“返回的值未被使用”字节编译宏时发出警告 [英] "value returned is unused" warning when byte-compiling a macro
问题描述
(defmacro foomacro(
`(defun foo(& optional arg)
(interactive,(concat shiftcodep))
(messagearg is%iarg))
'(defun bar(& optional arg)
(interactive,(concat shiftcodeNenter a number:))
(messagearg is%iarg)))
;;为Emacs提供向后兼容22
(if(fboundp'handle-shift-selection)
(foomacro^)
(foomacro))
这是我得到的警告:
$ emacs -Q --batch --eval'(byte-compile-filefoo.el)'
在foomacro中:
foo.el :1:21:警告:从(concat shiftcodep)返回的值未使用
如果我摆脱 bar
,警告消失:
<$
(交互式,(concat shiftcodep))
(消息arg是%i arg)))
;;为Emacs提供向后兼容22
(if(fboundp'handle-shift-selection)
(foomacro^)
(foomacro))
我使用的是GNU Emacs 24.2.1。
这是因为你忘了将宏体包裹在一个progn中:
(defmacro foomacro(shiftcode)
`(progn
(defun foo(& optional arg)
(interactive,(concat shiftcodep))
(messagearg is%iarg))
(defun bar(& optional arg)
(interactive,(concat shiftcodeNenter a number:))
(messagearg is%iarg))))
想想宏如何工作。当您调用(foomacro...)
时,lisp引擎会识别 foomacro
是一个宏, >扩展它,即调用它提供的参数。宏的返回值与预期的 ,第二个 defun
形式;而第一个 defun
表单已被弃用。然后,lisp引擎评估返回值(这是第二个 defun
形式)。因此,在 progn
无版本中,只定义 bar
,而不是 foo
。
为了理解这个过程,你需要意识到宏只是代码转换工具;他们真的不做任何事情。因此,编译器(或解释器)只能看到它们的返回值。
Why does byte-compiling the the following produce a warning?
(defmacro foomacro (shiftcode)
`(defun foo (&optional arg)
(interactive ,(concat shiftcode "p"))
(message "arg is %i" arg))
`(defun bar (&optional arg)
(interactive ,(concat shiftcode "Nenter a number: "))
(message "arg is %i" arg)))
;; provide backward compatibility for Emacs 22
(if (fboundp 'handle-shift-selection)
(foomacro "^")
(foomacro ""))
This is the warning I get:
$ emacs -Q --batch --eval '(byte-compile-file "foo.el")'
In foomacro:
foo.el:1:21:Warning: value returned from (concat shiftcode "p") is unused
If I get rid of bar
, the warning goes away:
(defmacro foomacro (shiftcode)
`(defun foo (&optional arg)
(interactive ,(concat shiftcode "p"))
(message "arg is %i" arg)))
;; provide backward compatibility for Emacs 22
(if (fboundp 'handle-shift-selection)
(foomacro "^")
(foomacro ""))
I'm using GNU Emacs 24.2.1.
That's because you forgot to wrap the macro body in a progn:
(defmacro foomacro (shiftcode)
`(progn
(defun foo (&optional arg)
(interactive ,(concat shiftcode "p"))
(message "arg is %i" arg))
(defun bar (&optional arg)
(interactive ,(concat shiftcode "Nenter a number: "))
(message "arg is %i" arg))))
Think about how macros work. When you call (foomacro "...")
, the lisp engine recognizes that foomacro
is a macro and expands it, i.e., calls it on the argument(s) supplied. the return value of the macro is, as expected, the second defun
form; while the first defun
form is discarded. Then the lisp engine evaluates the return value (which is the second defun
form). Thus in your progn
-less version only bar
is defined, not foo
.
To understand the process, you need to realise that macros are merely "code transformation" tools; they don't really do anything. Thus only their return values are seen by the compiler (or the interpreter).
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