用堆栈而不是递归实现Lisp-eval

我正在研究如何编写程序.blogspot.com/2010/11/lisp-inintereter-in-90-lines-of-c.html，我想看看如何在没有递归的情况下实现eval（）函数，而是作为一个使用堆栈的迭代函数。

以下是来自文章的eval（）代码：

////////////////////// eval
cell eval(cell x, environment * env)
{
    if (x.type == Symbol)
        return env->find(x.val)[x.val];
    if (x.type == Number)
        return x;
    if (x.list.empty())
        return nil;
    if (x.list[0].type == Symbol) {
        if (x.list[0].val == "quote")       // (quote exp)
            return x.list[1];
        if (x.list[0].val == "if")          // (if test conseq [alt])
            return eval(eval(x.list[1], env).val == "#f" ? (x.list.size() < 4 ? nil : x.list[3]) : x.list[2], env);
        if (x.list[0].val == "set!")        // (set! var exp)
            return env->find(x.list[1].val)[x.list[1].val] = eval(x.list[2], env);
        if (x.list[0].val == "define")      // (define var exp)
            return (*env)[x.list[1].val] = eval(x.list[2], env);
        if (x.list[0].val == "lambda") {    // (lambda (var*) exp)
            x.type = Lambda;
            // keep a reference to the environment that exists now (when the
            // lambda is being defined) because that's the outer environment
            // we'll need to use when the lambda is executed
            x.env = env;
            return x;
        }
        if (x.list[0].val == "begin") {     // (begin exp*)
            for (size_t i = 1; i < x.list.size() - 1; ++i)
                eval(x.list[i], env);
            return eval(x.list[x.list.size() - 1], env);
        }
    }
                                            // (proc exp*)
    cell proc(eval(x.list[0], env));
    cells exps;
    for (cell::iter exp = x.list.begin() + 1; exp != x.list.end(); ++exp)
        exps.push_back(eval(*exp, env));
    if (proc.type == Lambda) {
        // Create an environment for the execution of this lambda function
        // where the outer environment is the one that existed* at the time
        // the lambda was defined and the new inner associations are the
        // parameter names with the given arguments.
        // *Although the environmet existed at the time the lambda was defined
        // it wasn't necessarily complete - it may have subsequently had
        // more symbols defined in that environment.
        return eval(/*body*/proc.list[2], new environment(/*parms*/proc.list[1].list, /*args*/exps, proc.env));
    }
    else if (proc.type == Proc)
        return proc.proc(exps);
    std::cout << "not a functionn";
    exit(1);
}

我遇到的第一个问题是想知道如何用堆栈实现"if"逻辑。如果您将条件（第一个单元格）放在堆栈上并进行下一次迭代，您如何知道如何回到您决定是分支到"then"还是"else"单元格/节点的点？

不过，这同样适用于任何其他逻辑：例如，如果我将"define"的单元格放在堆栈上，然后进行下一次迭代，一旦进行了评估，我怎么知道要回到同一个地方？