评估和词汇变量 [英] Eval and lexical variables

问题描述

我正在做一个小项目，只是为了好玩，我添加了对它的eval支持以简化调试.但是后来我发现了一个问题:

I'm doing a small project just for fun, and I added eval support for it to make debug easier. But later I found a problem:

(let ((x 1))
    (eval (1+ x)))

(defun foo (x form)
    (eval form))
(foo 1 '(1+ x))

上面的代码不起作用.有人可以解释为什么以及如何解决吗?非常感谢.

Code above won't work. Could someone please explain why and how to work it around? Thanks very much.

推荐答案

首先，

(let ((x 1))
  (eval (1+ x)))

看起来确实可以工作(它确实可以做某事)，可能不起作用，您打算做什么. eval是常规函数，因此它接收由调用方评估的参数.有效地，您正在使用2的整数值调用eval，然后将其求值"(因为整数是自引号)，结果结果为2.

may look like it does work (it certainly does something), it is likely not doing, what you intend it to do. eval is a regular function, so it receives its arguments evaluated by the caller. Effectively, you are calling eval with an integer value of 2 -- which is then "evaluated" (since integers are self-quoting) to a result value of 2.

在

(defun foo (x form)
  (eval form))

更容易诊断故障.运行时词法绑定不是一流的对象，而是由后台的解释器/编译器维护的东西.常规函数(如eval)无法访问在其调用位置定义的词法变量.

it's easier to diagnose the failure. Run-time lexical bindings are not first-class objects, but something maintained by the interpreter/compiler behind the scenes. Regular functions (like eval) cannot access lexical variables defined at their call-sites.

一种解决方法是使用特殊变量:

One work-around would be to use special variables:

(defun foo (x form)
  (declare (special x))
  (eval form))

该声明告诉您的Lisp实现，x应该在其范围内动态绑定.

The declaration tells your lisp implementation, that x should be dynamically bound within its scope.

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