这个功能对顶级有影响吗？

Question

我想知道这个函数是否对顶级符号有影响？

(defun test (gallon)
  (setq cup '(D D D D))
  (pop cup) )

如果答案是肯定的，那是哪一个？

想知道如何修复此功能以防止这种副作用？ 谢谢你回答！

Answer 1

如果编译此函数，您将收到警告：

WARNING: in TEST : CUP is neither declared nor bound,
         it will be treated as if it were declared SPECIAL.
WARNING: in TEST : CUP is neither declared nor bound,
         it will be treated as if it were declared SPECIAL.
WARNING: in TEST : CUP is neither declared nor bound,
         it will be treated as if it were declared SPECIAL.
WARNING: in TEST : CUP is neither declared nor bound,
         it will be treated as if it were declared SPECIAL.
WARNING: in TEST : variable GALLON is not used.
         Misspelled or missing IGNORE declaration?

这告诉您，您没有绑定cup而没有使用gallon。

正在发生的是您正在创建（和修改）全局变量cup。

为了避免它，你应该在本地绑定它：

(defun test1 (gallon)
  (let ((cup '(D D D D)))
    (pop cup)))

注意您仍然忽略gallon参数。

要查看差异，请观察

cup
==> undefined variable error
(test)
==> D
cup
==> (D D D)
(setq cup 10)
==> 10
(test1)
==> D
cup 
==> 10

即，我的test1 不修改全局绑定。