How do I convert a list to a tuple in Haskell?

Question

How can I best convert a list to a tuple in Haskell:

[1,2,3,4,5,6] -> (1,2,3,4,5,6) 

Answer 1

In a general way, you can't. Each size of tuple is a distinct type, whereas lists of any length are a single type. Thus, there's no good way to write a function that takes a list and returns a tuple of the same length--it wouldn't have a well-defined return type.

For instance, you could have functions like:

tuplify2 :: [a] -> (a,a) tuplify2 [x,y] = (x,y)  tuplify3 :: [a] -> (a,a,a) tuplify3 [x,y,z] = (x,y,z) 

...but not one that does the job of both.

You can write a generic version using various kinds of meta-programming, but you'd rarely want to.

Note that the same problem applies to other things, such as writing class instances for various tuples--take a look at the source code for Data.Tuple from the standard libraries!

Answer 2

Template Haskell is as close as you can get due to type checking if you want to extract a variable number of elements, since (a,b) and (a,b,c) have different types.

{-# LANGUAGE TemplateHaskell #-} import Language.Haskell.TH tuple :: Int -> ExpQ tuple n = do     ns <- replicateM n (newName "x")     lamE [foldr (\x y -> conP '(:) [varP x,y]) wildP ns] (tupE $ map varE ns) 

Then:

$(tuple 6) [1,2,3,4,5,6] == (1,2,3,4,5,6) $(tuple 3) "abc" == ('a','b','c') 

But in general, if you need this answer, then you're asking the wrong question somewhere.

If you just want flat random access, perhaps the better option would be to use an Array.