recursion - Erlang-递归删除

Question

我正在用 erlang 编写一个递归函数，给定一个元素 X 和一个列表，从列表中删除元素 X 并返回新列表。我相信我已经正确编写了它，但是，当我对其进行测试时，我陷入了无限循环..

delete(_,[]) -> [];
delete(X,[X|_]) -> [];
delete(X,[Y|YS]) -> 
    if X == Y -> YS;
    true -> [Y] ++ delete(X,[YS]) % I believe the infinite loop is a result of this line..
end.

我对 erlang 非常陌生(这是我使用该语言的第二个项目)，因此故障排除对我来说有点困难，但如果有人可以提供一些指导，我将不胜感激。预先感谢您!

Answer 1

delete(_,[]) -> []; %% ok removing anything from an empty list gives an empty list
delete(X,[X|_]) -> []; %% big mistake. If you find the element you want to remove on top
                       %% of the list, you must remove it and continue with the rest of the list
delete(X,[Y|YS]) -> 
    if X == Y -> YS;   %% this will never occurs since you already test this case
                       %% in the previous clause. An the result should be delete(X,YS), not YS.
    true -> [Y] ++ delete(X,[YS]) %% correct
end.

我不明白哪里有无限循环，但第二个子句会使递归调用过早停止。

所以你的代码应该是:

delete(_,[]) -> [];
delete(X,[X|Rest]) -> delete(X,Rest);
delete(X,[Y|YS]) -> [Y] ++ delete(X,[YS]).

但是我建议使用列表理解来实现非常短的代码和快速的执行(这是lists:filter/2中使用的代码):

delete(X,L) -> [Y || Y <- L, Y =/= X].
%              ^        ^       ^
%              |        |       |_ when Y different from X
%              |        |_________ with all the elements Y from L
%              |__________________ make a list

在 shell 中定义函数，你会得到:

1> D = fun D(_,[]) -> [];       
1>         D(X,[X|R]) -> D(X,R);
1>         D(X,[Y|R]) -> [Y] ++ D(X,R) end.
#Fun<erl_eval.36.90072148>
2> D(4,[1,2,3,4,5,6]).
[1,2,3,5,6]
3> D1 = fun(X,L) -> [Y || Y <- L, Y =/= X] end.
#Fun<erl_eval.12.90072148>
4> D1(4,[1,2,3,4,5,6]).                        
[1,2,3,5,6]
5>