Haskell >> operator with two lists

Question

For a college assignment I am learning Haskell and when reading about the do-notation and sequencing with >>= and >> I came across this behaviour that I did not expect.

[1,2,3] >> [1] -- returns [1,1,1]

Can anyone explain why every element of the first array is replaced by the elements of the second array? It seems like the list is concatenated in some way, while I expected the result of the first expression to be completely ignored, thus I would expect [1] as the result.

Thanks a lot in advance.

Answer 1

The “result” is in this case the values contained in [1,2,3], which are indeed ignored. What >> does not ignore is the context, which for the list monad is the shape (i.e. length) of the list. This can't be ignored, because we must have x >>= pure ≡ x, i.e.

Prelude> [1,2,3] >>= pure
[1,2,3]
Prelude> [1,2,3] >>= \n -> [n]
[1,2,3]
Prelude> [1,2,3] >>= \n -> [1]
[1,1,1]
Prelude> [1,2,3] >>= \_ -> [1]
[1,1,1]
Prelude> [1,2,3] >> [1]
[1,1,1]

An example with length>1 on the RHS:

[1,2,3] >>= \n -> [n, n+10]
[1,11,2,12,3,13]
Prelude> [1,2,3] >>= \n -> [100, 10]
[100,10,100,10,100,10]
Prelude> [1,2,3] >> [100, 10]
[100,10,100,10,100,10]