How to understand an expression where the (.) operator is used as an argument?-CodePudding

I'm new to Haskell and even though I've done the basic, I still don't understand why the following result gives 4:

d f x = f x x
main = print ((d (.)) (d ( )) 1)

CodePudding user response：

d (.) will result in a function \x -> (.) x x which takes as input x a function and will thus take a variable y and applies x two times, so \x y -> x (x y).

The d ( ) expression. d ( ) is again a function that looks like \z -> ( ) z z, or less verbose \z -> z z, this is thus a function that takes as input a number z, and will return z z.

We use this function \z -> z z as variable x in the \x y -> x (x y) expression, so d (.) (d ( )) is equivalent to \y -> (\z -> z z) ((\z -> z z) y).

Finally we use that function with 1 as parameter, the result is thus (\z -> z z) ((\z -> z z) 1) which is equivalent to:

  (\z -> z   z) ((\z -> z   z) 1)
→ (\z -> z   z) (1   1)
→ (\z -> z   z) 2
→ (2   2)
→ 4