Recently I have been learning about compositions in haskell and am now a bit confused concerning this example.
(const . min) 3 0 4
As a result I get 3, so internally it must be computed like:
const (min 3) 0 4
But I thought since min takes two arguments, it should look like this:
const (min 3 0) 4
So apparently the composition only takes this one argument, in this case 3, instead of all the arguments for min like I expected it to. Does that mean compositions only take one argument per default or what am I not getting here?
You can answer your question by manually evaluating the initial expression.
(const . min) 3 0 4
⇒ (\x -> const (min x)) 3 0 4 * defn of (.)
⇒ (const (min 3)) 0 4 * function application
⇒ ((\x y -> x) (min 3)) 0 4 * defn of const
⇒ (\y -> min 3) 0 4 * function application
⇒ (min 3) 4 * function application
⇒ 3 * because 3 is less than 4
It is worth noting that function application is left-associative, so, fxy
means the same as (fx) y
.
It is also worth noting that \xy -> z
means the same as \x -> \y -> z
.
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