Getting infinite type error when making function that runs its input on itself-CodePudding

So I'm new to Haskell and trying to figure out why this code is not working as intended.

I want to make a function "run" that takes as input some function and returns that function applied to itself.

run::(t1->t2)->t2
run a = a a

The types included are supposed to be general because I want this to work on anything, but I'm getting and infinite type error when I try and I'm not sure why.

CodePudding user response：

You say a :: t1 -> t2. This means its argument should have type t1. Then you apply it to itself, a a. This means its argument is a.

So now we have an equation: the argument should have both the type t1 (because it is an argument to a) and the type t1 -> t2 (because it is a):

t1 ~ t1 -> t2
   ~ (t1 -> t2) -> t2
   ~ ((t1 -> t2) -> t2) -> t2
   ~ (((t1 -> t2) -> t2) -> t2) -> t2
   ~ ...

Each line follows from the previous by replacing t1 by the thing it's equal to, t1 -> t2. No finite types satisfy the given equality.

CodePudding user response：

You write

run::(t1->t2)->t2
run a = a a

Because a is the argument of the function,

a :: t1 -> t2

Now you're applying a to a. The argument of a has type t1, so we must have

a :: t1

The only this can make sense is if t1 is the same as t1 -> t2. So that would mean

t1
= t1 -> t2
= (t1 -> t2) -> t2
= ((t1 -> t2) -> t2) -> t2
= ...

That's probably where your infinite type error comes from. I'm not sure why you don't get a simpler error saying that t1 is not the same as t1 -> t2.

So ... you can't *exactly call a function with itself in Haskell. You can do some similar things, with some care. But how you do it will depend on exactly what you're trying to accomplish.