如何在Erlang中使用变量作为传递引用

Question

为什么我的输出没有反映在Lst1中？

-module(pmap). 
-export([start/0,test/2]). 
test(Lst1,0) ->
{ok, [Temp]} = io:fread( "Input the edge weight  ", "~d" ),
lists:append([Lst1,[Temp]]),
io:fwrite("~w~n",[Lst1]);
test(Lst1,V) ->
{ok, [Temp]} = io:fread( "Input the edge weight  ", "~d" ),
lists:append([Lst1,[Temp]]),
test(Lst1, V-1).
start() -> 
{ok, [V]} = io:fread( "Input the number of vertices your graph has  ", "~d" ),
Lst1 = [],
test(Lst1,V).

因此，我的Lst1正在打印[]，而我希望它打印，假设，如果我提供输入1,2,3，[1,2,3]。

Answer 1

因为Erlang变量是不可变的，根本无法更改。lists:append返回一个您丢弃的新列表。

Answer 2

正如@Alexey Romanov正确指出的那样，您没有使用lists:append/2的结果。

这就是我将如何修复你的代码…

-module(pmap). 
-export([start/0,test/2]). 
test(Lst1,0) ->
{ok, [Temp]} = io:fread( "Input the edge weight  ", "~d" ),
Lst2 = lists:append([Lst1,[Temp]]),
io:fwrite("~w~n",[Lst2]),
Lst2;
test(Lst1,V) ->
{ok, [Temp]} = io:fread( "Input the edge weight  ", "~d" ),
Lst2 = lists:append([Lst1,[Temp]]),
test(Lst2, V-1).
start() -> 
{ok, [V]} = io:fread( "Input the number of vertices your graph has  ", "~d" ),
Lst1 = [],
test(Lst1,V).

但实际上，一个更为惯用的代码可以实现相同的结果，那就是…

-module(pmap). 
-export([start/0,test/2]). 
test(Lst1,0) ->
{ok, [Temp]} = io:fread( "Input the edge weight  ", "~d" ),
Lst2 = lists:reverse([Temp|Lst1]),
io:fwrite("~w~n",[Lst2]),
Lst2;
test(Lst1,V) ->
{ok, [Temp]} = io:fread( "Input the edge weight  ", "~d" ),
test([Temp | Lst1], V-1).
start() -> 
{ok, [V]} = io:fread( "Input the number of vertices your graph has  ", "~d" ),
Lst1 = [],
test(Lst1,V).