我在 Prolog 中编写了一个解析器，它接受一个标记化列表，并且应该返回一个表达式，其中变量与评估方程的值统一：
Tokens = ['(', is, v('X',3),'(', +, 1, 2, ')', ')' ]
Expr = (3 is 1 + 2)
目前，我的解析器正在返回以下内容：
Expr [is, _G32432, '(', +, 1, 2, ')'|_G19343]
有谁知道我如何修复这个解析器？我已包含以下代码：
%Definite Clause Grammar (DCG) for Lisp s-expressions
expression(N) --> atom(N).
expression(N) --> integer(N).
expression(N) --> variable(N).
expression(N) --> list(N).
list(N) --> ['('], sequence(N), [')'].
sequence(_) --> [].
sequence([H|T]) --> expression(H), sequence(T).
%atom(_) --> [].
atom(N) --> [N],{atom(N)}.
%variable(_) --> [].
variable(N) --> [v(_,N)],{var(N)}.
%integer(_) --> [].
integer(N) --> [N],{integer(N)}.
evaluate(String, Expr):-
tokenize(String, Tokens),
expression(Expr,Tokens,[]),
write('Expression: '), write_term(Expr, [ignore_ops(true)]).
编辑：下面是我的解析器工作版本：
expression(N) --> atom(N). %an atom is a type of expression
expression(N) --> integer(N). %an integer is a type of expression
expression(N) --> variable(N). %a variable is a type of expression
expression(M) --> list(N),{M=..N}.
list(N) --> ['('], sequence(N), [')']. %a sequence within parens is a type of list
sequence([]) --> []. %a sequence can be empty
sequence([H|T]) --> expression(H), sequence(T). %a sequence can be composed of an expression
% sequence([]) --> []. %and a sequence atom(_) --> [].
atom(N) --> [N],{atom(N),N \= '(', N \= ')'}. %parens are not atoms, but all other Prolog atoms
% If N is a variable and it is within the v(Label,X) data structure,
% then it is a var in this grammar
variable(N) --> [v(_,N)],{var(N)}.
%variable(_) --> [].
%integer(_) --> [].
integer(N) --> [N],{integer(N)}.