What is the definition of "top-level-form" in Racket - racket

The Racket Reference sections 11.9 Expanding Top-Level Forms and 13.2 Evaluation and Compilation use the term "top-level-form" and the descriptions of the functions (eval top-level-form [nm]) and (expand top-level-form) in the reference manual have "top-level-form" as their function argument, but I'm unclear about the definition of "top-level-form". What is the meaning of the term "top-level-form" within the Racket language?

The intuition here is that these functions all deal with "top-level forms" as opposed to a form that depends on a lexical environment. As a semi-obvious example, eval can only deal with top-level forms, which is why this:
(let ([x 10])
(eval '(* 3 x)))
doesn't work. The usual use of just "forms" is talking about any forms, such as inputs to macros -- which of course can have such references.

I have now found the precise definition of top-level-form in a kind of Backus-Naur format in the Racket Reference manual section 1.2.3.1
http://docs.racket-lang.org/reference/syntax-model.html#(part._fully-expanded)

Related

Are Lisp macros just syntactic sugar? [duplicate]

This question already has answers here:
What makes Lisp macros so special?
(15 answers)
Closed 3 months ago.
I keep reading that Lisp macros are one of the most powerful features of the language. But reading over the specifications and manuals, they are just functions whose arguments are unevaluated.
Given any macro (defmacro example (arg1 ... argN) (body-forms)) I could just write (defun example (arg1 ... argN) ... (body-forms)) with the last body-form turned into a list and then call it like (eval (example 'arg1 ... 'argN)) to emulate the same behavior of the macro. If this were the case, then macros would just be syntactic sugar, but I doubt that syntactic sugar would be called a powerful language feature. What am I missing? Are there cases where I cannot carry out this procedure to emulate a macro?
I can't talk about powerful because it can be a little bit subjective, but macros are regular Lisp functions that work on Lisp data, so they are as expressive as other functions. This isn't the case with templates or generic functions in other languages that rely more on static types and are more restricted (on purpose).
In some way, yes macros are simple syntactic facilities, but you are focused in your emulation on the dynamic semantics of macros, ie. how you can run code that evaluates macros at runtime. However:
the code using eval is not equivalent to expanded code
the preprocessing/compile-time aspect of macros is not emulated
Lexical scope
Function, like +, do not inherit the lexical scope:
(let ((x 30))
(+ 3 4))
Inside the definition of +, you cannot access x. Being able to do so is what "dynamic scope" is about (more precisely, see dynamic extent, indefinite scope variables). But nowadays it is quite the exception to rely on dynamic scope. Most functions use lexical scope, and this is the case for eval too.
The eval function evaluates a form in the null lexical environment, and it never has access to the surrounding lexical bindings. As such, it behaves like any regular function.
So, in you example, calling eval on the transformed source code will not work, since arg1 to argnN will probably be unbound (it depends on what your macro does).
In order to have an equivalent form, you have to inject bindings in the transformed code, or expand at a higher level:
(defun expand-square (var)
(list '* var var))
;; instead of:
(defun foo (x) (eval (expand-square 'x))) ;; x unbound during eval
;; inject bindings
(defun foo (x) (eval `(let ((z ,x)) (expand-square z))))
;; or expand the top-level form
(eval `(defun foo (x) ,(expand-square 'x)))
Note that macros (in Common Lisp) also have access to the lexical environment through &environment parameters in their lambda-list. The use of this environment is implementation dependent, but can be used to access the declarations associated with a variable, for example.
Notice also how in the last example you evaluate the code when defining the function, and not when running it. This is the second thing about macro.
Expansion time
In order to emulate macros you could locally replace a call to a macro by a form that emulates it at runtime (using let to captures all the bindings you want to see inside the expanded code, which is tedious), but then you would miss the useful aspect of macros that is: generating code ahead of time.
