doc: Use fashionable terminology for macros.

[bpt/guile.git] / doc / ref / api-macros.texi

@c -*-texinfo-*-
@c This is part of the GNU Guile Reference Manual.
@c Copyright (C)  1996, 1997, 2000, 2001, 2002, 2003, 2004, 2009, 2010, 2011
@c   Free Software Foundation, Inc.
@c See the file guile.texi for copying conditions.

@node Macros
@section Macros

At its best, programming in Lisp is an iterative process of building up a
language appropriate to the problem at hand, and then solving the problem in
that language. Defining new procedures is part of that, but Lisp also allows
the user to extend its syntax, with its famous @dfn{macros}.

@cindex macros
@cindex transformation
Macros are syntactic extensions which cause the expression that they appear in
to be transformed in some way @emph{before} being evaluated. In expressions that
are intended for macro transformation, the identifier that names the relevant
macro must appear as the first element, like this:

@lisp
(@var{macro-name} @var{macro-args} @dots{})
@end lisp

@cindex macro expansion
@cindex domain-specific language
@cindex embedded domain-specific language
@cindex DSL
@cindex EDSL
Macro expansion is a separate phase of evaluation, run before code is
interpreted or compiled. A macro is a program that runs on programs, translating
an embedded language into core Scheme@footnote{These days such embedded
languages are often referred to as @dfn{embedded domain-specific
languages}, or EDSLs.}.

@menu
* Defining Macros::             Binding macros, globally and locally.
* Syntax Rules::                Pattern-driven macros.
* Syntax Case::                 Procedural, hygienic macros.
* Defmacros::                   Lisp-style macros.
* Identifier Macros::           Identifier macros.
* Eval When::                   Affecting the expand-time environment.
* Internal Macros::             Macros as first-class values.
@end menu

@node Defining Macros
@subsection Defining Macros

A macro is a binding between a keyword and a syntax transformer. Since it's
difficult to discuss @code{define-syntax} without discussing the format of
transformers, consider the following example macro definition:

@example
(define-syntax when
  (syntax-rules ()
    ((when condition exp ...)
     (if condition
         (begin exp ...)))))

(when #t
  (display "hey ho\n") 
  (display "let's go\n"))
@print{} hey ho
@print{} let's go
@end example

In this example, the @code{when} binding is bound with @code{define-syntax}.
Syntax transformers are discussed in more depth in @ref{Syntax Rules} and
@ref{Syntax Case}.

@deffn {Syntax} define-syntax keyword transformer
Bind @var{keyword} to the syntax transformer obtained by evaluating
@var{transformer}.

After a macro has been defined, further instances of @var{keyword} in Scheme
source code will invoke the syntax transformer defined by @var{transformer}.
@end deffn

One can also establish local syntactic bindings with @code{let-syntax}.

@deffn {Syntax} let-syntax ((keyword transformer) ...) exp...
Bind @var{keyword...} to @var{transformer...} while expanding @var{exp...}.

A @code{let-syntax} binding only exists at expansion-time. 

@example
(let-syntax ((unless
              (syntax-rules ()
                ((unless condition exp ...)
                 (if (not condition)
                     (begin exp ...))))))
  (unless #t
    (primitive-exit 1))
  "rock rock rock")
@result{} "rock rock rock"
@end example
@end deffn

A @code{define-syntax} form is valid anywhere a definition may appear: at the
top-level, or locally. Just as a local @code{define} expands out to an instance
of @code{letrec}, a local @code{define-syntax} expands out to
@code{letrec-syntax}.

@deffn {Syntax} letrec-syntax ((keyword transformer) ...) exp...
Bind @var{keyword...} to @var{transformer...} while expanding @var{exp...}.

In the spirit of @code{letrec} versus @code{let}, an expansion produced by
@var{transformer} may reference a @var{keyword} bound by the
same @var{letrec-syntax}.

