Question

该函数应该计算2个整数N和M之间的整数之和，其中N <= M。

sum(N, M) ->
    if
        N+1 =< M -> N + sum((N+1), M);
        N =< M -> N;
        N > M -> 'N must be <= M'
    end.

有更好的方法吗？我可以把第二个陈述放在第一个吗？

Answer 1

这个怎么样？

sum(N, M) when N > M ->
    {error, 'N must be <= M'};
sum(N, M) ->
    List = lists:seq(N, M),
    lists:foldl(fun(X, Sum) -> X + Sum end, 0, List).

最好使用尾递归，并保证在最佳实践中验证某些内容。

Answer 2

一些评论和主张

-module (ex).

-compile([export_all]).

%% some comments on your proposal
sum1(N, M) ->
    if  %% if statement is not very popular in Erlang since it is not realy an if like in java or C
        N+1 =< M -> N + sum1((N+1), M);
        N =< M -> N;
        N > M -> 'N must be <= M' %% don't do this, it creates useless atoms that are limited
                                  %% in number and cannot be destroyed during the VM life
    end.


%% second solution, tail recursive
sum2(N,M) when N > M -> {error,"N must be <= M"}; %% instead use a guard an return a tuple with
                                                  %% standard atom + a string for the reason
sum2(N,M) -> sum2(N,M,0). %% use an accumulator to create a tail recursive function
                          %% not really important for this example but it is a good usage
                          %% to avoid stack overflow :o)

sum2(M,M,R) -> M+R;
sum2(N,M,R) -> sum2(N+1,M,R+N).  %% like this it is tail recursive.


%% third solution, without recursion: don't use recursive function if it is not necessary
sum3(M,M) -> M;
sum3(0,M) -> sum3(1,M);
sum3(N,M) when N =< M ->  %% don't be defensive unless your function is likely subject to
                          %% receive bad argument from user intreface
    (M*(M+1)-(N-1)*N) div 2.

两个整数之间的因子计算

2 个答案: