I am confused about one particular example of the const function. So the type declaration const :: a -> b->a states that the function accepts two parameters of type a and b and returns a type a. For example:
const 5 3 => 5
const 1 2 => 1
This makes sense based on the declaration. However, I ran into this specific example:
const (1+) 5 3 => 4
This makes me question my understanding of the function declaration. I know this function only takes two parameters because I tried:
const 1 5 3
Now this reassures to me that it only takes 2 parameters. So how does this work? Is the (1+) not a parameter? If not, what is it?
const (1+) 5 3 => 4
I know this function only takes two parameters (…)
Every function in Haskell takes one parameter. Indeed, if you write:
const 5 1
then this is short for:
(const 5) 1
The type signature const :: a -> b -> a is a more compact form of const :: a -> (b -> a).
So const 5 will create a function that ignores the parameter (here 1) and returns the value that it was given 5.
Now for const (1+) 5 3 thus thus means that we wrote:
((const (1+)) 5) 3
const (1+) will thus construct a function that ignores the parameter, and returns (1+), hence const (1+) 5 is (1+). We thus then calculate:
(1+) 3
which is 4.