Haskell functions will not run

Yunfei Chen 0
Viewed 72

I am a beginner in Haskell, I looked at a lot of resources online but I still cannot figure out why the following snippet will not run:

isDivisor :: Int -> Int -> Bool
  isDivisor a b =
    b % a == 0

The code is supposed to take two integers a and b and it needs to figure out if a is a divisor of b. Now Haskell gives me the following:

• In the type ‘Int -> Int -> Bool isDivisor a b’
  In a pattern type signature: Int -> Int -> Bool isDivisor a b
  In the pattern: isDivisor :: Int -> Int -> Bool isDivisor a b

I have no idea why this happens. If someone can explain it to me it would be great!

Edit: Some people suggested I do the following:

isDivisor :: Int -> Int -> Bool
isDivisor a b =
  b % a == 0

Now it gives me:

Variable not in scope: (%) :: Int -> Bool -> Bool

After reading some suggestions I arrived at:

isDivisor :: Int -> Int -> Bool
isDivisor a b = b mod a == 0

It gives me:

• Couldn't match expected type ‘(a0 -> a0 -> a0) -> Int -> a1’
              with actual type ‘Int’
• The function ‘b’ is applied to two value arguments,
    but its type ‘Int’ has none
  In the first argument of ‘(==)’, namely ‘b mod a’
  In the expression: b mod a == 0
haskell
1 Answers

Your code is equivalent to

isDivisor :: Int -> Int -> Bool isDivisor a b =
    b % a == 0

You can't indent the first line of the definition; it treats that line as a continuation of the annotation. a and b in the annotation are type variables, and the result definition uses undefined variables a and b.

Instead, write

isDivisor :: Int -> Int -> Bool
isDivisor a b = 
   b `mod` a == 0

((%) is not defined as the modulo operator in Haskell; you are looking for the mod function. % is defined by Data.Ratio for use in constructing values of type Ratio a.)

chepner0
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