小编Has*_*bug的帖子

我是prolog的新手,所以这对我来说是一个很大的挑战.我应该在Prolog中实现一个简单的C语言.

the ultimate goal is to be able to execute something like this:
?- run([begin,a,:=,10,while,a,>,5,begin,write,a,a,:=,a,-,1,end,end]).

and get: 
10 
9
8
7
6 
yes

但是,我陷入了第一步.这是我迄今取得的成就.超出本地堆栈!

statement(Vars,_Vars) --> assign(Vars,_Vars).
statement(Vars,Vars2) --> statement(Vars,Vars1), statement(Vars1,Vars2). 

assign(Vars,_Vars) --> default_assign(Vars,_Vars).
assign(Vars,_Vars) --> simple_assign(Vars,_Vars).

% a //default value 0
default_assign(Vars,_Vars) --> 
    var_name(Var_Name),
    {update_vars([Var_Name,0],Vars,_Vars)}.

% a = 0
simple_assign(Vars,_Vars) --> 
    var_name(Var_Name),[=],var_value(Var_Value),
    {update_vars([Var_Name,Var_Value],Vars,_Vars)}.

% a = b
simple_assign(Vars,_Vars) --> 
    var_name(Var_Name1),[=],var_name(Var_Name2),
    {
    update_vars([Var_Name1,Var_Value],Vars,_Vars)
    }.

var_name(Var_Name) --> [Var_Name],{\+number(Var_Name2)}.    
var_value(Var_Value) -->[Var_Value],{number(Var_Value)}.

% found match, update
update_vars(Var,Vars,_Vars):-
    member(Var,Vars),
    update(Var,Vars,_Vars),
    _Vars\==[].
% no match, …