The last example above shows how you can quote defun and wrap it in eval, and basically you would need to do that for all functions if you wanted to emulate the preprocessing work done by macros.
The macro system is a way to integrate this preprocessing step in the language in a way that is simple to use.
Conclusion
Macros themselves are a nice way to abstract things when functions can't. For example you can have a more human-friendly, stable syntax that hides implementation details. That's how you define pattern-matching abilities in Common Lisp that make it look like they are part of the language, without too much runtime penalty or verbosity.
They rely on simple term-rewriting functions that are integrated in the language, but you can emulate their behavior either at compile-time or runtime yourself if you want. They can be used to perform different kinds of abstraction that are usually missing or more cumbersome to do in other languages, but are also limited: they don't "understand" code by themselves, they don't give access to all the facilities of the compiler (type propagation, etc.). If you want more you can use more advanced libraries or compiler tools (see deftransform), but macros at least are portable.
Macros are not just functions whose arguments are unevaluated. Macros are functions between programming languages. In other words a macro is a function whose argument is a fragment of source code of a programming language which includes the macro, and whose value is a fragment of source code of a language which does not include the macro (or which includes it in a simpler way).
In very ancient, very rudimentary, Lisps, before people really understood what macros were, you could simulate macros with things called FEXPRs combined with EVAL. A FEXPR was simply a function which did not evaluate its arguments. This worked in such Lisps only because they were completely dynamically scoped, and the cost of it working was that compilation of such things was not possible at all. Those are two enormous costs.
In any modern Lisp, this won't work at all. You can write a toy version of FEXPRs as a macro (this may be buggy):
(defmacro deffex (fx args &body body)
(assert (every (lambda (arg)
(and (symbolp arg)
(not (member arg lambda-list-keywords))))
args)
(args) "not a simple lambda list")
`(defmacro ,fx ,args
`(let ,(mapcar (lambda (argname argval)
`(,argname ',argval))
',args (list ,#args))
,#',body)))
So now we could try to write a trivial binding construct I'll call with using this thing:
(deffex with (var val form)
(eval `(let ((,var ,val)) ,form)))
And this seems to work:
> (with a 1 a)
1
Of course, we're paying the cost that no code which uses this construct can ever be compiled so all our programs will be extremely slow, but perhaps that is a cost we're willing to accept (it's not, but never mind).
Except, of course, it doesn't work, at all:
> (with a 1
(with b 2
(+ a b)))
Error: The variable a is unbound.
Oh dear.
Why doesn't it work? It doesn't work because Common Lisp is lexically scoped, and eval is a function: it can't see the lexical bindings.
So not only does this kind of approach prevent compilation in a modern Lisp, it doesn't work at all.
People often, at this point, suggest some kind of kludge solution which would allow eval to be able to see lexical bindings. The cost of such a solution is that all the lexical bindings need to exist in compiled code: no variable can ever be compiled away, not even its name. That's essentially saying that no good compilers can ever be used, even for the small part of your programs you can compile at all in a language which makes extensive use of macros like CL. For instance, if you ever use defun you're not going to be able to compile the code in its body. People do use defun occasionally, I think.
So this approach simply won't work: it worked by happenstance in very old Lisps but it can't work, even at the huge cost of preventing compilation, in any modern Lisp.
More to the point this approach obfuscates the understanding of what macros are: as I said at the start, macros are functions between programming languages, and understanding that is critical. When you are designing macros you are implementing a new programming language.

Is it possible to write a function that would take any macro and turn it into a function so that it can be passed as an argument to another function?

