common-lisp - 在 Common Lisp 中以宏列表作为参数的宏

Question

在 Common Lisp 中，如何定义一个“元宏”，它将宏列表(和其他参数)作为参数，并组合这些宏以生成所需的代码。

这个问题相当于编写一个“高阶宏”，它从其他宏的变量列表中定义一个宏。

提示问题的具体情况对我来说是一个 CLSQL 的实验，我想从 CLSQL-testsuite 中重新表达员工类

(clsql:def-view-class employee ()
  ((employee-id
    :db-kind :key
    :db-constraints (:not-null)
    :type integer)
   (first-name
    :accessor employee-first-name
    :type (string 30)
    :initarg :first-name)
   (last-name
    :accessor employee-last-name
    :type (string 30)
    :initarg :last-name)
   (email
    :accessor employee-email
    :type (string 100)
    :initarg :email)
   (company-id
     :type integer
     :initarg :company-id)
   (company
    :accessor employee-company
    :db-kind :join
    :db-info (:join-class company
              :home-key companyid
              :foreign-key companyid
              :set nil))
   (manager-id
    :type integer
    :nulls-ok t
    :initarg :manager-id)
   (manager
    :accessor employee-manager
    :db-kind :join
    :db-info (:join-class employee
              :home-key managerid
              :foreign-key emplid
              :set nil))))

作为

(def-view-class-with-traits employee ()
  (trait-mapsto-company trait-mapsto-manager)
  ((employee-id
    :db-kind :key
    :db-constraints (:not-null)
    :type integer)
   (first-name
    :accessor employee-first-name
    :type (string 30)
    :initarg :first-name)
   (last-name
    :accessor employee-last-name
    :type (string 30)
    :initarg :last-name)
   (email
    :accessor employee-email
    :type (string 100)
    :initarg :email)))

在定义复杂的数据库模式时，使用这种技术将有利于保持一致性和简洁性。

我将我需要的两个特征定义为

(defmacro trait-mapsto-company (class super slots &rest cl-options)
  (declare (ignore super slots cl-options))
  (let ((company-accessor-name
          (intern (concatenate 'string (symbol-name class) "-COMPANY"))))
    `((company-id
       :type integer
       :initarg :company-id)
      (company
       :accessor ,company-accessor-name
       :db-kind :join
       :db-info (:join-class company
                 :home-key companyid
                 :foreign-key companyid
                 :set nil)))))

(defmacro trait-mapsto-manager (class super slots &rest cl-options)
  (declare (ignore super slots cl-options))
  (let ((manager-accessor-name
          (intern (concatenate 'string (symbol-name class) "-MANAGER"))))
    `((manager-id
       :type integer
       :initarg :manager-id)
      (manager
       :accessor ,manager-accessor-name
       :db-kind :join
       :db-info (:join-class manager
                 :home-key managerid
                 :foreign-key emplid
                 :set nil)))))

但是，我编写 def-view-class-with-traits 的尝试失败了。

(defmacro def-view-class-with-traits (class super traits slots &rest cl-options)
  (let ((actual-slots
          (reduce (lambda (trait ax) (append (apply trait class super slots cl-options) ax))
                  traits
                  :initial-value slots)))
    `(clsql:def-view-class ,class ,super ,actual-slots ,@cl-options)))

在用于归约的 lambda 中，trait 代表宏，而我对 apply 的使用对 Lisp 没有任何意义——这是正确的! – 但希望将我的意图传达给其他程序员。

如何让def-view-class-with-traits以合适的方式处理traits的宏列表？

Answer 1

如果您将特征定义为类本身并使用普通继承，我会发现这并不令人惊讶:

(def-view-class trait-mapsto-company ()
  ((company-id
    :type integer
    :initarg :company-id)
   (company
    :accessor company
    :db-kind :join
    :db-info (:join-class company
              :home-key company-id
              :foreign-key company-id
              :set nil))))

(def-view-class trait-mapsto-manager ()
  ((manager-id
    :type integer
    :initarg :manager-id)
   (manager
    :accessor manager
    :db-kind :join
    :db-info (:join-class manager
              :home-key managerid
              :foreign-key emplid
              :set nil)))

(def-view-class employee (trait-mapsto-company trait-mapsto-manager)
  ((employee-id
    :db-kind :key
    :db-constraints (:not-null)
    :type integer)
   (first-name
    :accessor employee-first-name
    :type (string 30)
    :initarg :first-name)
   (last-name
    :accessor employee-last-name
    :type (string 30)
    :initarg :last-name)
   (email
    :accessor employee-email
    :type (string 100)
    :initarg :email)))

这当然不会使访问器名称依赖于继承类的名称，但你真的想要吗？我的观点是，这种写法表明这实际上会破坏解耦原则。