'作为变量的符号值是无效的：'在 lambda 内部的 defun 上使用参数时

Question

我想让这个功能更短一些。

(defun defkey-arg2 ()
  (exwm-input-set-key (kbd "s-g")
                      (lambda ()
                        (interactive)
                        (start-process-shell-command gkamus nil gkamus))))

然后我用 2 个参数（按键绑定和应用程序名称）编写较短的函数

(defun defkey-arg2 (key command) (...)

当我尝试将密钥作为参数时，它会起作用

(defun defkey-arg2 (key)
  (exwm-input-set-key (kbd key)
                      (lambda ()
                        (interactive)
                        (start-process-shell-command gkamus nil gkamus))))

(defkey-arg2 "s-g")

但是，当我尝试编写这样的函数时

(defun defkey-arg2 (key command)

或

(defun defkey-arg2 (command)
  (exwm-input-set-key (kbd "s-g")
                      (lambda ()
                        (interactive)
                        (start-process-shell-command command nil command)))

(defkey-arg2 "gkamus")

它引发错误：

Symbol's value as variable is void:' when using parameter on defun

Answer 1

lambda 的正文未被评估。使用反引号，command 的值可以代入结果表达式。

(defun defkey-arg2 (command)
  (define-key (current-local-map)
    (kbd "s-g")
    `(lambda ()
       (interactive)
       (start-process-shell-command ,command nil ,command))))

'作为变量的符号值是无效的：'在 lambda 内部的 defun 上使用参数时

'Symbol's value as variable is void:' when using parameter on defun inside lambda

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