CODEWITHSUNDEEP
haskellrecursion
Akash
Akash

Reputation: 969

error parsing in haskell code

My haskell code goes as follows :

module Lain where
let value = 0 
divideBy ::(Ord a, Num a)=>a -> a -> a
divideBy _ 0 = 0
divideBy num den
    | (num - den) >= 0 = do 
        value = value + 1
        return divideBy (num-den) den
    | otherwise = value

The error goes on loading the haskell Lain.hs file is :

app/Lain.hs:18:1: error: parse error (possibly incorrect indentation or mismatched brackets) Failed, modules loaded: none.

Not able to interpret where I am going wrong, seems to me more of a logical mistake. My code wants to print the quotient given numerator and denominator. Please help as to what is exactly my mistake. Thanks in advance.

Upvotes: 0

Views: 45

Answers (1)

Jon Purdy
Jon Purdy

Reputation: 54981

value = value + 1 isn’t a valid statement in a do block. The only valid statements are:

  • A variable binding: let pattern = expression

  • A monadic binding: pattern <- expression

  • An expression that evaluates to a monadic action

However, you don’t need a do block because you don’t need monadic side effects to implement this function. Furthermore, return isn’t like return in an imperative language—it’s not a keyword that returns from the current function, but a function that creates an action that returns a value. It looks like your intent with value = value + 1 was to mutate value, but there are no mutable variables in Haskell. (There are mutable reference types such as IORef, but you don’t need them here.)

So one solution is to simply use an expression:

divideBy :: (Ord a, Num a) => a -> a -> a
divideBy _ 0 = 0
divideBy num den
    | num - den >= 0 = 1 + divideBy (num - den) den
    | otherwise = 0

This says exactly what it means: the quotient of num and den is 0 if den is 0; if num - den >= 0, then it’s 1 more than the quotient of num - den and den; otherwise it’s 0.

Upvotes: 1

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