[Haskell-beginners] Equivalence (or not) of lists

[Sylvain Henry]

Hi,


 > length' 1:2:3:[]

is equivalent to:


 > (length' 1):2:3:[]

hence the error.


Try:

 > length' (1:2:3:[])


Sylvain


On 12/11/2016 21:20, Lawrence Wickert wrote:
> Hello all,
>
> I am a rank beginner to functional languages.  Working through 
> Lipovaca's book, up to Chapter 3.
>
> Ok, setup this function in editor and compiled:
>
> length' :: (Num b) => [a] -> b
> length' [] = 0
> length' (_:xs) = 1 + length' xs
>
>
> skippy at skippy:~$ ghci
> GHCi, version 7.10.3: http://www.haskell.org/ghc/  :? for help
> Prelude> :l baby
> [1 of 1] Compiling Main             ( baby.hs, interpreted )
> Ok, modules loaded: Main.
> *Main> length' [1,2,3]
> 3
> *Main> 1:2:3:[]
> [1,2,3]
> *Main> length' 1:2:3:[]
>
> <interactive>:5:9:
>     Could not deduce (Num [a0]) arising from the literal ‘1’
>     from the context (Num a)
>       bound by the inferred type of it :: Num a => [a]
>       at <interactive>:5:1-16
>     The type variable ‘a0’ is ambiguous
>     In the first argument of ‘length'’, namely ‘1’
>     In the first argument of ‘(:)’, namely ‘length' 1’
>     In the expression: length' 1 : 2 : 3 : []
> *Main>
>
>
> Obviously, there is something I don't understand about the apparent 
> non-equivalence of the lists [1,2,3] and 1:2:3:[]I am guessing that 
> the solution is contained in that error message but I can't quite 
> decipher it.
>
> Thanks for any help.
>
>
>
>
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