Haskell: Understanding the map function when used with lambda function

Question

I have the following two Haskell expressions:

map (\f x -> f x 5) [(-),(+),(*)]
map (\f x -> f 5 x) [(-),(+),(*)]

And I'm trying to figure out whether either expression above is equivalent to the following expression:

map ($ 5) [(-),(+),(*)]

I am trying to understand what the difference between the first two expressions is. Since for both expressions, there is only one parameter passed to the lambda function (e.g. the operator), the function will be partially applied.

Is it correct to say that the elements in the result list from the first expression will be:

(1) - x 5 = (- x) 5

(2) + x 5 = (+ x) 5

(3) * x 5 = (* x) 5

And for the second expression:

(1) - 5 x = (- 5) x

(2) + 5 x = (+ 5) x

(3) * 5 x = (* 5) x

However, I don't think that either expression is equivalent to map ($ 5) [(-),(+),(*)]. This is because (- x) 5 (where x is a number) gives an error in GHCI and is an invalid expression. Similarly (- 5) x also gives an error. On the other hand, map ($5) [(-)], results in a function that takes a number and subtracts it from 5. Is this reasoning correct? Any insights are appreciated.

duplode duplode · Accepted Answer · 2019-06-14T02:26:40

(- 5) 5 gives out an error because prefix minus is a special case in the language syntax: (- 5) means minus five, the number, and not a function that subtracts five (see also: Currying subtraction). That being so, I will focus on the (+) case, which is not exceptional.

In your second expression, map (\f x -> f 5 x) [(-),(+),(*)], the second element of the result list will be:

(\f x -> f 5 x) (+)

When evaluating such a thing by hand, it is important to be careful to not mix up prefix, infix and sectioned uses of operators. Application here gives out...

\x -> (+) 5 x  -- Prefix syntax (note the parentheses around the operator)

... which is equivalent to...

\x -> 5 + x  -- Infix syntax

... and to:

\x -> (5 +) x  -- Left section
\x -> (+ x) 5  -- Right section

(5 +)          -- Left section, pointfree

So the sections, which are patterned after infix usage of the operators, should be the other way around relative to your question. As for map ($ 5) [(-),(+),(*)], it is equivalent to map (\f x -> f 5 x) [(-),(+),(*)], your second expression. You can confirm that by using the fact that ($) f x = f x to figure out what the ($ 5) right section is.

Haskell: Understanding the map function when used with lambda function

1 Answers