了解带函数保护的Erlang Basic递归

rxztt3cl 于 2022-12-08 发布在 Erlang

关注(0)|答案(1)|浏览(124)

I've been looking at some Erlang recursion examples asked earlier here on stackoverflow.
Specifically with this question Erlang basic recursion with guards
But I can't quite understand how the code works. So I have created a module to see the results it returns with a simple list with 3 elements.

-module(recursion). 
-export([start/0]). 

start() -> 
    List = [1,2,3],
    Final = delete(1,List),
    Final.
   

delete(_, []) -> 
    io:format("return []~n~n~n"),
    [];
delete(Del, [Del|Xs]) ->
    io:format("~p =:= ~p~n",[Del,Del]),
    io:format("delete(~p, ~p)~n~n~n~n",[Del,Xs]),
    delete(Del, Xs);
delete(Del, [X|Xs]) ->
    io:format("~p =/= ~p~n",[Del,X]),
    io:format(" [~p|delete(~p, ~p)]~n~n~n~n",[X,Del,Xs]),
    [X|delete(Del, Xs)].

And this is the result of the log

1> recursion:start().
1 =:= 1
delete(1, [2,3])


1 =/= 2
 [2|delete(1, [3])]                


1 =/= 3
 [3|delete(1, [])]               

return []    The result of the 'Final' variable of the main function, shouldn't it be []?
             bacause [3|delete(3, [])] in the last call matches with delete(_, []) -> []
             
             or is it this way?  [2,[3,[]]] -> [2,3]

[2,3]
2>

My question is: Every time the program calls delete(Del, [X|Xs]) ->, is the function returning the value to the previous call? Is it being stored somewhere? or is it just something like that? [2,[3,[]]] -> [2,3]
edit: I think I have found the solution in this link, about how the final result is built https://learnyousomeerlang.com/starting-out-for-real#lists where

13> List = [2,3,4].
[2,3,4]
14> NewList = [1|List].
[1,2,3,4]

So [2|[3|[]]] -> [2,3]
is that so?

erlang

来源：https://stackoverflow.com/questions/65567714/understanding-erlang-basic-recursion-with-function-guards

1条答案

按热度按时间

xfb7svmp1#

Yes.
This is how the function works:

If the head of the list matches the first argument, 1 in this case, then the head of the list isn't saved anywhere, i.e. it's skipped, and the function is called again with the tail of the list:

delete(Del, [Del|Xs]) ->
     delete(Del, Xs);

If the head of the list does NOT match the first argument, then the head of the list is saved by adding it to a result list:

[X|delete(Del, Xs)].

When the list is empty, the function returns [] , which is very important when cons'ing elements together:

[3 | f(X) ]

if f(X) does not return a list at some point, then the list won't be a proper list. A proper list, such as:

[1, 2, 3]

is equivalent to:

[1 | [2 | [3 | [] ]]]

as you can see here:

2> [1 | [2 | [3 | [] ]]].
    [1,2,3]

When you write:

[X|delete(Del, Xs)]

that's a little be tricky, and you need some experience to know how that works. You can understand things better, by writing out by hand what is happening:

delete(1, [1,2,3])
        |
        V
   delete(1, [2, 3])
           |
           V
      [2 | delete(1, [3]) ]  %% Can't know the result here without determining the return value of delete(1, [3])
                    |
                    V
                [3 | delete(1, []) ]  %% Can't know the result here without determining the return value of delete(1, [])
                          |
                          V
                          []

Once, you've got the return value [] , because there are no more function calls in the result, now you can move upwards substituting:

delete(1, [1,2,3])
        |
        V
   delete(1, [2, 3])
           |
           V
      [2 | delete(1, [3]) ]
                    |
                    V
                [3 | [] ]

And substituting again:

delete(1, [1,2,3])
        |
        V
   delete(1, [2, 3])
           |
           V
      [2 | [3 | [] ] ]

which is equivalent to:

[2, 3]

Here is a conceptually simpler version of delete() :

start() -> 
    List = [1,2,3],
    Final = delete(1,List),
    Final.
   

delete(Term, List) ->
    delete(Term, List, _Acc=[]).

delete(_, [], Acc) -> 
    Acc;
delete(Term, [Term|Xs], Acc) ->
    delete(Term, Xs, Acc);
delete(Term, [X|Xs], Acc) ->
    delete(Term, Xs, [X|Acc]).

However, the result is:

[3, 2]

So, when you use an accumulator variable, you need to reverse the final result:

delete(_, [], Acc) -> 
    lists:reverse(Acc);

赞(0）回复(0）举报 2022-12-08

我来回答

了解带函数保护的Erlang Basic递归

1条答案

相关问题

热门标签

最新问答