[Haskell-beginners] Variable re-use

[Alexey Shmalko]

I'm sorry, my example should've been:

[1,2,3] >>= \a -> [a+1] >>= \a -> return a

On Tue, Apr 28, 2015 at 1:12 PM Grzegorz Milka <grzegorzmilka at gmail.com>
wrote:

>  Hi,
>
> The variable is not reassigned, but hidden. The do notation is a syntactic
> sugar for:
>
> [1, 2, 3] >>= (\a ->
> [a + 1] >>= (\a ->
> return a))
>
>
> The second 'a' is inside a nested lambda function and therefore inside a
> nested scope. The first 'a' still exists, but the value to which it refers
> is hidden inside the second lambda function, where 'a' is bound to a
> different value.
>
> On 28.04.2015 11:53, Shishir Srivastava wrote:
>
> Hi,
>
> Please can anyone explain how does 'a' get re-used in the code below. My
> understanding so far of haskell is that variables are not allowed to mutate
> or re-assigned.
>
> ---
> do
> a <- [1,2,3]
> a <- [a+1]
> return a
>
> [2,3,4]
> ---
>
> Thanks,
> Shishir
>
>
>
>
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