Static result in a recursive Haskell function?

Question

I am a beginner in Haskell programming and have gotten an assignment to create a function that checks if a given string is the substring of another string and if it is it returns the position of the substring in the main string. This is what I have so far:

findString :: String -> String -> Integer
findString mainstring substring 
   | length substring > length mainstring = (-1)
   | take (length substring) mainstring == substring  = 0
   | otherwise = 1 + findString (drop 1 mainstring) substring

The instructions for this assignment explicitly state that I have to use findString :: String -> String -> Integer and that if the substring is not a substring of the mainstring it should return (-1). Right now the recursive part of the function interferes with the result of| length substring > length mainstring = (-1) by adding +1 to it for each recursion but I just want a static (-1). I feel like I'm very close here but I've been stuck on this one for quite some time now. Any help would be appreciated!

Alexey Romanov · Accepted Answer · 2019-11-28 17:17:11Z

2

Instead of just doing otherwise = 1 + findString (drop 1 mainstring) substring, compare the value with -1 first. Look up let and where in your study materials, they will help. I think that's enough given it's an assignment.

answered Nov 28, 2019 at 17:17

Alexey Romanov

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Pawan Kumar · Accepted Answer · 2019-11-28 18:27:55Z

1

Don't fix your thoughts on the type signatures, you can have define functions inside a function as well

findString :: String -> String -> Integer
findString mainstring substring = helper mainstring substring 0
  where 
    helper mainstring substring len
      | length substring > length mainstring = (-1)
      | take (length substring) mainstring == substring = len
      | otherwise = helper (drop 1 mainstring) substring (len + 1)

answered Nov 28, 2019 at 18:27

Pawan Kumar

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1 Comment

RedLine Over a year ago

Thank you very much! I was thinking something like this, a separate variable for counting) just did'nt know what that woud look like in haskell. I didn't really know the syntax for where so I will have to look into that properly.

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