AND and OR are macros and since macros aren't first class in scheme/racket they cannot be passed as arguments to other functions. A partial solution is to use and-map or or-map. Is it possible to write a function that would take arbitrary macro and turn it into a function so that it can be passed as an argument to another function? Are there any languages that have first class macros?
In general, no. Consider that let is (or could be) implemented as a macro on top of lambda:
(let ((x 1))
(foo x))
could be a macro that expands to
((lambda (x) (foo x)) 1)
Now, what would it look like to convert let to a function? Clearly it is nonsense. What would its inputs be? Its return value?
Many macros will be like this. In fact, any macro that could be routinely turned into a function without losing any functionality is a bad macro! Such a macro should have been a function to begin with.
I agree with #amalloy. If something is written as a macro, it probably does something that functions can't do (e.g., introduce bindings, change evaluation order). So automatically converting arbitrary macro into a function is a really bad idea even if it is possible.
Is it possible to write a function that would take arbitrary macro and turn it into a function so that it can be passed as an argument to another function?
No, but it is somewhat doable to write a macro that would take some macro and turn it into a function.
#lang racket
(require (for-syntax racket/list))
(define-syntax (->proc stx)
(syntax-case stx ()
[(_ mac #:arity arity)
(with-syntax ([(args ...) (generate-temporaries (range (syntax-e #'arity)))])
#'(λ (args ...) (mac args ...)))]))
((->proc and #:arity 2) 42 12)
(apply (->proc and #:arity 2) '(#f 12))
((->proc and #:arity 2) #f (error 'not-short-circuit))
You might also be interested in identifier macro, which allows us to use an identifier as a macro in some context and function in another context. This could be used to create a first class and/or which short-circuits when it's used as a macro, but could be passed as a function value in non-transformer position.
On the topic of first class macro, take a look at https://en.wikipedia.org/wiki/Fexpr. It's known to be a bad idea.
Not in the way you probably expect
To see why, here is a way of thinking about macros: A macro is a function which takes a bit of source code and turns it into another bit of source code: the expansion of the macro. In other words a macro is a function whose domain and range are source code.
Once the source code is fully expanded, then it's fed to either an evaluator or a compiler. Let's assume it's fed to a compiler because it makes the question easier to answer: a compiler itself is simply a function whose domain is source code and whose range is some sequence of instructions for a machine (which may or may not be a real machine) to execute. Those instructions might include things like 'call this function on these arguments'.
So, what you are asking is: can the 'this function' in 'call this function on these arguments' be some kind of macro? Well, yes, it could be, but whatever source code it is going to transform certainly can not be the source code of the program you are executing, because that is gone: all that's left is the sequence of instructions that was the return value of the compiler.
