string "9".

The numeric library has the (rather clunky) showIntAtBase, but I haven't been able to find the opposite.

解决方案

Here is more or less what you were looking for from Prelude. From Numeric:

(NB: readInt is the "dual" of showIntAtBase, and readDec is the "dual" of showInt. The inconsistent naming is a historical accident.)

import Data.Char  (digitToInt)
import Data.Maybe (listToMaybe)
import Numeric    (readInt)

readBin :: Integral a => String -> Maybe a
readBin = fmap fst . listToMaybe . readInt 2 (`elem` "01") digitToInt
-- readBin "1001" == Just 9

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