What is happening when I compose * with + in Haskell?

Question

I\'m trying to understand the result of

(*) . (+) 

in Haskell. I know that the composition operator is just the standard composition of ma

Answer 1

There are good answers here, but let me quickly point out a few steps where you went wrong.

First, the correct definition of function composition is

(f . g) x = f (g x)

you omitted the x on the LHS. Next, you should remember that in Haskell h x y is the same as (h x) y. So, contrary to what you expected,

((*) . (+)) 1 2 = (((*) . (+)) 1) 2 = ((*) ((+) 1)) 2 = ((+) 1) * 2,

and now you see why that fails. Also,

((*) . (+)) 1 (\x -> x + 1) 1

does not work, because the constraint Num (Int -> Int) is not satisfied.