[erlang-questions] Lists in recursion

harsha sri harsha.pic@REDACTED
Fri Mar 27 18:55:11 CET 2015


Hi Jesper,

Thank you so much, lists:map helped me for my requirement.

Regards,
Harsha


On Fri, Mar 27, 2015 at 6:30 PM, Jesper Louis Andersen <
jesper.louis.andersen@REDACTED> wrote:

>
> On Fri, Mar 27, 2015 at 1:42 PM, harsha sri <harsha.pic@REDACTED> wrote:
>
>> Is it possible to add dynamically add elements to List in recursion?
>>
>
> There is some indication, you may want to spend time with a good book on
> Erlang. For instance learnyousomeerlang.com or perhaps Joe Armstrong's
> book on Erlang. I'm not entirely sure what you want to do, but I can
> explain to you what happens in the system for the lines below.
>
>
>> ex: List = [1,2,3]
>>
>>
> This will create a list [1,2,3] and bind that list to the name List.
>
>
>> lists:foreach...
>>
>>
> calling, for instance, lists:foreach(fun(Item) -> io:format("~B~n",
> [Item]) end, List) will not be able to change List. 'foreach/2' is executed
> for its side-effect only, in this case printing elements. You can use a
> map/2: List2 = lists:map(fun(Item) -> Item + 3 end, List), but this will
>
> * Add three to each element of the list and bind the result to List2,
> i.e., List2 = [4,5,6]
> * Not be able to add or remove elements from L
> * Not change L in any way. The language uses persistent data structures,
> not ephemeral data structures.
>
>
>> .....
>> List = [4 | List]
>>
>
> The expression [4 | List] evaluates to [4,1,2,3] because it prepends
> (cons'es) 4 in front of the list. If you write
>
> List = [4 | List]
>
> then the Erlang semantics are:
>
> * build up the list [4 | List] --> [4,1,2,3]
> * Since 'List' is already bound, assert that the current contents of the
> list is [4,1,2,3]. But List is already bound to [1,2,3], so the assertion
> is failing.
> * Crash the program with a badmatch error since the assertion is a failure
>
> Finally i am expecting in List: [1,2,3,4]
>>
>
> One way to solve this is to append two lists: List ++ [4]. This will build
> the single-element list [4] and then append it as a new suffix. If you want
> to give it a name, it has to be one that hasn't been used already in the
> scope, e.g.,
>
> NewList = List ++ [4].
>
> For small lists, this is a viable strategy. For large lists, append can
> end up being costly in execution speed, but I'd focus on solving problems
> first and then worrying about performance later. Especially when you are
> trying to learn what is going on in the first place.
>
> --
> J.
>
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