If you consider the type of the operator *(:)* you have:
Prelude> :t (:)
(:) :: a -> [a] -> [a]
So it takes an element of type a and a list. So if you write 1:[2,3,4] the
type is correct, because you give an integer 1 and a list of integers
[2,3,4]. You will obtain the list of integers [1,2,3,4]. Similarly, writing
1:[] is correct and gives you [1] as result.
Then, if you write
0 : 1 : []
(as in your example), is the same as
0 : (1 : [])
so it means 0 : [1], which is [0,1]. So, the operator (:) is right
associative.
If it was left associative, your example would give an error. Indeed
(0 : 1) : []
is not correct in Haskell.
Furthermore, your final examples are both false:
Prelude> [] == [] : []
False
[[], []] == [] : []
False
The following is True:
Prelude> [[]] == [] : []
True
Indeed if you write [] : [] youy mean you want to build a list whose first
element (head) is [] and whose "tail" (i.e. the rest of the list) is the
empty list.
So, if 1:[] is [1], then []:[] is [[]].
Ut
2018-08-18 11:13 GMT+02:00 trent shipley
Why does Haskell so often seem to treat [] as a general null.
For example I know 0 : 1 : [] gives [0, 1].
But shouldn't it produce a type fault in a consistent world?
Int:Int:List isn't properly a list. It mixes types.
I expect something like:
Let GSN mean general_scalar_null.
1 : 2 : GSN -- works
1 : 2 : [] -- type fault, cannot mix int and empty list in the same list.
And why does [] == [] : [] instead of [[], []] == [] : []
What sorts of nullity are there in core Haskell?
Trent.
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