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我编写了一个程序，它将解析树作为输入并对其进行评估。它在树中的标识符不再出现时起作用。我希望评估者接受多个语句，并在我解决后面的语句时使用分配给标识符的值。这是我的程序。

add_op --> ['+'].
assign_op --> ['='].
ident(ident(Ident)) --> [Ident], {atom(Ident)}. /* True if an identifier */
int(int(Int)) --> [Int], {integer(Int)}. /* True if an integer */
semicolon --> [';'].

evaluator(statements(Assignment),VariablesOut):-
    evaluator(Assignment,VariablesOut).

evaluator(statements(Assignment,Statements),VariablesOut):-
    evaluator(Assignment,List1),
    evaluator(Statements,List2),
    merge(List1,List2,VariablesOut).

evaluator(assignment(ident(Ident),assign_op,int(Int),add_op,int(Int2),semicolon),Answer5):-
    Answer2 is Int + Int2,
    atom_concat(Ident,' = ',Answer1),
    atom_concat(Answer1,Answer2,Answer3),
    read_term_from_atom(Answer3, (Answer4), []),    /*removes quotation marks*/
    Answer5 = [Answer4].

以下执行使用解析树 a = 2 + 3 ；b = 3 + 4 ; 此执行有效，因为我的程序仅评估 a 和 b 的答案，而 b 的评估不包含 a。

evaluator(statements(assignment(ident(a), assign_op, int(2), add_op, int(3), semicolon), statements(assignment(ident(b), assign_op, int(3), add_op, int(4), semicolon))), Answer).
Answer = [a=5, b=7].

以下使用解析树 a = 2 + 3 ；b = a + 4 ;
这个不起作用，因为 b 必须是 4 添加到 a 的结果，并且 a 没有在程序中分配给值 5。

evaluator(statements(assignment(ident(a), assign_op, int(2), add_op, int(3), semicolon), statements(assignment(ident(b), assign_op, int(a), add_op, int(4), semicolon))), Answer).

ERROR: Arithmetic: `a/0' is not a function
ERROR: In:
ERROR:   [11] _2952 is a+4
...

更新

我找到了一个类似的解决方案，但我不确定它是如何工作的。我不知道在哪里进行任何评估，也看不到 maplist 和 WithSym =.. 行是如何需要的。

exp_symbols(Symbols, Expr, WithSym) :-
    Expr =.. [F|Args],   /*F is sent to the first element of Expr which in my case would be a = 5 */
    ( member(F=V, Symbols) -> G = V ; G = F ), /* if F followed by a colon and some element V is in the equation, set G to be assign to both V and F */
    maplist(exp_symbols(Symbols), Args, ArgsSWithSym), /*perform the current function on all elements left **/
    WithSym =.. [G|ArgsSWithSym]. /** Set WithSym to be the first element in ArgsSWithSym /**

evaluate(Exp, LstVars, Val) :-
    exp_symbols(LstVars, Exp, NewExp),
    Val is NewExp.
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