Why does this Haskell code produce the "infinite type" error?

Posted by Charlie Flowers on Stack Overflow See other posts from Stack Overflow or by Charlie Flowers
Published on 2009-04-27T21:16:38Z Indexed on 2010/05/27 13:41 UTC
Read the original article Hit count: 236

Filed under:

I am new to Haskell and facing a "cannot construct infinite type" error that I cannot make sense of.

In fact, beyond that, I have not been able to find a good explanation of what this error even means, so if you could go beyond my basic question and explain the "infinite type" error, I'd really appreciate it.

Here's the code:

intersperse :: a -> [[a]] -> [a]

-- intersperse '*' ["foo","bar","baz","quux"] 
--  should produce the following:
--  "foo*bar*baz*quux"

-- intersperse -99 [ [1,2,3],[4,5,6],[7,8,9]]
--  should produce the following:
--  [1,2,3,-99,4,5,6,-99,7,8,9]

intersperse _ [] = []
intersperse _ [x] = x
intersperse s (x:y:xs) = x:s:y:intersperse s xs

And here's the error trying to load it into the interpreter:

Prelude> :load ./chapter.3.ending.real.world.haskell.exercises.hs [1 of 1] Compiling Main ( chapter.3.ending.real.world.haskell.exercises.hs, interpreted )

chapter.3.ending.real.world.haskell.exercises.hs:147:0: Occurs check: cannot construct the infinite type: a = [a] When generalising the type(s) for `intersperse' Failed, modules loaded: none.

Thanks.

EDIT: Thanks to the responses, I have corrected the code and I also have a general guideline for dealing with the "infinite type" error in Haskell:

Corrected code

intersperse _ [] = []
intersperse _ [x] = x
intersperse s (x:xs) =  x ++ s:intersperse s xs

What the problem was:

My type signature states that the second parameter to intersperse is a list of lists. Therefore, when I pattern matched against "s (x:y:xs)", x and y became lists. And yet I was treating x and y as elements, not lists.

Guideline for dealing with the "infinite type" error:

Most of the time, when you get this error, you have forgotten the types of the various variables you're dealing with, and you have attempted to use a variable as if it were some other type than what it is. Look carefully at what type everything is versus how you're using it, and this will usually uncover the problem.

© Stack Overflow or respective owner

Related posts about haskell