1

I'm trying to write this function applying composition and partial application with Haskell:

function m n = (m^2) + n

I tried this with:

function m = (m^2).(+)
Willem Van Onsem
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cde33
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1 Answers1

4

The problem with that approach is that (+) is a binary operator. Since you put it at the right of the dot ., it will not be applied to the left operand. So you have written:

function :: Num a => a -> a -> a
function m = (.) (m^2) (+)  -- wrong

This is short for:

function m = \n -> ((m^2) ((+) n))

So that means that (+) n will result in a function (n+) and we will apply that function to the result of (m^2), which does not make much sense.

You can however simply use:

function :: Num a => a -> a -> a
function m = (+) (m^2)

Or:

function :: Num a => a -> a -> a
function m = ((m^2) +)

Given function m = (+) (m^2), if we apply n on that function, we will obtain:

((+) (m^2)) n
-> (+) (m^2) n
-> (m^2) + n

You can further modify the function and drop the m argument as well, with:

function :: Num a => a -> a -> a
function = (+) . (^ 2)

Which is syntactical sugar for:

function :: Num a => a -> a -> a
function = (.) (+) (^2)

If we now apply m on the function, it will evaluate to:

((.) (+) (^2)) m
-> (\x -> (+) ((^2) x)) m
-> (+) ((^2) m)
-> (+) (m^2)

So we obtain the state like in the previous command.

Willem Van Onsem
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