My TXR Lisp dialect allows a symbol to be simultaneously a macro and function. Moreover, certain special operators are also backed by functions.
I put a bit of thought into the design, and haven't run into any problems. It works very well and is conceptually clean.
Common Lisp is the way it is for historic reasons.
Here is a brief rundown of the system:
When a global macro is defined for symbol X with defmacro, the symbol X does not become fboundp. Rather, what becomes fboundp is the compound function name (macro X).
The name (macro X) is then known to symbol-function, trace and in other situations. (symbol-function '(macro X)) retrieves the two-argument expander function which takes the form and an environment.
It's possible to write a macro using (defun (macro X) (form env) ...).
There are no compiler macros; regular macros do the job of compiler macros.
A regular macro can return the unexpanded form to indicate that it's declining to expand. If a lexical macrolet declines to expand, the opportunity goes to a more lexically outer macrolet, and so on up to the global defmacro. If the global defmacro declines to expand, the form is considered expanded, and thus is necessarily either a function call or special form.
If we have both a function and macro called X, we can call the function definition using (call (fun X) ...) or (call 'X ...), or else using the Lisp-1-style dwim evaluator (dwim X ...) that is almost always used through its [] syntactic sugar as [X ...].
For a sort of completeness, the functions mboundp, mmakunbound and symbol-macro are provided, which are macro analogs of fboundp, fmakunbound and symbol-function.
The special operators or, and, if and some others have function definitions also, so that code like [mapcar or '(nil 2 t) '(1 0 3)] -> (1 2 t) is possible.
Example: apply constant folding to sqrt:
1> (sqrt 4.0)
2.0
2> (defmacro sqrt (x :env e :form f)
(if (constantp x e)
(sqrt x)
f))
** warning: (expr-2:1) defmacro: defining sqrt, which is also a built-in defun
sqrt
3> (sqrt 4.0)
2.0
4> (macroexpand '(sqrt 4.0))
2.0
5> (macroexpand '(sqrt x))
(sqrt x)
However, no, (set (second x) 42) is not implemented via a macro definition for second. That would not work very well. The main reason is that it would be too much of a burden. The programmer may want to have, for a given function, a macro definition which has nothing to do with implementing assignment semantics!
Moreover, if (second x) implements place semantics, what happens when it is not embedded in an assignment operation, such that the semantics is not required at all? Basically, to hit all the requirements would require concocting a scheme for writing macros whose complexity would equal or exceed that of existing logic for handling places.
TXR Lisp does, in fact, feature a special kind of macro called a "place macro". A form is only recognized as a place macro invocation when it is used as a place. However, place macros do not implement place semantics themselves; they just do a straightforward rewrite. Place macros must expand down to a form that is recognized as a place.
Example: specify that (foo x), when used as a place, behaves as (car x):
1> (define-place-macro foo (x) ^(car ,x))
foo
2> (macroexpand '(foo a)) ;; not a macro!
(foo a)
3> (macroexpand '(set (foo a) 42)) ;; just a place macro
(sys:rplaca a 42)
If foo expanded to something which is not a place, things would fail:
4> (define-place-macro foo (x) ^(bar ,x))
foo
5> (macroexpand '(foo a))
(foo a)
6> (macroexpand '(set (foo a) 42))
** (bar a) is not an assignable place
