Question

I want to know what the Haskell operator % does. It is hard to find on google, I could not find it in the Haskell Report either.

I saw it used in this piece of code:

 fi=zz.bu
 bu=zz.(:).(++"zz")
 []#zz=zz;zz#__=zz
 zZ%zz=zZ zz$zZ%zz
 zz=(([[],[]]++).)
 zZ=zipWith;z=zZ((#).show)[1..]$zZ(++)(bu%"Fi")(fi%"Bu")

taken from: https://codegolf.stackexchange.com/questions/88/obfuscated-fizzbuzz-golf/110#110

Answer 1

Here is the relevant section of the Haskell Report:

Haskell provides special syntax to support infix notation. An operator is a function that can be applied using infix syntax (Section 3.4), or partially applied using a section (Section 3.5).

An operator is either an operator symbol, such as + or $$, or is an ordinary identifier enclosed in grave accents (backquotes), such as `op`. For example, instead of writing the prefix application op x y, one can write the infix application x `op` y. If no fixity declaration is given for `op` then it defaults to highest precedence and left associativity (see Section 4.4.2).

Dually, an operator symbol can be converted to an ordinary identifier by enclosing it in parentheses. For example, (+) x y is equivalent to x + y, and foldr (*) 1 xs is equivalent to foldr (\x y -> x * y) 1 xs.

That is to say, there is nothing special about "operators" in Haskell other than their syntax. A function whose name is made from symbols defaults to infix, a function whose name is alphanumeric defaults to prefix, and either can be used in the other style with a bit of extra syntax.

Incidentally, since it’s often impossible to search based on operator names using Google, to find operators that are declared in the standard libraries there are two search engines specifically for finding things on Hackage.