“value returned is unused” warning when byte-compiling a macro

Question

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.

Answer 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).