Defining variables inside a function Haskell

Question

I am a huge newbie to Haskell, I actually just started 10 minutes ago. I am trying to figure out how to define a variable inside a function. Lets say I have the function

foo :: Int -> Int
foo a = 
    b = a * 2
    b
-- Yes, I know, it doesn't do anything interesting

When I run it in GHCi I get a syntax error! How can you define a variable inside a function?

Related: http://stackoverflow.com/q/993124 – Robert Harvey Jul 17 '12 at 21:48 — Robert Harvey, Jul 17 '12 at 21:48
Read [Learn You a Haskell](http://learnyouahaskell.com). – Cat Plus Plus Jul 17 '12 at 21:56 — Cat Plus Plus, Jul 17 '12 at 21:56

score 31 · Answer 1 · answered Jul 17 '12 at 22:17

31

There are two ways to do this:

foo a = b where b = a * 2
foo a = let b = a * 2 in b

In most cases, the choice between them is an aesthetic rather than technical one. More precisely, where may only be attached to definitions, whereas let ... in ... may be used anywhere an expression is allowed. Both where and let introduce blocks, making multiple internal variables convenient in both cases.

answered Jul 17 '12 at 22:17

Daniel Wagner

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Hey I was wondering how you would define multiple helper variables inside a function, would you do something like `where b = a * 2, c = a * 3` or would you use an entirely different "where" statement? – Eben Cowley Mar 04 '17 at 19:58
@EbenCowley A single `where`/`let` will do. Use `;` to separate declarations on the same line, or read about the indentation rules to use multiple lines. (This is the meaning of the last sentence of my post.) – Daniel Wagner Mar 04 '17 at 20:11
@DanielWagner : I am very new to haskell. so one basic question : does `foo a = b where b = a * 2` basically translates to `foo a=(/b->b)a * 2` ? – rahulaga-msft Dec 06 '19 at 12:31
@rahulaga_dev That's a decent first approximation, but as always the truth is a bit more complicated. One difference of between lambdas and let/where blocks that's important here is that the former make their variable monomorphic and the latter don't. For example, `foo :: (Int, Double); foo = (x, x) where x = 3` works fine, but `foo :: (Int, Double); foo = (\x -> (x, x)) 3` does not. – Daniel Wagner Dec 06 '19 at 15:28

score 3 · Answer 2 · answered Jul 17 '12 at 21:55

3

Disregarding technical correctness, the answer is "sort of".

I think it's better to think of a variable as a function of zero arguments evaluating to a given value.

module Main where
import System.IO

foo :: Integer -> Integer
foo a =
  b where
    b = a * 2

main = do
  putStrLn $ show $ foo 10

answered Jul 17 '12 at 21:55

Paul Nathan

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It is not a function of zero arguments. Every function in Haskell takes precisely one argument. – Magnus Kronqvist Jul 17 '12 at 21:59
2

For more on what Magnus said, see Conal Elliott's blog post ["Everything is a function" in Haskell?](http://conal.net/blog/posts/everything-is-a-function-in-haskell). – Antal Spector-Zabusky Jul 17 '12 at 23:47
@MagnusKronqvist: Would you mind providing a document giving the documentation for that? – Paul Nathan Jul 17 '12 at 23:48
4

Functions are the values that have arrow type. Some values do, some values don't. But if it doesn't have an arrow type it's not a function. – augustss Jul 18 '12 at 00:00
@PaulNathan http://www.haskell.org/haskellwiki/Currying - if you allow something informal. Also you can google "Haskell currying". – Magnus Kronqvist Jul 18 '12 at 06:10
This is also detailed in Learn You a Haskell. – drumfire Dec 25 '14 at 15:08

Defining variables inside a function Haskell

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