[Haskell-beginners] Question about data structures

matthew coolbeth mac01021 at engr.uconn.edu
Wed Nov 24 17:19:44 EST 2010


May we see a code sample of what you're describing?

Generally, when you're writing haskell programs, you will not approach
the problem in the same way that you would appoach it in a language
with mutable objects.

If you can provide a reasonably simple example of a computing task
that someone would want to perform,  I'm sure someone will be able to
show you how it would be best approached in haskell, and how you need
not write too many lines of code, provided that you use the proper,
idiomatic style.

On 2010-11-24, Russ Abbott <russ.abbott at gmail.com> wrote:
> OK. So putting a Map at Leaf nodes doesn't solve the problem.   (Apparently
> I haven't been able to communicate what I see as the problem.)
>
> The problem that I'm trying to get to is the need to write excessive code
> for something that would require a lot less code in an OO world.  It's not a
> matter of execution time or space. It's a matter of the amount of code one
> is required to write.
> *
> -- Russ *
>
>
>
> On Wed, Nov 24, 2010 at 1:52 PM, Daniel Fischer
> <daniel.is.fischer at web.de>wrote:
>
>> On Wednesday 24 November 2010 22:12:37, Russ Abbott wrote:
>> > Cool. I wasn't aware of that notation.  It doesn't quite get to the
>> > issue though.
>> >
>> > The problem I'm concerned about is the need to define y in the first
>> > place. If one is chasing through a data structure and finds a need to
>> > change something buried within it, it seems necessary to rebuild
>> > everything that includes the changed thing.
>>
>> In general, values are immutable, so you can't "change something buried
>> within it". You have to build a new value containing some of the old stuff
>> and a new part. Building the new value usually consists mostly of copying
>> a
>> couple of pointers (plus building the new part of course), so isn't too
>> expensive normally.
>>
>> You can have mutable values in the IO or (ST s) monads, if you need them.
>>
>> > That is, I can't change a
>> > component of somethingNew without creating y. The point is there's
>> > nothing about x that changed,
>>
>> The thing with the changed component is not x anymore.
>>
>> > and there may be nothing about (var1 x)
>> > that changed, and there may be nothing about var11 . var1 $ x that
>> > changed, etc. Yet one is apparently forced to keep track of and
>> > reconstruct all those elements.
>>
>> The compiler takes care of that.
>>
>> >
>> > Another example is to imagine a Tree in which the leaves contain
>> > "objects." If I want to change a property of one of those leaf objects,
>>
>> You can't in general, the thing with a different property is a different
>> object.
>>
>> > I am forced to rebuild all the ancestor nodes of that leaf down to
>> > rebuilding the root.
>>
>> Yes (well, not you, the compiler does it), except if your tree contains
>> mutable objects (IORefs/STRefs for example).
>>
>> >
>> > One way to avoid that is for the leaves to refer to their objects
>> > through a Map. Then changing a leaf object requires only that the value
>> > associated with key represented by the leaf be (re-)inserted into the
>> > Map.  The Tree itself need not change at all.
>>
>> Oh, it will. If you change a Map, you get a new one, thus you get a new
>> tree containing the new Map.
>>
>> >
>> > But that trick isn't always available.  In the example we are talking
>> > about we can't make a Map where they keys are the instance variable
>> > names and the values are their values.  That would seem to do the job,
>> > but since the values are of different types, we can't create such a Map.
>> >
>> > So now what?
>>
>> Well, what's the problem with the compiler copying some nodes?
>> Normally, that doesn't cost very much performance, if it does in your
>> case,
>> we'd need to know more to suggest the best way to go.
>>
>> > *
>> > -- Russ *
>> >
>>
>


-- 
mac


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