Haskell operator vs function precedence

Question

I am trying to verify something for myself about operator and function precedence in Haskell. For instance, the following code

list = map foo $ xs

can be rewritten as

list = (map foo) $ (xs)

and will eventually be

list = map foo xs

My question used to be, why the first formulation would not be rewritten as

list = (map foo $) xs

since function precedence is always higher than operator precedence, but I think that I have found the answer: operators are simply not allowed to be arguments of functions (except of course, if you surround them with parentheses). Is this right? If so, I find it odd, that there is no mention of this mechanic/rule in RWH or Learn you a Haskell, or any of the other places that I have searched. So if you know a place, where the rule is stated, please link to it.

-- edit: Thank you for your quick answers. I think my confusion came from thinking that an operator literal would somehow evaluate to something, that could get consumed by a function as an argument. It helped me to remember, that an infix operator can be mechanically translated to a prefix functions. Doing this to the first formulation yields

($) (map foo) (xs)

where there is no doubt that ($) is the consuming function, and since the two formulations are equivalent, then the $ literal in the first formulation cannot be consumed by map.

Answer 1

Firstly, application (whitespace) is the highest precedence "operator".

Secondly, in Haskell, there's really no distinction between operators and functions, other than that operators are infix by default, while functions aren't. You can convert functions to infix with backticks

2 `f` x

and convert operators to prefix with parens:

(+) 2 3

So, your question is a bit confused.

Now, specific functions and operators will have declared precedence, which you can find in GHCi with ":info":

Prelude> :info ($)
($) :: (a -> b) -> a -> b   -- Defined in GHC.Base

infixr 0 $

Prelude> :info (+)

class (Eq a, Show a) => Num a where
  (+) :: a -> a -> a

infixl 6 +

Showing both precedence and associativity.

Answer 2

You are correct. This rule is part of the Haskell syntax defined by the Haskell Report . In particular note in Section 3, Expressions, that the argument to function application (an fexp ) must be an aexp . An aexp allows operators as part of sections, and also within a parenthesized expression, but not bare operators.

In map foo $ xs , the Haskell syntax means that this is parsed as two expressions which are applied to the binary operator $ . As sepp2k notes, the syntax (map foo $) is a left section and has a different meaning.

I have to confess I've never thought much about this and actually had to look it up in the Report to see why operators have the behavior they do.

Answer 3

In addition to the information provided by other answers already, note that different operators can have different precedences over other operators, as well as being left-/right- or non-associative. You can find these properties for the Prelude operators in the Haskell 98 Report fixity section .

+--------+----------------------+-----------------------+-------------------+
| Prec-  |   Left associative   |    Non-associative    | Right associative |
| edence |      operators       |       operators       |    operators      |
+--------+----------------------+-----------------------+-------------------+
| 9      | !!                   |                       | .                 |
| 8      |                      |                       | ^, ^^, **         |
| 7      | *, /, `div`,         |                       |                   |
|        | `mod`, `rem`, `quot` |                       |                   |
| 6      | +, -                 |                       |                   |
| 5      |                      |                       | :, ++             |
| 4      |                      | ==, /=, <, <=, >, >=, |                   |
|        |                      | `elem`, `notElem`     |                   |
| 3      |                      |                       | &&                |
| 2      |                      |                       | ||                |
| 1      | >>, >>=              |                       |                   |
| 0      |                      |                       | $, $!, `seq`      |
+--------+----------------------+-----------------------+-------------------+

Any operator lacking a fixity declaration is assumed to be left associative with precedence 9 .

Remember, function application has highest precedence (think of precedence 10 compared to the other precedences in the table) [1] .

Answer 4

The difference is that infix operators get placed between their arguments, so this

list = map foo $ xs

can be rewritten in prefix form as

list = ($) (map foo) xs

which, by the definition of the $ operator, is simply

list = (map foo) xs

Answer 5

Operators can be passed as function arguments if you surround them with parenthesis (ie map foo ($) xs , which would indeed be passed as (map foo ($)) xs ). However if you do not surround them with parenthesis, you are correct that they cannot be passed as argument (or assigned to variables).

Also note that the syntax (someValue $) (where $ could be any operator) actually means something different: it is equivalent to \\x -> someValue $ x , ie it partially applies the operator to its left operand (which in case of $ is a noop of course). Likewise ($ x) partially applies the operator to the right operand. So map ($ x) [f, g, h] would evaluate to [fx, gx, hx] .