Question

Can someone explain to me with simple words what those two operators do:

$

\

Answer 1

\ is not an operator, it is part of the literal syntax. More precisely, it is part of two literal syntaxes: it denotes a lambda literal and it serves as an escape character in string literals.

The operator $ is defined in the prelude as

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

In other words, it does exactly the same thing as whitespace does, namely just plain function application. However, while function application is left-associative and has high precedence (the highest, in fact), $ is right-associative and has low precedence.

This allows you to omit parentheses when you have chains like “f applied to g applied to h applied to x“, which without the $ operator would be written like

f (g (h x))

but with the operator can be written as

f $ g $ h x

It is also useful if you want to pass the function application operator itself as an argument to another function. Say, you have list of functions and a list of values and you want to apply every function in the list to the corresponding value in the other list:

zipWith ($) fs xs