Can someone please help me with the below,

applyTwice :: (a -> a) -> a -> a  
applyTwice f x = f (f x)

I do not understand how the above works. If we had something like ( 3) 10 surely it would produce 13? How is that f (f x). Basically I do not understand currying when it comes to looking at higher order functions.

So what I'm not understanding is if say we had a function of the form a -> a -> a it would take an input a then produce a function which expects another input a to produce an output. So if we had add 5 3 then doing add 5 would produce a function which would expect the input 3 to produce a final output of 8. My question is how does that work here. We take a function in as an input so does partial function application work here like it did in add x y or am I completely overcomplicating everything?

CodePudding user response：

That's not currying, that's partial application.

> :t ( )
( ) :: Num a => a -> a -> a

> :t ( ) 3
( ) 3 :: Num a =>    a -> a

The partial application ( ) 3 indeed produces a function ( 3)^(*) which awaits another numerical input to produce its result. And it does so, whether once or twice.

You example is expanded as

applyTwice ( 3) 10 = ( 3) (( 3) 10) 
                   = ( 3) (10 3)
                   =      (10 3) 3

That's all there is to it.

^(*)(actually, it's (3 ), but it's the same as ( 3) anyway)