在公共lisp (SBCL)中生成结构体名称

Question

我使用SBCL以以下方式生成一些类型

(defun subpennant-sym (i) (make-symbol (format nil "SUBPENNANT-~D" i)))
(defun pennant-type-name (ty i) (make-symbol (format nil "PENNANT-TYPE-~D" i)))
(defmacro def-pennant-internal (ty n)
"A pennant is a balanced tree of depth N. (def-pennant 5 integer) expands to

; ... define pennant-type-{0..4}
(struct pennant-type-5
(node :type integer)
(subpennant-0 :type pennant-type-0)
(subpennant-1 :type pennant-type-1)
(subpennant-2 :type pennant-type-2)
(subpennant-3 :type pennant-type-3)
(subpennant-4 :type pennant-type-4))"
(when (>= n 0)
`(,@(macroexpand `(def-pennant-internal ,ty ,(- n 1)))
(defstruct ,(pennant-type-name ty n)
(node nil :type ,ty)
,@(loop for i to (- n 1)
collect
`(,(subpennant-sym i) nil :type ,(pennant-type-name ty i)))))))
(defmacro def-pennant (ty n)
`(progn ,@(macroexpand `(def-pennant-internal ,ty ,n))))

但是当我求值时:

CL-USER> (def-pennant fixnum 1)
(PROGN
(DEFSTRUCT #:PENNANT-TYPE-0 (NODE NIL :TYPE FIXNUM))
(DEFSTRUCT #:PENNANT-TYPE-1
(NODE NIL :TYPE FIXNUM)
(#:SUBPENNANT-0 NIL :TYPE #:PENNANT-TYPE-0)))
T
CL-USER> (def-pennant fixnum 1)
[...]
; 
; compilation unit finished
;   Undefined type:
;     #:PENNANT-TYPE-0
;   caught 2 STYLE-WARNING conditions
CL-USER> (make-pennant-type-0 :node 1) ; weird that it works...
#S(#:PENNANT-TYPE-0 :NODE 1)

谁能解释一下这个"错误"是怎么回事?意味着什么?

编辑:虽然上面的错误看起来是良性的，但当试图使用defstruct创建的符号时，我们会得到一个正确的错误。所以上面的代码对下面的

不起作用

(defun subpennant-sym (i) (make-symbol (format nil "SUBPENNANT-~D" i)))
(defun pennant-type-name (ty i) (make-symbol (format nil "PENNANT-TYPE-~D" i)))
(defmacro def-pennant-internal (ty n)
"A pennant is a balanced tree of depth N. (def-pennant 5 integer) expands to

; ... define pennant-type-{0..4}
(struct pennant-type-5
(node :type integer)
(subpennant-0 :type pennant-type-0)
(subpennant-1 :type pennant-type-1)
(subpennant-2 :type pennant-type-2)
(subpennant-3 :type pennant-type-3)
(subpennant-4 :type pennant-type-4))"
(when (>= n 0)
(let ((pen-ty (pennant-type-name ty n)))
`(,@(macroexpand `(def-pennant-internal ,ty ,(- n 1)))
(defstruct ,pen-ty
(node nil :type ,ty)
,@(loop for i to (- n 1)
collect
`(,(subpennant-sym i) nil :type ,(pennant-type-name ty i))))
(setf (get ,pen-ty :order) ,n)))))
(defmacro def-pennant (ty n)
`(progn ,@(macroexpand `(def-pennant-internal ,ty ,n))))
(def-pennant fixnum 4)
(defmacro pennant-order (obj) (get (type-of obj) 'order))

这个得到

The variable PENNANT-TYPE-0 is unbound.
[Condition of type UNBOUND-VARIABLE]

编辑2:冒着跑题的风险(尽管我认为我正在接近问题的根源)，也许我不理解的一件相关的事情是为什么会发生以下情况:

CL-USER> (eq 'hello (make-symbol "hello"))
NIL
CL-USER> (eq 'hello (intern "hello"))
NIL
CL-USER> (eq 'hello 'hello)
T

Answer 1

我在宏展开中看到类似(setf (get foo :bar) 0)的东西。

FOO未绑定。你可以引用一下那个符号吗?

还有:为什么要在宏中调用MACROEXPAND ?为什么不直接生成代码并让Lisp为您宏展开代码呢?通常需要遵循一个清晰的宏编写模型:不要将宏展开与您自己的宏展开混合在一起。如果需要的话，最好有一个机器文档->否则没人能理解你的代码。