Haskell: example of function of type a -> a, besides the identity

Question

I\'ve just started playing a little with Haskell... I want to write a function of the same type of the identity. Obviously, not equivalent to it. That would be something lik

Answer 1

Let me formulate an answer that elaborates on dbaupp’s comment. Any function of type a -> a would also give rise to an function of type () -> (), so I will look at this subproblem first.

A usual semantics of Haskell types and functions would represent a type as a pointed chain-complete partial order, and functions as continuous functions. The type () is represented by the two element set {⊥,()} with the order ⊥⊏(). In plain set theory, there are 2^2=4 functions from this set onto itself, but only three of them are continuous:

• f1: ⊥ ↦ ⊥, () ↦ ⊥,
• f2: ⊥ ↦ ⊥, () ↦ (), and
• f3: ⊥ ↦ (), () ↦ ().

So in our semantic model, there are three different functions of type () -> (). But which of them can be implemented in Haskell? All of them!

• f1 _ = undefined (or f1 x = f1 x)
• f2 x = x (or f2 = id)
• f3 _ = () (or f3 = const ())

Looking at these definitions, you can see that f1 and f2 can also be used to define a function of type a -> a. As they do different things already on (), they are different. So we have at least two different functions of type a -> a.

In the above semantic model, there are many more functions of type a -> a, but these would not be expressible in Haskell (this is related to parametricity and Wadler’s Theorems for Free). A proper proof that f1 and f2 are the only such functions does not seem to be very easy, as it depends on what the Haskell language disallows (e.g. no pattern matching on the type of the argument).