I'm trying to learn and understand the Lisp programming language to a deep level. The function + evaluates its arguments in applicative order:
(+ 1 (+ 1 2))
(+ 1 2) will be evaluated and then (+ 1 3) will be evaluated, but the if function works differently:
(if (> 1 2) (not-defined 1 2) 1)
As the form (not-defined 1 2) isn't evaluated, the program doesn't break.
How can the same syntax lead to different argument evaluation? How is the if function defined so that its arguments aren't evaluated?
if is a special operator, not an ordinary function.
This means that the normal rule that the rest elements in the compound form are evaluated before the function associated with the first element is invoked is not applicable (in that it is similar to macro forms).
The way this is implemented in a compiler and/or an interpreter is that one looks at the compound form and decides what to do with it based on its first element:
if it is a special operator, it does its special thing;
if it is a macro, its macro-function gets the whole form;
otherwise it is treated as a function - even if no function is defined.
Note that some special forms can be defined as macros expanding to other special forms, but some special forms must actually be present.
E.g., one can define if in terms of cond:
(defmacro my-if (condition yes no)
`(cond (,condition ,yes)
(t ,no)))
and vice versa (much more complicated - actually, cond is a macro, usually expanding into a sequence of ifs).
PS. Note that the distinction between system-supplied macros and special operators, while technically crisp and clear (see special-operator-p and macro-function), is ideologically blurred because
An implementation is free to implement a Common Lisp special operator
as a macro. An implementation is free to implement any macro operator
as a special operator, but only if an equivalent definition of the
macro is also provided.
sds's answer answers this question well, but there are a few more general aspects that I think are worth mentioning. As that answer and others have pointed out, if, is built into the language as a special operator, because it really is a kind of primitive. Most importantly, if is not a function.
That said, the functionality of if can be achieved using just functions and normal function calling where all the arguments are evaluated. Thus, conditionals can be implemented in the lambda calculus, on which languages in the family are somewhat based, but which doesn't have a conditional operator.
In the lambda calculus, one can define true and false as functions of two arguments. The arguments are presumed to be functions, and true calls the first of its arguments, and false calls the second. (This is a slight variation of Church booleans which simply return their first or second argument.)
true = λ[x y].(x)
false = λ[x y].(y)
(This is obviously a departure from boolean values in Common Lisp, where nil is false and anything else is true.) The benefit of this, though, is that we can use a boolean value to call one of two functions, depending on whether the boolean is true or false. Consider the Common Lisp form:
(if some-condition
then-part
else-part)
If were were using the booleans as defined above, then evaluating some-condition will produce either true or false, and if we were to call that result with the arguments
(lambda () then-part)
(lambda () else-part)
then only one of those would be called, so only one of then-part and else-part would actually be evaluated. In general, wrapping some forms up in a lambda is a good way to be able delay the evaluation of those forms.
The power of the Common Lisp macro system means that we could actually define an if macro using the types of booleans described above:
(defconstant true
(lambda (x y)
(declare (ignore y))
(funcall x)))
(defconstant false
(lambda (x y)
(declare (ignore x))
(funcall y)))
(defmacro new-if (test then &optional else)
`(funcall ,test
(lambda () ,then)
(lambda () ,else)))
With these definitions, some code like this:
(new-if (member 'a '(1 2 3))
(print "it's a member")
(print "it's not a member"))))
expands to this:
(FUNCALL (MEMBER 'A '(1 2 3)) ; assuming MEMBER were rewritten
(LAMBDA () (PRINT "it's a member")) ; to return `true` or `false`
(LAMBDA () (PRINT "it's not a member")))
In general, if there is some form and some of the arguments aren't getting evaluated, then the (car of the) form is either a Common Lisp special operator or a macro. If you need to write a function where the arguments will be evaluated, but you want some forms not to be evaluated, you can wrap them up in lambda expressions and have your function call those anonymous functions conditionally.
This is a possible way to implement if, if you didn't already have it in the language. Of course, modern computer hardware isn't based on a lambda calculus interpreter, but rather on CPUs that have test and jump instructions, so it's more efficient for the language to provide if a primitive and to compile down to the appropriate machine instructions.
Lisp syntax is regular, much more regular than other languages, but it's still not completely regular: for example in
(let ((x 0))
x)
let is not the name of a function and ((x 0)) is not a bad form in which a list that is not a lambda form has been used in the first position.
There are quite a few "special cases" (still a lot less than other languages, of course) where the general rule of each list being a function call is not followed, and if is one of them. Common Lisp has quite a few "special forms" (because absolute minimality was not the point) but you can get away for example in a scheme dialect with just five of them: if, progn, quote, lambda and set! (or six if you want macros).
While the syntax of Lisp is not totally uniform the underlying representation of code is however quite uniform (just lists and atoms) and the uniformity and simplicity of representation is what facilitates metaprogramming (macros).
"Lisp has no syntax" is a statement with some truth in it, but so it's the statement "Lisp has two syntaxes": one syntax is what uses the reader to convert from character streams to s-expressions, another syntax is what uses the compiler/evaluator to convert from s-expressions to executable code.
It's also true that Lisp has no syntax because neither of those two levels is fixed. Differently from other programming languages you can customize both the first step (using reader macros) and the second step (using macros).
It would not make any sense to do so. Example: (if (ask-user-should-i-quit) (quit) (continue)). Should that quit, even though the user does not want to?
IF is not a function in Lisp. It is a special built-in operator. Lisp a several built-in special operators. See: Special Forms. Those are not functions.
The arguments are not evaluated as for functions, because if is a special operator. Special operators can be evaluated in any arbitrary way, that's why they're called special.