So you might say: OK, let's say we disallow compilers: can we do it now? Well, leaving aside that 'disallowing compilers' is kind of a serious limitation, this was, in fact, something that very old dialects of Lisp sort-of did, using a construct called a FEXPR, as mentioned in another answer. It's important to realise that FEXPRs existed because people had not yet invented macros. Pretty soon, people did invent macros, and although FEXPRs and macros coexisted for a while – mostly because people had written code which used FEXPRs which they wanted to keep running, and because writing macros was a serious pain before things like backquote existed – FEXPRs died out. And they died out because they were semantically horrible: even by the standards of 1960s Lisps they were semantically horrible.
Here's one small example of why FEXPRs are so horrible: Let's say I write this function in a language with FEXPRs:
(define (foo f g x)
(apply f (g x)))
Now: what happens when I call foo? In particular, what happens if f might be a FEXPR?. Well, the answer is that I can't compile foo at all: I have to wait until run-time and make some on-the-fly decision about what to do.
Of course this isn't what these old Lisps with FEXPRs probably did: they would just silently have assumed that f was a normal function (which they would have called an EXPR) and compiled accordingly (and yes, even very old Lisps had compilers). If you passed something which was a FEXPR you just lost: either the thing detected that, or more likely it fall over horribly or gave you some junk answer.
And this kind of horribleness is why macros were invented: macros provide a semantically sane approach to processing Lisp code which allows (eventually, this took a long time to actually happen) minor details like compilation being possible at all, code having reasonable semantics and compiled code having the same semantics as interpreted code. These are features people like in their languages, it turns out.
Incidentally, in both Racket and Common Lisp, macros are explicitly functions. In Racket they are functions which operate on special 'syntax' objects because that's how you get hygiene, but in Common Lisp, which is much less hygienic, they're just functions which operate on CL source code, where the source code is simply made up of lists, symbols &c.
Here's an example of this in Racket:
> (define foo (syntax-rules ()
[(_ x) x]))
> foo
#<procedure:foo>
OK, foo is now just an ordinary function. But it's a function whose domain & range are Racket source code: it expects a syntax object as an argument and returns another one:
> (foo 1)
; ?: bad syntax
; in: 1
; [,bt for context]
This is because 1 is not a syntax object.
> (foo #'(x 1))
#<syntax:readline-input:5:10 1>
> (syntax-e (foo #'(x 1)))
1
And in CL this is even easier to see: Here's a macro definition:
(defmacro foo (form) form)
And now I can get hold of the macro's function and call it on some CL source code:
> (macro-function 'foo)
#<Function foo 4060000B6C>
> (funcall (macro-function 'foo) '(x 1) nil)
1
In both Racket and CL, macros are, in fact, first-class (or, in the case of Racket: almost first-class, I think): they are functions which operate on source code, which itself is first-class: you can write Racket and CL programs which construct and manipulate source code in arbitrary ways: that's what macros are in these languages.
In the case of Racket I have said 'almost first-class', because I can't see a way, in Racket, to retrieve the function which sits behind a macro defined with define-syntax &c.
I've created something like this in Scheme, it's macro that return lambda that use eval to execute the macro:
(define-macro (macron m)
(let ((x (gensym)))
`(lambda (,x)
(eval `(,',m ,#,x)))))
Example usage:
;; normal eval
(define x (map (lambda (x)
(eval `(lambda ,#x)))
'(((x) (display x)) ((y) (+ y y)))))
;; using macron macro
(define x (map (macron lambda)
'(((x) (display x)) ((y) (+ y y)))))
and x in both cases is list of two functions.
another example:
(define-macro (+++ . args)
`(+ ,#args))
((macron +++) '(1 2 3))

