在Rebol 2中，对象上的位置PICK是什么，以及等效的Rebol 3是什么？(What does positional PICK on an object do in Rebol 2, and what's the equivalent Rebol 3?)

 调用(list (A 3))返回列表中的函数： 
 
user> (list (A 3) 0)
(#<user$A$fn__2934 user$A$fn__2934@2f09b4cb> 0)
 
 eval期望在列表中获得一个符号，它正在获得它自己的功能。 如果您引用(A 3)的电话，那么您将获得您寻求的结果 
 
user> (println (eval (list '(A 3) 0)))
3
nil
 
 在调用eval之前正在评估此代码的一部分，然后eval正在评估其余代码。 在引用的表格中使用的seval可能是一个术语或者有选择地不加引号（ ~ ）更常见。 
 
user> (eval '((A 3) 0))
3
 
 要么 
 
user> (def my-number 3)
#'user/my-number
user> (eval `((A ~my-number) 0))
3
 
 编辑：关于为什么它与零args一起工作的问题并且使用一个arg失败： 
 
 似乎两种形式都可以工作，如果你不将它们存储在变量中（即使用defn定义它们），而是手动内联它们： 
 
user> (def A (fn [b] (fn [a] (+ a b))))
#'user/A
user> (eval (list (A 3) 0))
; Evaluation aborted.
user> (eval (list (fn [b] (fn [a] (+ a b)) 3) 0))
3

Calling (list (A 3)) returns a function in a list:
 
user> (list (A 3) 0)
(#<user$A$fn__2934 user$A$fn__2934@2f09b4cb> 0)
 
eval expects to get a symbol in a list and it was getting the function it's self. if you quote the call to (A 3) then you will get the result you seek
 
user> (println (eval (list '(A 3) 0)))
3
nil
 
Part of this code is being evaluated before the call to eval and then eval is evaluating the rest. It is more common to se eval used on quoted forms with perhaps a term or to selectively unquoted (~). 
 
user> (eval '((A 3) 0))
3
 
or
 
user> (def my-number 3)
#'user/my-number
user> (eval `((A ~my-number) 0))
3
 
edit: on the question of why it works with zero args and fails with one arg:
 
it seems that both forms work if you don't store them in vars (ie define them with defn) and instead manually inline them:
 
user> (def A (fn [b] (fn [a] (+ a b))))
#'user/A
user> (eval (list (A 3) 0))
; Evaluation aborted.
user> (eval (list (fn [b] (fn [a] (+ a b)) 3) 0))
3