Consider e.g.
(if (not (= x 0))
(/ y x))
If the division was always evaluated, there could be a division by zero error which obviously was not intended.
If isn't a function, it's a special form. If you wanted to implement similar functionality yourself, you could do so by defining a macro rather than a function.
This answer applies to Common Lisp, but it'll probably the same for most other Lisps (though in some if may be a macro rather than a special form).
Related
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))
I'm familiar with Emacs Lisp, but not Common (or any other) Lisp. Some Lisp programmers suggest (e.g. A basic function for emacs) that it's good to use #' in front of function arguments in Lisp code. For example:
(mapc #'my-fun '(1 2 3))
In Emacs Lisp, I believe that this is equivalent to
(mapc 'my-fun '(1 2 3))
From the elisp manual, section 12.7.
The read syntax #' is a short-hand for using function. The
following forms are all equivalent:
(lambda (x) (* x x))
(function (lambda (x) (* x x)))
#'(lambda (x) (* x x))
and the help for function
function is a special form in eval.c.
(function ARG)
Like quote, but preferred for objects which are functions. In byte
compilation, function causes its argument to be compiled. quote
cannot do that.
So it seems like a potential optimization, but no more. Moreover, coming from an ML/Haskell background, it seems strange to treat a function differently from any other data.
Question:
Do you agree that #' should be used in emacs-lisp function arguments?
(A brief explanation of why they're needed in Common Lisp would be great as well.)
Notes:
I thought it may read a different cell when the #' is omitted (value vs function). But
this seems wrong because the documentation for function doesn't say anything about grabbing the function cell. This is achieved by using symbol-function.
Related questions are
The #' in common lisp
Writing lambda expressions in common lisp
Why #' is used before lambda in Common Lisp?
but they seem to suggest that the #' is unnecessary, at least in lambdas.
The quote character in #'foo has nothing to do with the one in 'foo.
#'foo is replaced at read time by (function foo). When that is compiled and executed, it looks up the functional definition named by foo (by defun, flet, labels or similar).
'foo is replaced at read time by (quote foo). When that is compiled and executed, it is simply replaced by the symbol foo.
Funcall and apply (and thus generally higher order functions) take function designators as arguments. A function designator can be a function or a symbol naming a function, so both #'foo and 'foo are function designators.
Therefore, the forms 'foo and #'foo seem interchangable at first glance. However, the lookup of the real function is done at different times—#'foo is looked up where it is invoked, while the function named by 'foo would only be looked up when it is finally applied.
If you use a function designator many times over, it is much more efficient to do the lookup just once, or even just at compile time. That can be a really big timesaver, and translates to a smoother user experience in an editor.
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.
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
Consider this piece of code:
(defvar lst '(1 1))
(defmacro get-x (x lst)
`(nth ,x ,lst))
(defun get-y (y lst)
(nth y lst))
Now let us assume that I want to change the value of the elements of the list called lst, the car with get-x and the cdr with get-y.
As I try to change the value with get-x (with setf) everything goes fine but if I try it with get-y it signals an error (shortened):
; caught STYLE-WARNING:
; undefined function: (SETF GET-STUFF)
Why does this happen?
I myself suspect that this happens because the macro simply expands and the function nth simply returns a reference to the value of an element in the list and the function on the other hand evaluates the function-call to nth and returns the value of the referenced value (sounds confusing).
Am I correct in my suspicions?
If I am correct then how will one know what is simply a reference to a value and an actual value?
The error does not happen with the macro version, because, as you assumed, the expression (setf (get-x some-x some-list) some-value) will be expanded (at compile-time) into something like (setf (nth some-x some-list) some-value) (not really, but the details of setf-expansion are complex), and the compiler knows, how to deal with that (i.e., there is a suitable setf expander defined for function nth).
However, in the case of get-y, the compiler has no setf expander, unless you provide one. The easiest way to do so would be
(defun (setf get-y) (new-value x ls) ; Note the function's name: setf get-y
(setf (nth x ls) new-value))
Note, that there are a few conventions regarding setf-expanders:
The new value is always provided as the first argument to the setf function
All setf functions are supposed to return the new value as their result (as this is, what the entire setf form is supposed to return)
There is, BTW, no such concept as a "reference" in Common Lisp (at least not in the C++ sense), though there once were Lisp dialects which had locatives. Generalized place forms (ie., setf and its machinery) work very differently from plain C++ style references. See the CLHS, if you are curious about the details.
SETF is a macro.
The idea is that to set and read elements from data structures are two operations, but usually require two different names (or maybe even something more complex). SETF now enables you to use just one name for both:
(get-something x)
Above reads a datastructure. The inverse then simply is:
(setf (get-something x) :foobar)
Above sets the datastructure at X with :FOOBAR.
SETF does not treat (get-something x) as a reference or something like that. It just has a database of inverse operations for each operation. If you use GET-SOMETHING, it knows what the inverse operation is.
How does SETF know it? Simple: you have to tell it.
For The NTH operation, SETF knows how to set the nth element. That's builtin into Common Lisp.
For your own GET-Y operation SETF does not have that information. You have to tell it. See the Common Lisp HyperSpec for examples. One example is to use DEFUN and (SETF GET-Y) as a function name.
Also note following style problems with your example:
lst is not a good name for a DEFVAR variable. Use *list* as a name to make clear that it is a special variable declared by DEFVAR (or similar).
'(1 2) is a literal constant. If you write a Common Lisp program, the effects of changing it are undefined. If you want to change a list later, you should cons it with LIST or something like COPY-LIST.