是否有类似 Python 中的 Haskell 等效的可迭代解包方式？

Question

I wonder if it is possible to create variables from an iterable of things in Haskell. I found this when I search for it but I couldn't adapt it for my case. Maybe it's not possible or I'm missing something since I'm a beginner. Basically, I'm wondering if something like this is possible in Haskell:

>>> list_of_lists = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
>>> a, b, c = list_of_lists
>>> print(a)
[1, 2, 3]

Answer 1

ghci> list_of_lists = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
ghci> [a,b,c] = list_of_lists
ghci> print a
[1,2,3]

Answer 2

The answer given by luqui works when the list has exactly three elements. It is, however, partial, which means that it'll fail at run-time for lists of any other size.

A more idiomatic Haskell solution, I think, would be a function like this:

listToTriple :: [a] -> Maybe (a, a, a)
listToTriple [a, b, c] = Just (a, b, c)
listToTriple _ = Nothing

You can safely call it with lists of any length:

*Q62157846> listToTriple [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
Just ([1,2,3],[4,5,6],[7,8,9])
*Q62157846> listToTriple [[1, 2, 3], [4, 5, 6]]
Nothing
*Q62157846> listToTriple [[1, 2, 3], [4, 5, 6], [7, 8, 9], [10, 11, 12]]
Nothing

If, in the first case, you only want the first of those three lists, you can pattern-match on the triple:

*Q62157846> fmap (\(a, _, _) -> a) $ listToTriple [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
Just [1,2,3]