Lisp: expand property name in macro

Question

Consider this property list:

(defvar *some-variable* (list :foo "fooval" :bar "barval"))

This simple call:

(getf *some-variable* :foo)

yields "fooval" as expected. I define a macro which is supposed to do the same except that I can pass the name of any property to retrieve:

(defmacro my-macro (property-name)
    `(getf *some-variable* :,property-name))

Unfortunately, calling it like this:

(my-macro 'foo)

results in FOO. Why?

Answer 1

Why don't you just check it out yourself:

(macroexpand-1 '(my-macro 'foo))
; ==> (getf *some-variable* :|| 'foo) ;
T

The documentation for getf says that if you give it a 4th argument it is the value when the key is not found. Since :|| (the empty symbol in the keyword package) doesn't exist it returns the supplied default foo.

So here is a function that does what you want:

(defun get-field (name)
 (getf *some-variable* 
       (intern (symbol-name name) "KEYWORD")))

(defparameter *test* 'foo)
(get-field *test*)
; ==> "fooval"

The only reason to make it a macro is to make it syntax and the main difference between syntax and a function is that the arguments are not evaluated.

(defmacro get-mfield (name)
  `(get-field ',name))

(get-mfield foo)
; ==> "fooval"

(get-mfield *test*)
; ==> nil

You get to come with literals bare, but you loose the feature that *test* is regarded as a variable and not the key :*test*