CODEWITHSUNDEEP
haskell
JJ23
JJ23

Reputation: 1

Accept a list of inputs

I want to accept 5 values from the user. However, in my code (below) if I enter a list like [1,2,3,4,5] I get an error. This code only accepts inputs of the form [999], for example.

Does anyone know how to fix this problem?

putStrLn("Enter 5 binary numbers [,,] : ")
    input<-getLine
    let n=(read input)::[Int]
    let result = convertionTO binaryToDec n
    putStrLn(show result)

In the above code, the line let n=(read input)::[Int] will just accept the user input, [999] or whatever, as one input. Is there a way to enter a list of values?

With the line let result = convertionTO binaryToDec n I'm trying to convert here list of binary values to decimal

Upvotes: 0

Views: 191

Answers (2)

pat
pat

Reputation: 12749

You could put a newtype wrapper around an Integer to give it its own Read and Show instances. That way, you can still use the list parser, but substitute your own parser for the elements:

{-# LANGUAGE GeneralizedNewtypeDeriving #-}

import Numeric ( readInt, showIntAtBase )
import Data.Char ( digitToInt, intToDigit )
import Control.Arrow ( first )

newtype BinNum = BinNum { unBinNum :: Integer }
  deriving ( Eq, Ord, Num )

main = do
  putStrLn "Enter 5 binary numbers [,,] :"
  input <- getLine
  let ns = read input :: [BinNum]
      result = map unBinNum ns
  putStrLn $ show result

isBinDigit :: Char -> Bool
isBinDigit c = c >= '0' && c <= '1'

readBinNum :: ReadS BinNum
readBinNum = map (first BinNum) . readInt 2 isBinDigit digitToInt

showBinNum :: BinNum -> ShowS
showBinNum = showIntAtBase 2 intToDigit . unBinNum

instance Read BinNum where
  readsPrec _ = readBinNum

instance Show BinNum where
  showsPrec _ = showBinNum

Upvotes: 0

dave4420
dave4420

Reputation: 47072

You're asking for the input to be converted into an [Int] using read, so it's being converted using the same rules as Haskell source code, which means they're being parsed as decimal numbers.

As you want to parse them as binary numbers, you will have to write your own String -> Int (or, to allow for errors, String -> Maybe Int) function to convert them into Ints from binary.

There is no distinction between decimal numbers and binary numbers. There are only numbers. Decimal or binary are artifacts of how numbers are represented as strings.

Upvotes: 2

Related Questions