result list of map is one longer [duplicate]

Question

in this Haskell example the result is sometimes one longer than the base list. Any clue for this?

Prelude> let x3 = [1,3..10]
Prelude> x3
[1,3,5,7,9]
Prelude> length x3
5
Prelude> length $ map (+1) x3
5
Prelude> length $ map (*3) x3
5
Prelude> length $ map (/2) x3
6
Prelude> length $ map (/1) x3
6
Prelude> length $ map (`div`1) x3
5
Prelude> map log x3
   [0.0,1.0986122886681098,1.6094379124341003
   ,1.9459101490553132,2.1972245773362196,2.3978952727983707]
Prelude> length it
6

Answer 1

As @duplode suggested, it has to do with how .. behaves with floating point vs. integers:

Prelude> let x3 = [1,3..10] :: [Double]
Prelude> length x3
6
Prelude> let x3 = [1,3..10] :: [Int]
Prelude> length x3
5

Update in response to the comment...

A definition like this in ghci:

let xs = [1,3..11]

is polymorphic. The .. is a shortcut for the enumFromThenTo function:

let xs = enumFromThenTo 1 3 11

The result is polymorphic - the above expression has type:

ghci> :t xs
x3 :: (Enum t, Num t) => [t]

If you just print it out, Haskell chooses to type is as [Integer], and you get:

[1,3,5,7,9]

Here the Integer version of enumFromThenTo is being used.

However, if you apply a floating point operation to the list, Haskell interprets it as [Double], and then it becomes one element longer for the reasons explained above:

ghci> map (+ 0.0) x3
[1.0,3.0,5.0,7.0,9.0,11.0]  -- length 6 now!

So the map operation "changes" the length of the list only because it changes the interpretation of its type which changes which enumFromThenTo function is called to construct it.

Upshot: A list defined like [1,3..11] does not have known length until its type is determined.