haskell - flip fix / fix

Question

>>>flip fix (0 :: Int) (\\a b -> putStrLn \"abc\")
Output: \"abc\"

This is a simplified version of using flip fix.
I saw t

Answer 1

fix is the fixed-point operator. As you probably know from it's definition, it computes the fixed point of a function. This means, for a given function f, it searches for a value x such that f x == x.

How to find such a value for an arbitrary function?

We can view x as the result of infinite term f (f (f ... ) ...)). Obviously, since it is infinite, adding f in front of it doesn't change it, so f x will be the same as x. Of course, we cannot express an infinite term, but we can define fix as fix f = f (fix f), which expresses the idea.

Does it makes sense?

Will it ever terminate? Yes, it will, but only because Haskell is a lazy language. If f doesn't need its argument, it will not evaluate it, so the computation will terminate, it won't loop forever. If we call fix on a function that always uses its argument (it is strict), it will never terminate. So some functions have a fixed point, some don't. And Haskell's lazy evaluation ensures that we compute it, if it exists.

Why is fix useful?

It expresses recursion. Any recursive function can be expressed using fix, without any additional recursion. So fix is a very powerful tool! Let's say we have

fact :: Int -> Int
fact 0 = 1
fact n = n * fact (n - 1)

we can eliminate recursion using fix as follows:

fact :: Int -> Int
fact = fix fact'
  where
    fact' :: (Int -> Int) -> Int -> Int
    fact' _ 0 = 1
    fact' r n = n * r (n - 1)

Here, fact' isn't recursive. The recursion has been moved into fix. The idea is that fact' accepts as its first argument a function that it will use for a recursive call, if it needs to. If you expand fix fact' using the definition of fix, you'll see that it does the same as the original fact.

So you could have a language that only has a primitive fix operator and otherwise doesn't permit any recursive definitions, and you could express everything you can with recursive definitions.

Back to your example

Let's view flip fix (0 :: Int) (\a b -> putStrLn "abc"), it is just fix (\a b -> putStrLn "abc") (0 :: Int). Now let's evaluate:

fix (\a b -> putStrLn "abc") =
(\a b -> putStrLn "abc") (fix (\a b -> putStrLn "abc")) =
\b -> putStrLn "abc"

So the whole expression evaluates to (\b -> putStrLn "abc") (0 :: Int) which is just putStrLn "abc". Because function \a b -> putStrLn "abc" ignores its first argument, fix never recurses. It's actually used here only to obfuscate the code.