haskell - Why does this function work even though the argument is missing? -

i'm trying understand following piece of code:

import data.char (ord)  encodeinteger :: string -> integer encodeinteger = read . concatmap ch     ch c = show (ord c)

but don't see how can work when encodeinteger defined function takes string, in second line, function implemented without string argument.

also, concatmap (according hoogle), takes function , list, function ch provided.

why code still work? argument somehow magically passed? has currying?

edit: , why doesn't work change this:

encodeinteger :: string -> integer encodeinteger = read . concatmap ch     ch c = show (ord c)

basically defining function

f = g

is same defining function

f x = g x

in specific case, can use

encodeinteger = (read . concatmap ch)

to define function. parentheses needed, otherwise parsed as

encodeinteger = (read) . (concatmap ch a)

and concatmap ch a not function , can not composed. @ write

encodeinteger = read (concatmap ch a) -- or encodeinteger = read $ concatmap ch

about "why concatmap ch takes 1 argument?". partial application, common in haskell. if have

f x y z = x+y+z

you can call f fewer arguments, , obtain result function of remaining arguments. e.g., f 1 2 function taking z , returning 1+2+z.

concretely, currying, there's no such thing function taking 2 or more arguments. every function takes 1 argument. when have function like

foo :: int -> bool -> string

then foo takes 1 argument, int. returns function, takes bool , returns string. can visualize writing

foo :: int -> (bool -> string)

anyway, if currying , partial application, find plenty of examples.

Brent