[Haskell-beginners] Variable re-use
Grzegorz Milka
grzegorzmilka at gmail.com
Tue Apr 28 10:12:10 UTC 2015
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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