Confused by Lisp Quoting

I have a question concerning evaluation of lists in lisp.
Why is (a) and (+ a 1) not evaluated,
(defun test (a) (+ a 1))
just like (print 4) is not evaluated here
(if (< 1 2) (print 3) (print 4))
but (print (+ 2 3)) is evaluated here
(test (print (+ 2 3)))
Does it have something to do with them being standard library functions? Is it possible for me to define functions like that in my lisp program?
As you probably know, Lisp compound forms are generally processed from the outside in. You must look at the symbol in the first position of the outermost nesting to understand a form. That symbol completely determines the meaning of the form. The following expressions all contain (b c) with completely different meaning; therefore, we cannot understand them by analyzing the (b c) part first:
;; Common Lisp: define a class A derived from B and C
(defclass a (b c) ())
;; Common Lisp: define a function of two arguments
(defun a (b c) ())
;; add A to the result of calling function B on variable C:
(+ a (b c))
Traditionally, Lisp dialects have divided forms into operator forms and function call forms. An operator form has a completely arbitrary meaning, determined by the piece of code which compiles or interprets that functions (e.g. the evaluation simply recurses over all of the function call's argument forms, and the resulting values are passed to the function).
From the early history, Lisp has allowed users to write their own operators. There existed two approaches to this: interpretive operators (historically known as fexprs) and compiling operators known as macros. Both hinge around the idea of a function which receives the unevaluated form as an argument, so that it can implement a custom strategy, thereby extending the evaluation model with new behaviors.
A fexpr type operator is simply handed the form at run-time, along with an environment object with which it can look up the values of variables and such. That operator then walks the form and implements the behavior.
A macro operator is handed the form at macro-expansion time (which usually happens when top-level forms are read, just before they are evaluated or compiled). Its job is not to interpret the form's behavior, but instead to translate it by generating code. I.e. a macro is a mini compiler. (Generated code can contain more macro calls; the macro expander will take care of that, ensuring that all macro calls are decimated.)
The fexpr approach fell out of favor, most likely because it is inefficient. It basically makes compilation impossible, whereas Lisp hackers valued compilation. (Lisp was already a compiled language from as early as circa 1960.) The fexpr approach is also hostile toward lexical environments; it requires the fexpr, which is a function, to be able to peer into the variable binding environment of the form in which its invoked, which is a kind of encapsulation violation that is not allowed by lexical scopes.
Macro writing is slightly more difficult, and in some ways less flexible than fexprs, but support for macro writing improved in Lisp through the 1960's into the 70's to make it close to as easy as possible. Macro originally had receive the whole form and then have to parse it themselves. The macro-defining system developed into something that provides macro functions with arguments that receive the broken-down syntax in easily digestible pieces, including some nested aspects of the syntax. The backquote syntax for writing code templates was also developed, making it much easier to express code generation.
So to answer your question, how can I write forms like that myself? For instance if:
;; Imitation of old-fashioned technique: receive the whole form,
;; extract parts from it and return the translation.
;; Common Lisp defmacro supports this via the &whole keyword
;; in macro lambda lists which lets us have access to the whole form.
;;
;; (Because we are using defmacro, we need to declare arguments "an co &optional al",
;; to make this a three argument macro with an optional third argument, but
;; we don't use those arguments. In ancient lisps, they would not appear:
;; a macro would be a one-argument function, and would have to check the number
;; of arguments itself, to flag bad syntax like (my-if 42) or (my-if).)
;;
(defmacro my-if (&whole if-form an co &optional al)
(let ((antecedent (second if-form)) ;; extract pieces ourselves
(consequent (third if-form)) ;; from whole (my-if ...) form
(alternative (fourth if-form)))
(list 'cond (list antecedent consequent) (list t alternative))))
;; "Modern" version. Use the parsed arguments, and also take advantage of
;; backquote syntax to write the COND with a syntax that looks like the code.
(defmacro my-if (antecedent consequent &optional alternative)
`(cond (,antecedent ,consequent) (t ,alternative))))
This is a fitting example because originally Lisp only had cond. There was no if in McCarthy's Lisp. That "syntactic sugar" was invented later, probably as a macro expanding to cond, just like my-if above.
if and defun are macros. Macros expand a form into a longer piece of code. At expansion time, none of the macro's arguments are evaluated.
When you try to write a function, but struggle because you need to implement a custom evaluation strategy, its a strong signal that you should be writing a macro instead.
Disclaimer: Depending on what kind of lisp you are using, if and defun might technically be called "special forms" and not macros, but the concept of delayed evaluation still applies.
Lisp consists of a model of evaluation of forms. Different Lisp dialects have different rules for those.
Let's look at Common Lisp.
data evaluates to itself
a function form is evaluated by calling the function on the evaluated arguments
special forms are evaluated according to rules defined for each special operator. The Common Lisp standard lists all of those, defines what they do in an informal way and there is no way to define new special operators by the user.
macros forms are transformed, the result is evaluated
How IF, DEFUN etc. works and what they evaluated, when it is doen and what is not evaluated is defined in the Common Lisp standard.

What are the practical differences between special forms and macros?

