Do IO actions *have* to be glued by the do syntax?
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. Thanks, - Olumide
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
-- Thanks, Quanyang
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
Hi Olumide, Yes, do blocks are syntactic sugar. Yes you are right in that answers here should be .. a bit friendlier. But also in my opinion your last post is a bit.. arrogant sounding... M. Dne Út, 22. březen 2016 v 12:36 h uživatel Olumide napsal:
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
_______________________________________________ Beginners mailing list Beginners@haskell.org http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
Can I also add that this sort of elitist, unhelpful quip of a pseudo-answer is the reason that I've largely avoided asking Haskell beginner questions on stackoverflow. Haskell is a wonderful language with a steep learning curve and I believe that it is the intention of this group grow the user base of the language by create a place where beginners will not be afraid to ask questions. And hasty ill-thought, poorly worded answers like Quanyang's won't help the language get there sooner. - Olumide On 22/03/2016 11:36, Olumide 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
Hi Olumide, I didn't mean quip... If that offended you, I apologize. But if you learn to find the answer out yourself, it will be more efficient, since you don't need to wait for others... For your question, a single query with 'haskell do' as keyword should be enough. On Tue, Mar 22 2016 at 19:46:00 +0800, Olumide wrote:
Can I also add that this sort of elitist, unhelpful quip of a pseudo-answer is the reason that I've largely avoided asking Haskell beginner questions on stackoverflow.
Haskell is a wonderful language with a steep learning curve and I believe that it is the intention of this group grow the user base of the language by create a place where beginners will not be afraid to ask questions. And hasty ill-thought, poorly worded answers like Quanyang's won't help the language get there sooner.
- Olumide
-- Thanks, Quanyang
Quanyang, Thanks for your apology. I apologize too. I do try to find out answers by myself but was unsuccessful in this case probably because the answer lay in a topic/chapter of the book (Monads) that I am yet to read. (I am reading LYH chapter-by-chapter.) Sometimes our best attempts to find an answer will _fail_ ... and that's why this mailing list exists. We should give each other the benefit of the doubt and assume that someone that's asked a question has made some sort of effort to understand before asking. Regards, - Olumide On 22/03/2016 12:06, Quanyang Liu wrote:
Hi Olumide,
I didn't mean quip... If that offended you, I apologize.
But if you learn to find the answer out yourself, it will be more efficient, since you don't need to wait for others...
For your question, a single query with 'haskell do' as keyword should be enough.
On Tue, Mar 22 2016 at 19:46:00 +0800, Olumide wrote:
Can I also add that this sort of elitist, unhelpful quip of a pseudo-answer is the reason that I've largely avoided asking Haskell beginner questions on stackoverflow.
Haskell is a wonderful language with a steep learning curve and I believe that it is the intention of this group grow the user base of the language by create a place where beginners will not be afraid to ask questions. And hasty ill-thought, poorly worded answers like Quanyang's won't help the language get there sooner.
- Olumide
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@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
_______________________________________________ Beginners mailing list Beginners@haskell.org http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
Sorry, autocorrect changed learnhaskell to something else. On 22-Mar-2016 5:16 pm, "Sumit Sahrawat, Maths & Computing, IIT (BHU)" < sumit.sahrawat.apm13@iitbhu.ac.in> 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@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
_______________________________________________ Beginners mailing list Beginners@haskell.org http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
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@web.de mailto:50295@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
_______________________________________________ Beginners mailing list Beginners@haskell.org mailto:Beginners@haskell.org http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
Hello Sumit,
What is the Leavenworth guide? Googling for ‘leavenworth guide haskell’ provides no hint at all. Well, it does give a lot of travel guides!
Andrew
On 22/03/2016, 22:46, "Beginners on behalf of Sumit Sahrawat, Maths & Computing, IIT (BHU)"
Is talking about the learnhaskell guide. The autocorrect showed its magic there :) It's a huge brr Chris Allen, who's currently writing a huge book about Haskell. Regards, Sumit
Again, I meant 'guide by'. On 22-Mar-2016 7:32 pm, "Sumit Sahrawat, Maths & Computing, IIT (BHU)" < sumit.sahrawat.apm13@iitbhu.ac.in> wrote:
Is talking about the learnhaskell guide. The autocorrect showed its magic there :)
It's a huge brr Chris Allen, who's currently writing a huge book about Haskell.
Regards, Sumit
Hello Olumide, The simple answer is "no", do syntax is an entirely optional piece of syntactic sugar that is just there to make the code easier to read in some cases. But, IO actions do have to be glued together by something, and that something is the >>= operator (pronounced "bind").
= takes an IO action on the left, (let's use getLine, which reads a line from the user, for our example) and a function on the right which takes the result of that action and returns the next IO action to perform. For example:
getLine >>= (\line -> putStrLn line)
Here we can see the getLine action on the left. The result of that is passed, through the mechanism provided by >>=, to the function on the right, which is a lambda function taking that line as a parameter and calling the action "putStrLn" to print it back out again. This is equivalent to the following do notation:
do line <- getLine putStrLn line
What if you want to string together more than two actions, though? For example, say we wanted to get two lines and concatenation them in the output, like so:
do line1 <- getLine line2 <- getLine putStrLn (line1 ++ line2)
Since >>= can only take two parameters, what do we do? Well, since the result of >>= is itself an IO action, we can use it again in the function on the right! Like this:
getLine >>= (\line1 -> (getLine >>= (\line2 -> putStrLn (line1 ++ line2))))
Now, this may be starting to look a little bit like lisp with all the parentheses! In fact, because of the associativity of the >>= operator, the brackets are all optional, but they may be helpful for you to see what's going on. Without the brackets, it looks like this:
getLine >>= \line1 -> getLine >>= \line2 -> putStrLn (line1 ++ line2)
Which, if you were to lay it out a little differently, looks like this:
getLine >>= \line1 -> getLine >>= \line2 -> putStrLn (line1 ++ line2)
...Starting to look a little more like the do syntax version above, isn't it? What if you don't care about the value of the parameter? Well, you could ignore it using the _ syntax like this:
getLine >>= \_ -> putStrLn "42"
But Haskell actually provides an operator to save you the effort, called >>, so you can do:
getLine >> putStrLn "42"
Do notation is a purely syntactic transformation that emits code using these operators. Incidentally, the function on the right doesn't have to be a lambda function! Any function taking a value of whatever type is returned by the action on the left will do, and in the case of getLine/putStrLn they match! So the first example could be rewritten:
getLine >>= putStrLn
Which many would argue looks nicer than the equivalent do notation. Hope that helps, sorry if it was a little verbose! -Dani. PS- I know you said not to mention monads, but if you followed everything above you're 90% of the way there anyway -- a monad is just a type class providing the >>= operator, and the "return" function, which takes a value and turns it into an action returning that value (at least in terms of the IO monad).
On Mar 22, 2016, at 19:52, Olumide <50295@web.de> 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.
Thanks,
- Olumide _______________________________________________ Beginners mailing list Beginners@haskell.org http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
participants (6)
-
Andrew Bernard -
Daniel Wright -
Martin Vlk -
Olumide -
Quanyang Liu -
Sumit Sahrawat, Maths & Computing, IIT (BHU)