[Haskell-beginners] let (x',seed') = x seed

[Johan Tibell]

Hi Amy,

On Thu, Apr 29, 2010 at 5:45 PM, Amy de Buitléir <amy at nualeargais.ie> wrote:

> I'm confused by the syntax this function definition I found in a tutorial:
>
> bind :: (a → StdGen → (b,StdGen)) → (StdGen → (a,StdGen)) → (StdGen →
> (b,StdGen))
> bind f x seed = let (x',seed') = x seed in f x' seed'
>
> The part that confuses me is "let (x',seed') = x seed". The left side
> is a tuple, and the right side is two separate values, so how does the
> binding work? I gather from the context of the example (a tutorial on
> monads) that the value of x is bound to x', and the value of seed is
> bound to seed', so that "bind f" is a function that acts on a tuple
> and produces a new tuple. Is that correct?
>
>
The names and parenthesis are a bit confusing. First, the type signature

    bind :: (a → StdGen → (b,StdGen)) → (StdGen → (a,StdGen)) → (StdGen →
(b,StdGen))

is equivalent to

    bind :: (a → StdGen → (b,StdGen)) → (StdGen → (a,StdGen)) → StdGen →
(b,StdGen)

as → associates to the right

So in

    bind f x seed

* f is a function of type (a → StdGen → (b,StdGen)),
* x is a function of type (StdGen → (a,StdGen)),
* and seed is a value of type StdGen.

We can now see how let

    (x',seed') = x seed

makes sense: x is applied to seed, producing a value of type (a,StdGen), a
tuple, which components are assigned to the variables x' and seed'. It would
perhaps have been clearer if x was called g or some other name that would
make it clear that it's not of the same type as x'.

Hope this helps.

-- Johan
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