[Haskell-beginners] Do IO actions *have* to be glued by the do syntax?
Olumide
50295 at web.de
Tue Mar 22 11:56:21 UTC 2016
Sumit,
I do, and did, look up other resources but none emphasized that IO
actions have to be bound, and that's why I asked here. I'm sure it will
become clearer when I reach the chapter on monads.
@Martin: I realize that I am equally responsible for the culture of the
list and will also try to be friendlier ...
- Olumide
On 22/03/2016 11:46, Sumit Sahrawat, Maths & Computing, IIT (BHU) wrote:
> Olumide, do blocks are indeed syntactic sugar. They are a piece of
> syntax that converts to a specific type of expressions, namely those
> using the bind (>>=) operator.
>
> It is really helpful, when learning to be able to come up with answers
> in your own. Learning Haskell has always involved combining many
> different perspectives on the same topic, told differently in different
> places. It would only help you if you took the time and dug further for
> answers. It's really satisfying too!
>
> Also, if you haven't, search for the Leavenworth guide. It takes this
> very question in one of the miscellaneous discussions.
>
> Be nice, be respectful. Have fun learning Haskell :)
>
> On 22-Mar-2016 5:07 pm, "Olumide" <50295 at web.de <mailto:50295 at web.de>>
> wrote:
>
> Quanyang,
>
> do bocks are *not* sugar, and please refer to Daniel's answer for an
> example of how to answer a beginner's question.
>
> - Olumide
>
> On 22/03/2016 11:06, Quanyang Liu wrote:
>
> On Tue, Mar 22 2016 at 18:52:41 +0800, Olumide wrote:
>
> Hello List,
>
> Do IO actions *have* to be glued by the do syntax? Many, if
> not all,
> the examples that I've come across in LYH seem to suggest
> so. And if
> so, why?
>
> BTW, if possible I'd appreciate an explanation that does not
> resort to
> monads. I haven't studied them yet and I'm sure I'd struggle to
> understand any explanation that resorts to monads.
>
>
> I think you should google that first... do blocks are just sugar...
>
> Thanks,
>
> - Olumide
>
>
>
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