@example
(letrec-syntax ((my-or
                 (syntax-rules ()
                   ((my-or)
                    #t)
                   ((my-or exp)
                    exp)
                   ((my-or exp rest ...)
                    (let ((t exp))
                      (if exp
                          exp
                          (my-or rest ...)))))))
  (my-or #f "rockaway beach"))
@result{} "rockaway beach"
@end example
@end deffn

@node Syntax Rules
@subsection Syntax-rules Macros

@code{syntax-rules} macros are simple, pattern-driven syntax transformers, with
a beauty worthy of Scheme.

@deffn {Syntax} syntax-rules literals (pattern template)...
Create a syntax transformer that will rewrite an expression using the rules
embodied in the @var{pattern} and @var{template} clauses.
@end deffn

A @code{syntax-rules} macro consists of three parts: the literals (if any), the
patterns, and as many templates as there are patterns.

When the syntax expander sees the invocation of a @code{syntax-rules} macro, it
matches the expression against the patterns, in order, and rewrites the
expression using the template from the first matching pattern. If no pattern
matches, a syntax error is signalled.

@subsubsection Patterns

We have already seen some examples of patterns in the previous section:
@code{(unless condition exp ...)}, @code{(my-or exp)}, and so on. A pattern is
structured like the expression that it is to match. It can have nested structure
as well, like @code{(let ((var val) ...) exp exp* ...)}. Broadly speaking,
patterns are made of lists, improper lists, vectors, identifiers, and datums.
Users can match a sequence of patterns using the ellipsis (@code{...}).

Identifiers in a pattern are called @dfn{literals} if they are present in the
@code{syntax-rules} literals list, and @dfn{pattern variables} otherwise. When
building up the macro output, the expander replaces instances of a pattern
variable in the template with the matched subexpression.

@example
(define-syntax kwote
  (syntax-rules ()
    ((kwote exp)
     (quote exp))))
(kwote (foo . bar))
@result{} (foo . bar)
@end example

An improper list of patterns matches as rest arguments do:

@example
(define-syntax let1
  (syntax-rules ()
    ((_ (var val) . exps)
     (let ((var val)) . exps))))
@end example

However this definition of @code{let1} probably isn't what you want, as the tail
pattern @var{exps} will match non-lists, like @code{(let1 (foo 'bar) . baz)}. So
often instead of using improper lists as patterns, ellipsized patterns are
better. Instances of a pattern variable in the template must be followed by an
ellipsis.

@example
(define-syntax let1
  (syntax-rules ()
    ((_ (var val) exp ...)
     (let ((var val)) exp ...))))
@end example

This @code{let1} probably still doesn't do what we want, because the body
matches sequences of zero expressions, like @code{(let1 (foo 'bar))}. In this
case we need to assert we have at least one body expression. A common idiom for
this is to name the ellipsized pattern variable with an asterisk:

@example
(define-syntax let1
  (syntax-rules ()
    ((_ (var val) exp exp* ...)
     (let ((var val)) exp exp* ...))))
@end example

A vector of patterns matches a vector whose contents match the patterns,
including ellipsizing and tail patterns.

@example
(define-syntax letv
  (syntax-rules ()
    ((_ #((var val) ...) exp exp* ...)
     (let ((var val) ...) exp exp* ...))))
(letv #((foo 'bar)) foo)
@result{} foo
@end example

Literals are used to match specific datums in an expression, like the use of
@code{=>} and @code{else} in @code{cond} expressions.

@example
(define-syntax cond1
  (syntax-rules (=> else)
    ((cond1 test => fun)
     (let ((exp test))
       (if exp (fun exp) #f)))
    ((cond1 test exp exp* ...)
     (if test (begin exp exp* ...)))
    ((cond1 else exp exp* ...)
     (begin exp exp* ...))))

(define (square x) (* x x))
(cond1 10 => square)
@result{} 100
(let ((=> #t))
  (cond1 10 => square))
@result{} #<procedure square (x)>
@end example

A literal matches an input expression if the input expression is an identifier
with the same name as the literal, and both are unbound@footnote{Language
lawyers probably see the need here for use of @code{literal-identifier=?} rather
than @code{free-identifier=?}, and would probably be correct. Patches
accepted.}.

