erlang的递归不起作用。获取函数子句匹配错误

Posted 2021-04-13

所以我的递归出了问题。我得到一个匹配错误的函数子句。当每个学生收到相同数量的糖果时，递归停止。我确定我的递归退出条件正确。怎么了？

-module(assignment3). 
-export([main/0, students/4, givecandy/4, teachergivecandy/4]).

main() ->
    io:fwrite("Hello
"),
    students(4, 2, 2, 1). %Define how many candy each students get

givecandy(Student1, Student2, Student3, Turn) when (Student1 == Student2) and (Student2 == Student3) -> 1;
% recursivecandy(Student1, Student2, Student3, Turn) when (Student1 /= Student2) and (Student2 /= Student3) -> givecandy().

givecandy(Student1, Student2, Student3, Turn) when (Student1 /= Student2) and (Student2 /= Student3) ->
    io:format("Turn123"),
    Student1Total = (Student1 div 2) + (Student3 div 2),
    Student2Total = (Student2 div 2) + (Student1 div 2),
    Student3Total = (Student3 div 2) + (Student2 div 2),
    TurnCount = Turn + 1,
    teachergivecandy(Student1Total, Student2Total, Student3Total, TurnCount).

students(X, Y, Z, Turn) ->
    Student1 = X,
    Student2 = Y,
    Student3 = Z,
    TurnCount = Turn,
    io:format("Turn: ~p~n" , [TurnCount]),
    io:format("Student 1 Total: ~p~n" , [Student1]),
    io:format("Student 2 Total: ~p~n" , [Student2]),
    io:format("Student 3 Total: ~p~n" , [Student3]),
    givecandy(Student1, Student2, Student3, TurnCount).

teachergivecandy(Student1Total, Student2Total, Student3Total, TurnCount) ->
    io:format("hello1"),
    if 
        Student1Total rem 2 == 1 -> 
            Student1FinalTotal = Student1Total + 1; %Student1FinalTotal = Student1Total + 1,
        true ->
            Student1FinalTotal = Student1Total
    end,
    if 
        Student2Total rem 2 == 1 -> 
            Student2FinalTotal = Student2Total + 1; %Student1FinalTotal = Student1Total + 1,
        true ->
            Student2FinalTotal = Student2Total
    end,
    if 
        Student3Total rem 2 == 1 -> 
            Student3FinalTotal = Student3Total + 1; %Student1FinalTotal = Student1Total + 1,
        true ->
            Student3FinalTotal = Student3Total
    end,
    io:format("Student 1 Total: ~p~n" , [Student1FinalTotal]),
    io:format("Student 2 Total: ~p~n" , [Student2FinalTotal]),
    io:format("Student 3 Total: ~p~n" , [Student3FinalTotal]),
    io:format("Turn: ~p~n" , [TurnCount]),
    givecandy(Student1FinalTotal, Student2FinalTotal, Student3FinalTotal, TurnCount).

答案

你的givecandy/4没有功能子句匹配。看来你想要检查所有3个学生是否有相同的糖果，所以你可以改变如下：

givecandy(Student1, Student1, Student1, _Turn) -> 1;
givecandy(Student1, Student2, Student3, Turn) ->
    io:format("Turn123"),
    Student1Total = (Student1 div 2) + (Student3 div 2),
    Student2Total = (Student2 div 2) + (Student1 div 2),
    Student3Total = (Student3 div 2) + (Student2 div 2),
    TurnCount = Turn + 1,
    teachergivecandy(Student1Total, Student2Total, Student3Total, TurnCount).

shell中的结果：

1>assignment3:main().
Hello
Turn: 1
Student 1 Total: 4
Student 2 Total: 2
Student 3 Total: 2
Turn123hello1Student 1 Total: 4
Student 2 Total: 4
Student 3 Total: 2
Turn: 2
Turn123hello1Student 1 Total: 4
Student 2 Total: 4
Student 3 Total: 4
Turn: 3
1

我有一些小代码来解决下面这个问题：

-module(foo).
-compile(export_all).

start()->
    io:format("A   B   C   Turns~n"),
    d(4, 2, 2, 0).
d(_A, _A, _A, B) ->
    io:format("~p   ~p   ~p   ~p~n",[_A, _A, _A, B]);
d(A, B, C, D)->
    io:format("~p   ~p   ~p   ~p~n", [A, B, C, D]),
    Na = check(A, B),
    Nb = check(B, C),
    Nc = check(C, A),
    d(Na, Nb, Nc, D + 1).

even(X) when X >= 0 -> (X band 1) == 0.
check(A, B)->
    N = trunc((A + B)/2),
    case even(N) of
        true -> N;
        false -> N + 1
    end.

shell中的结果：

> foo:start().
A   B   C   Turns
4   2   2   0
4   2   4   1
4   4   4   2
ok

另一答案

你有givecandy的两个条款，保护条件(Student1 == Student2) and (Student2 == Student3)和(Student1 /= Student2) and (Student2 /= Student3)。但可能是两名警卫都失败了，例如givecandy(1,2,2)。

可能其中一个递归调用最终看起来像那样（“可能”因为你没有在问题中包含错误信息，它应该确切地说明问题是什么函数和参数）。因此，您需要添加另一个子句来处理此案例或更改警卫。

最后一个条款没有警卫是好的做法，所以它可以处理所有遗失的案件。当然，如果没有更好的事情可能会抛出异常！