Are there any practical differences between special forms and macros? In what do they differ?
The terms aren't quite synonymous, but they aren't exclusive either (this answer assumes Scheme):
A special form (also known as a syntax in the Scheme Reports) is an expression that's not evaluated according to the default rule for function application. (The default rule, just to be explicit, is to eval all of the subexpressions, and then apply the result of the first one to the list of the results of the others.)
The macro system is a language feature that allows definition of new special forms within the language itself. A macro is a special form defined using the macro system.
So you could say that "special form" is a term that pertains to interface or semantics, whereas "macro" is a term that pertains to implementation. "Special form" means "these expressions are evaluated with a special rule," while "macro" means "here's an implementation of a special rule for evaluating some expressions."
Now one important thing is that most Scheme special forms can be defined as macros from a really small core of primitives: lambda, if and macros. A minimal Scheme implementation that provides only these can still implement the rest as macros; recent Scheme Reports have made that distinction by referring to such special forms as "library syntax" that can be defined in terms of macros. In practice, however, practical Scheme systems often implement a richer set of forms as primitives.
Semantically speaking, the only thing that matters about an expression is what rule is used to evaluate it, not how that rule is implemented. So in that sense, it's not important whether a special form is implemented as a macro or a primitive. But on the other hand, the implementation details of a Scheme system often "leak," so you may find yourself caring about it...
Lisp has certain language primitives, which make up Lisp forms:
literal data: numbers, strings, structures, ...
function calls, like (sin 2.1) or like ((lambda (a b) (+ a b 2)) 3 4)
special operators used in special forms. These are the primitive built-in language elements. See Special Operators in Common Lisp. These need to be implemented in the interpreter and compiler. Common Lisp provides no way for the developer to introduce new special operators or to provide your own version of these. A code parsing tool will need to understand these special operators; these tools are usually called 'code walkers' in the Lisp community. During the definition of the Common Lisp standard, it was made sure that the number is very small and that all extensions otherwise are done via new functions and new macros.
macros: macros are functions which are transforming source code. The transformation will happen recursively until no macro is left in the source code. Common Lisp has built-in macros and allows the user to write new ones.
So the most important practical difference between special forms and macros is this: special operators are built-in syntax and semantics. They can't be written by the developer. Macros can be written by the developer.
In contrast to special forms, macro forms can be macroexpanded:
CL-USER(1): (macroexpand '(with-slots (x y z)
foo
(format t "~&X = ~A" x)))
(LET ((#:G925 FOO))
(DECLARE (IGNORABLE #:G925))
(DECLARE (SB-PCL::%VARIABLE-REBINDING #:G925 FOO))
#:G925
(SYMBOL-MACROLET ((X (SLOT-VALUE #:G925 'X))
(Y (SLOT-VALUE #:G925 'Y))
(Z (SLOT-VALUE #:G925 'Z)))
(FORMAT T "~&X = ~A" X)))
T
For me the most practical difference has been in the debugger: Macros don't show up in the debugger; instead, the (typically) obscure code from the macro's expansion shows up in the debugger. It is a real pain to debug such code and a good reason to ensure your macros are rock solid before you start relying upon them.
the super short answer for the lazy
You can write your own macroes any time you want, though you can't add special forms without recompiling clojure.

Why the function/macro dichotomy?

