convert string of digits to int in Haskell - haskell

Question

I am trying to write this function in Haskell called scanString, which takes a string and convert it into int if it's composed of only digits and should return 0 otherwise.
For example, scanString "123" = 123 but scanString "12a" = 0.
Here's my implementation so far:
scanChar :: Char -> Int
scanChar c
| 48 <= fromEnum c && fromEnum c <= 57 = (fromEnum c) - fromEnum '0'
| otherwise = 0
scanString :: String -> Int
scanString str = case str of
[] -> 0
x:xs
| 48 <= fromEnum x && fromEnum x <= 57 ->
((scanChar x) * (10 ^ ((length str) -1 ))) + scanString xs
| otherwise -> 0
This code does not do the right thing as scanString "3a" would give 30.
Is there a way (like in Java or Python) where one can simply terminate a function and return a value? Of course, advice on the implementation on this function would be awesome!
Thanks in advance!

Answer 1

The main problem here I think is that you let scanChar :: Char -> Int return both a zero for the zero character ('0') as well as for other characters. As a result the scanString has to include extra logic and this makes it only more complex.
So we can clean the scanChar by for instance returning a -1 (or we could let it return a Maybe Int and let it return Nothing, regardless how you exactly specify it, the key is to try to encapsulate the checking logic in one function, such that we no longer have to care about it). So for example:
scanChar :: Char -> Int
scanChar c | '0' <= c && c <= '9' = fromEnum c - fromEnum '0'
| otherwise = -1
So now we can encapsulate all the digit parsing logic in scanChar. Now we still need to implement scanString :: String -> Int. This can be done by writing an extra function that works with an accumulator. For example:
scanString :: String -> Int
scanString = go 0
where go a s = ...
So here go acts as a function to emulate some sort of while loop. The a parameter is the accumulator, a parameter we pass through recursive calls and each time we can update it with more data. Initially we set it to zero.
The go function has basically three cases:
the end of the string is reached, we can return the accumulator;
the first character of the string is not a digit, we return 0; and
the first character of the string is a digit, we multiply the accumulator with 10, add the parsed value, and perform recursion on the tail of the string.
We can thus implement those three cases like:
scanString :: String -> Int
scanString = go 0
where go a [] = a
go a (x:xs) | 0 <= sc && sc <= 9 = go (10*a+sc) xs
| otherwise = 0
where sc = scanChar x

Answer 2

So you're limited by the specification of the problem that the outermost question be of type String -> Int, but that doesn't mean that your helper function scanChar can't return Maybe Int.
So let's look at doing that:
scanChar :: Char -> Maybe Int
scanChar c
| 48 <= fromEnum c && fromEnum c <= 57 = Just $ (fromEnum c) - fromEnum '0'
| otherwise = Nothing
Now, using the approach in the other answer:
scanString :: String -> Int
scanString = go 0
where go a [] = a
go a (x:xs) = case (scanChar x) of
Nothing -> 0
Just d -> go (10*a + d) xs
where sc = scanChar x

Answer 3

Why not
scanString :: String -> Int
scanString x = if all (`elem` "0123456789") x
then read x :: Int
else 0
Note: it will not read negative integers.
Or:
import Data.Char (isDigit)
scanString' :: String -> Int
scanString' x = if all isDigit x
then read x :: Int
else 0
Also a simple solution here using readMaybe.

Answer 4

You can make the following into a function with one parameter that is a string. This function will take only digits out of the string and pack them into another string which is then then converted.
[read [d | d <- "12a", elem d "1234567890"] :: Int] !! 0
Yields 12