Function application in Haskell

Question

OK, it's been a long day and my brain may not function at Haskell level, but I just cannot understand one example from 'Learn You a Haskell'.

The section is called Function Application with $, and there is example of how $ may be defined:

($) :: (a -> b) -> a -> b
f $ x = f x

So far everything is clear. I understand all examples in the section, except for the last one:

ghci> map ($ 3) [(4+), (10*), (^2), sqrt]
[7.0,30.0,9.0,1.7320508075688772]

Here we map ($ 3) across list of functions and get result of application of those functions to 3. But how is this possible?

From the first code snippet it's clear that first argument is a function, we can even write:

*Main> ($) sqrt 4
2.0

Now ($ 3) is a partial application of function $, but 3 goes on function's position! So 3 is supposed to be a function or what?

There is another mystery: what the heck is (4+)? I know that (+4) is a partial application of function +, so (4+) should be partial application of function 4? Nonsense. What sort of trick works here?

Answer 1

($ 3) and (+ 4) aren't partial applications - they're operator sections. A partial application would look like (($) 3) or ((+) 4).

An operator section of the form (? x) (where ? stands for an arbitrary infix operator) binds the right operand of the operator, i.e. it is equivalent to \y -> y ? x. Likewise the operator section (x ?) binds the left operand and is thus equivalent to partial application.

Answer 2

I think what's tripping you up is operator sections. These let you partially apply an operator with either one of its arguments, so you can have the operators (+4) and (4+), where 4 is the the second then the first argument to + respectively. A more clear example might be ("Hello" ++) versus (++ "world"), the former prepends "Hello" onto the front of a string, while the latter appends "world" onto the end of a string.

This is contrasted with using operators in prefix form with just parens around it. In this form, the following are equivalent:

> let join = (++)
> join "Hello, " "world"
"Hello, world"
> (++) "Hello, " "world"
"Hello, world"

In prefix form, you treat the operator as a normal function and it accepts its first then second argument in order. In operator sections, it matters which side of the operator the argument is on.

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So in your example, you have the partial application of ($ 3), you can reduce it as

map ($ 3) [(4+), (10*), (^2), sqrt]
[($ 3) (4+), ($ 3) (10 *), ($ 3) (^ 2), ($ 3) sqrt]
[4 + 3, 10 * 3, 3 ^ 2, sqrt 3]
[7.0, 30.0, 9.0, 1.7320508075688772]