CODEWITHSUNDEEP
haskellrecursionmultiplication
Gonzalo_Arcos
Gonzalo_Arcos

Reputation: 150

Recursive functions with multiple arguments Haskell

A really simple question: I want to implement the multiplication of 2 integers in Haskell. What I wrote doesn't compile:

mult :: Int -> Int -> Int

mult x 1 = x

mult 1 y = y

mult x y = x + (mult x-1 y)

The problem is the last statement. I've tried writing it as:

mult x y = x + (mult x-1 y)

and also

mult x y = x + (mult(x-1,y))

The error I get is:

   Couldn't match expected type `Int' with actual type `Int -> Int'
    In the return type of a call of `mult'

I don't know why the compiler would say that mult returns an Int -> Int when it clearly returns an Int.

Upvotes: 1

Views: 4382

Answers (3)

duplode
duplode

Reputation: 34378

In

mult x y = x + (mult x-1 y)

the expression within the parentheses parses as:

(mult x) - (1 y)

And so the compiler thinks the first argument to (-) is mult x, which is an Int -> Int function because just one argument (rather than two) is being passed. Instead, you want:

mult x y = x + mult (x-1) y

Upvotes: 4

kaan
kaan

Reputation: 796

You have to put x-1 into brackets! Like so

mult x y = x + (mult (x-1) y)

By the way, this does not compute the multiplication of x and y :-) Try some examples... it's a little mistake only.

Upvotes: 4

cassandracomar
cassandracomar

Reputation: 1519

It's a simple parser issue. The compiler is reading mult x-1 y as ((-) (mult x) y) when what you mean is mult (x-1) y. Function application binds very tightly in Haskell, so it's sometimes better to use too many parentheses rather too few, especially while you're still learning the basics of the language.

The error occurs because the type of (mult x) is Int -> Int but the type of y is Int, and you can't subtract those two things.

Upvotes: 0

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