CODEWITHSUNDEEP
haskellinputioprogram-entry-point
Greg Peckory
Greg Peckory

Reputation: 8058

Scope of input haskell

I'm having trouble taking in an input. Running it through functions. And outputting it. I have tried to do this in 2 different ways but neither work. Looking around online I see everyone only using the variable of the input inside the main. This would explain why I'm getting a "not in scope" error. But how would this be possible? Here are my two attempts.

result = lcm 3 inp

main = do
   inp <- getLine
   putStr result

and this:

main = do
   inp <- getLine
   putStr result
      where
      result = lcm 3 inp

Upvotes: 1

Views: 91

Answers (2)

Tomo
Tomo

Reputation: 3521

Let's see the types of the things you use:

*Main> :t lcm
lcm :: Integral a => a -> a -> a
*Main> let result inp = lcm 3 inp
*Main> :t result
result :: Integral a => a -> a

But you read in a String:

*Main> :t getLine
getLine :: IO String

So, you need to convert a String to something like an integer, and convert an Inegral the result returned back to Striing for printing.

main = do
    inp <- getLine
    putStr $ show $ result (read inp)

Upvotes: 0

Benesh
Benesh

Reputation: 3428

inp exists only within the scope of the do expression, which explains why you get an error in the first version.

As for the second version. it can be rewritten to:

main = e
  where
    e = do
        inp <- getLine
        putStr result
    result = lcm 3 inp

The two where bindings have different scopes, which is why a local binding from one expression is inaccessible from the other.

This, on the other hand, should work:

main = do
  inp <- getLine
  let result = lcm 3 inp
  putStr result

result is now defined within the do notation scope, so it can use inp. If you still want to use the where clause, result will need to accept inp as an argument:

main = do
  inp <- getLine
  putStr result
  where     
    result inp = lcm 3 inp

Upvotes: 4

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