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<  Erlang  ~  Changing element in list?
pottageb
Posted: Sun Jan 23, 2011 7:58 pm Reply with quote
Joined: 23 Jan 2011 Posts: 1
Is there any way of editing a specific element in a list, say the 2nd one?

Thanks
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rvirding
Posted: Mon Jan 24, 2011 11:08 am Reply with quote
User Joined: 30 Aug 2006 Posts: 452 Location: Stockholm, Sweden
For a discussion on this see the following question in Stack Overflow:

http://stackoverflow.com/questions/4776033/how-to-change-an-element-in-a-list-in-erlang

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ultranewb
Posted: Mon Jan 24, 2011 2:50 pm Reply with quote
User Joined: 10 Jan 2011 Posts: 10
Something I find odd is on the Erlang course on the website http://www.erlang.org/course/sequential_programming.html :

* Used to store a variable number of items.
* Lists are dynamically sized.

Yet apparently there cannot be true if I understand the language correctly. Once you have constructed a list, it cannot be changed, no?

Further down on the same web page:

Variables can only be bound once! The value of a variable can never be changed once it has been set (bound).
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rvirding
Posted: Mon Jan 31, 2011 5:53 pm Reply with quote
User Joined: 30 Aug 2006 Posts: 452 Location: Stockholm, Sweden
Well, yes and no depending on how you look at it. A tuple like

{a,b,c,d,e}

is one object so if you want to change one of the elements or add a new element you have to build a completely new tuple containing the old and new elements. A list like

[a,b,c,d,e]

is actually a linked chain of pairs, cons cells, where each one contains one of the elements of the list and a link to the next cons cell. Each cell is immutable but you can step down the list and add new cells to the beginning of the list. It is in this that lists are dynamically sized.

The erlang syntax for a cons cell is [Head|Tail] where the Head is used for the list element and the Tail for the next cons cell. So the above list is really:

[a|[b|[c|[d|[e|[]]]]]]

([] is the empty list, nil, which marks the end of the list). Seeing lists are used so much and the above syntax is cumbersome there is a short form for this which is as I first wrote the list. But they are exactly equivalent, it is purely syntactic sugar, but a very useful one. So the syntax [Head|Tail] can either mean "construct a new list taking the list in the variable Tail and adding a new element Head to it" or if it is a pattern "match the value for a cons cell and put the first element in the variable Head and the rest of the list in the variable Tail".

Immutability of variables and immutability of data are really two separate things, Erlang has both. In Erlang variables are really just references to values (data) and aren't real objects as such. All it means that in a function clause once a Variable references a value, is bound to it, it will always reference that value, it cannot be bound to another value.

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niglas
Posted: Wed Mar 02, 2011 1:59 pm Reply with quote
User Joined: 31 Jan 2011 Posts: 12
If I got it right, there is no way you can achieve this kinda operation, even if you are using lists instead of tuples:

Code:
1> A = [a,b,c,d,e].     % A = [a,b,c,d,e]
2> lists:add(f,A).      % A = [a,b,c,d,e,f]

Nor:
Code:
1> B = [b,c,d,e].       % B =   [b,c,d,e]
2> lists:push(a,B).     % B = [a,b,c,d,e]

Right?
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rvirding
Posted: Sat Mar 05, 2011 12:03 am Reply with quote
User Joined: 30 Aug 2006 Posts: 452 Location: Stockholm, Sweden
I think there is some confusion here. Neither of these two functions exist in lists and even if they did they would both neither modify the original list nor variables referring to them:
niglas wrote:
Code:
1> A = [a,b,c,d,e].     % A = [a,b,c,d,e]
2> lists:add(f,A).      % A = [a,b,c,d,e,f]
Here the intent seems to be to add an element to the end of the list. If this was done a new list would be created (as data can't be changed) and A would still point to the old list.
niglas wrote:
Code:
1> B = [b,c,d,e].       % B =   [b,c,d,e]
2> lists:push(a,B).     % B = [a,b,c,d,e]
Here an element would be prepended to the list without rebuilding the list but B would still refer to the old list. Once a variable is bound to a value it can not be rebound.

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niglas
Posted: Mon Mar 07, 2011 9:15 am Reply with quote
User Joined: 31 Jan 2011 Posts: 12
Great, so I had it right
niglas wrote:
...there is no way you can achieve this kinda operation...
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Posted: Tue Jun 12, 2012 8:56 am Reply with quote
User Joined: 28 Mar 2012 Posts: 32
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