Erlang 不会让我结束函数

Question

这是我的RPS游戏代码。每当我试图把"；结束"它不会让我结束这个功能。前面写着语法错误："end"。

-module(project).
-export([rps/0]).
rps() ->
Computer = rand:uniform(3),
{ok,Player} = io:read("Rock, Paper, or Scissors?"),
Win = fun() -> io:fwrite("You Win!"),
Lose = fun() -> io:fwrite("You Lose."),
% player chooses rock (win, lose)
if
Player == "Rock" ->
if
Computer == 3 ->
Win();
Computer == 2 ->
Lose()
end;
% player chooses paper (win, lose)
if
Player == "Paper" ->
if
Computer == 1 ->
Win();
Computer == 3 ->
Lose()
end;
% player chooses scissors (win, lose)
if
Player == "Scissors" ->
if
Computer == 2 ->
Win();
Computer == 1 ->
Lose()
end;
end.  <--------- (this is where error occurs)

Answer 1

Artee的回答解释了语法错误发生的位置和方式。让我建议一种不同的写作方式，这可以被认为是一种更多的"Erlang-like"；，并且顺便消除了语法错误。

首先，让我们让Computer变量包含计算机的选择，其格式与玩家的相同：

Computer = element(rand:uniform(3), {"Rock", "Paper", "Scissors"}),

然后，创建一个函数，接受玩家和计算机的选择，并返回谁获胜：

who_wins("Rock", "Scissors") ->
player;
who_wins("Rock", "Paper") ->
computer;
who_wins("Paper", "Scissors") ->
computer;
who_wins("Paper", "Rock") ->
player;
who_wins("Scissors", "Rock") ->
computer;
who_wins("Scissors", "Paper") ->
player;
who_wins(Same, Same) ->
tie.

(在最后一个子句中，由于我们对两个参数使用相同的变量名，因此只有当参数相同时，该子句才会匹配。(

然后，回到rps函数中，我们可以将嵌套的if表达式替换为：

case who_wins(Player, Computer) of
player ->
Win();
computer ->
Lose();
tie ->
io:fwrite("It's a tie!")
end.

Answer 2

这里有几个错误：

• 没有乐趣"结束"；(x2(Win=fun((->io：fwrite("你赢了！"(结束
• if语法

正确代码：

-module(project).
-export([rps/0]).
rps() ->
Computer = rand:uniform(3),
{ok,Player} = io:read("Rock, Paper, or Scissors?"),
Win = fun() -> io:fwrite("You Win!") end,
Lose = fun() -> io:fwrite("You Lose.") end,
% player chooses rock (win, lose)
if
Player == "Rock" ->
if
Computer == 3 ->
Win();
Computer == 2 ->
Lose()
end;
% player chooses paper (win, lose)
Player == "Paper" ->
if
Computer == 1 ->
Win();
Computer == 3 ->
Lose()
end;
% player chooses scissors (win, lose)
Player == "Scissors" ->
if
Computer == 2 ->
Win();
Computer == 1 ->
Lose()
end
end.