Why is the function/macro dichotomy present in Common Lisp?
What are the logical problems in allowing the same name representing both a macro (taking precedence when found in function position in compile/eval) and a function (usable for example with mapcar)?
For example having second defined both as a macro and as a function would allow to use
(setf (second x) 42)
and
(mapcar #'second L)
without having to create any setf trickery.
Of course it's clear that macros can do more than functions and so the analogy cannot be complete (and I don't think of course that every macro shold also be a function) but why forbidding it by making both sharing a single namespace when it could be potentially useful?
I hope I'm not offending anyone, but I don't really find a "Why doing that?" response really pertinent... I'm looking for why this is a bad idea. Imposing an arbitrary limitation because no good use is known is IMO somewhat arrogant (sort of assumes perfect foresight).
Or are there practical problems in allowing it?
Macros and Functions are two very different things:
macros are using source (!!!) code and are generating new source (!!!) code
functions are parameterized blocks of code.
Now we can look at this from several angles, for example:
a) how do we design a language where functions and macros are clearly identifiable and are looking different in our source code, so we (the human) can easily see what is what?
or
b) how do we blend macros and functions in a way that the result is most useful and has the most useful rules controlling its behavior? For the user it should not make a difference to use a macro or a function.
We really need to convince ourselves that b) is the way to go and we would like to use a language where macros and functions usage looks the same and is working according to similar principles. Take ships and cars. They look different, their use case is mostly different, they transport people - should we now make sure that the traffic rules for them are mostly identical, should we make them different or should we design the rules for their special usage?
For functions we have problems like: defining a function, scope of functions, life-time of functions, passing functions around, returning functions, calling functions, shadowing of functions, extension of functions, removing the definition a function, compilation and interpretation of functions, ...
If we would make macros appear mostly similar to functions, we need to address most or all above issues for them.
In your example you mention a SETF form. SETF is a macro that analyses the enclosed form at macro expansion time and generates code for a setter. It has little to do with SECOND being a macro or not. Having SECOND being a macro would not help at all in this situation.
So, what is a problem example?
(defmacro foo (a b)
(if (and (numberp b) (zerop b))
a
`(- ,a ,b)))
(defun bar (x list)
(mapcar #'foo (list x x x x) '(1 2 3 4)))
Now what should that do? Intuitively it looks easy: map FOO over the lists. But it isn't. When Common Lisp was designed, I would guess, it was not clear what that should do and how it should work. If FOO is a function, then it was clear: Common Lisp took the ideas from Scheme behind lexically scoped first-class functions and integrated it into the language.
But first-class macros? After the design of Common Lisp a bunch of research went into this problem and investigated it. But at the time of Common Lisp's design, there was no wide-spread use of first-class macros and no experience with design approaches. Common Lisp is standardizing on what was known at the time and what the language users thought necessary to develop (the object-system CLOS is kind of novel, based on earlier experience with similar object-systems) software with. Common Lisp was not designed to have the theoretically most pleasing Lisp dialect - it was designed to have a powerful Lisp which allows the efficient implementation of software.
We could work around this and say, passing macros is not possible. The developer would have to provide a function under the same name, which we pass around.
But then (funcall #'foo 1 2) and (foo 1 2) would invoke different machineries? In the first case the function fooand in the second case we use the macro foo to generate code for us? Really? Do we (as human programmers) want this? I think not - it looks like it makes programming much more complicated.
From a pragmatic point of view: Macros and the mechanism behind it are already complicated enough that most programmers have difficulties dealing with it in real code. They make debugging and code understanding much harder for a human. On the surface a macro makes code easier to read, but the price is the need to understand the code expansion process and result.
Finding a way to further integrate macros into the language design is not an easy task.
readscheme.org has some pointers to Macro-related research wrt. Scheme: Macros
What about Common Lisp
Common Lisp provides functions which can be first-class (stored, passed around, ...) and lexically scoped naming for them (DEFUN, FLET, LABELS, FUNCTION, LAMBDA).
Common Lisp provides global macros (DEFMACRO) and local macros (MACROLET).
Common Lisp provides global compiler macros (DEFINE-COMPILER-MACRO).
With compiler macros it is possible to have a function or macro for a symbol AND a compiler macro. The Lisp system can decide to prefer the compiler macro over the macro or function. It can also ignore them entirely. This mechanism is mostly used for the user to program specific optimizations. Thus it does not solve any macro related problems, but provides a pragmatic way to program global optimizations.
I think that Common Lisp's two namespaces (functions and values), rather than three (macros, functions, and values), is a historical contingency.
Early Lisps (in the 1960s) represented functions and values in different ways: values as bindings on the runtime stack, and functions as properties attached to symbols in the symbol table. This difference in implementation led to the specification of two namespaces when Common Lisp was standardized in the 1980s. See Richard Gabriel's paper Technical Issues of Separation in Function Cells and Value Cells for an explanation of this decision.
Macros (and their ancestors, FEXPRs, functions which do not evaluate their arguments) were stored in many Lisp implementations in the symbol table, in the same way as functions. It would have been inconvenient for these implementations if a third namespace (for macros) had been specified, and would have caused backwards-compatibility problems for many programs.
See Kent Pitman's paper Special Forms in Lisp for more about the history of FEXPRs, macros and other special forms.
(Note: Kent Pitman's website is not working for me, so I've linked to the papers via archive.org.)
Because then the exact same name would represent two different objects, depending on the context. It makes the programme unnecessarily difficult to understand.
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