If a pattern is not a list, vector, or an identifier, it matches as a literal,
with @code{equal?}.

@example
(define-syntax define-matcher-macro
  (syntax-rules ()
    ((_ name lit)
     (define-syntax name
       (syntax-rules ()
        ((_ lit) #t)
        ((_ else) #f))))))

(define-matcher-macro is-literal-foo? "foo")

(is-literal-foo? "foo")
@result{} #t
(is-literal-foo? "bar")
@result{} #f
(let ((foo "foo"))
  (is-literal-foo? foo))
@result{} #f
@end example

The last example indicates that matching happens at expansion-time, not
at run-time.

Syntax-rules macros are always used as @code{(@var{macro} . @var{args})}, and
the @var{macro} will always be a symbol. Correspondingly, a @code{syntax-rules}
pattern must be a list (proper or improper), and the first pattern in that list
must be an identifier. Incidentally it can be any identifier -- it doesn't have
to actually be the name of the macro. Thus the following three are equivalent:

@example
(define-syntax when
  (syntax-rules ()
    ((when c e ...)
     (if c (begin e ...)))))

(define-syntax when
  (syntax-rules ()
    ((_ c e ...)
     (if c (begin e ...)))))

(define-syntax when
  (syntax-rules ()
    ((something-else-entirely c e ...)
     (if c (begin e ...)))))
@end example

For clarity, use one of the first two variants. Also note that since the pattern
variable will always match the macro itself (e.g., @code{cond1}), it is actually
left unbound in the template.

@subsubsection Hygiene

@code{syntax-rules} macros have a magical property: they preserve referential
transparency. When you read a macro definition, any free bindings in that macro
are resolved relative to the macro definition; and when you read a macro
instantiation, all free bindings in that expression are resolved relative to the
expression.

This property is sometimes known as @dfn{hygiene}, and it does aid in code
cleanliness. In your macro definitions, you can feel free to introduce temporary
variables, without worrying about inadvertently introducing bindings into the
macro expansion.

Consider the definition of @code{my-or} from the previous section:

@example
(define-syntax my-or
  (syntax-rules ()
    ((my-or)
     #t)
    ((my-or exp)
     exp)
    ((my-or exp rest ...)
     (let ((t exp))
       (if exp
           exp
           (my-or rest ...))))))
@end example

A naive expansion of @code{(let ((t #t)) (my-or #f t))} would yield:

@example
(let ((t #t))
  (let ((t #f))
    (if t t t)))
@result{} #f
@end example

@noindent
Which clearly is not what we want. Somehow the @code{t} in the definition is
distinct from the @code{t} at the site of use; and it is indeed this distinction
that is maintained by the syntax expander, when expanding hygienic macros.

This discussion is mostly relevant in the context of traditional Lisp macros
(@pxref{Defmacros}), which do not preserve referential transparency. Hygiene
adds to the expressive power of Scheme.

@subsubsection Shorthands

One often ends up writing simple one-clause @code{syntax-rules} macros.
There is a convenient shorthand for this idiom, in the form of
@code{define-syntax-rule}.

@deffn {Syntax} define-syntax-rule (keyword . pattern) [docstring] template
Define @var{keyword} as a new @code{syntax-rules} macro with one clause.
@end deffn

Cast into this form, our @code{when} example is significantly shorter:

@example
(define-syntax-rule (when c e ...)
  (if c (begin e ...)))
@end example

@subsubsection Further Information

For a formal definition of @code{syntax-rules} and its pattern language, see
@xref{Macros, , Macros, r5rs, Revised(5) Report on the Algorithmic Language
Scheme}.

@code{syntax-rules} macros are simple and clean, but do they have limitations.
They do not lend themselves to expressive error messages: patterns either match
or they don't. Their ability to generate code is limited to template-driven
expansion; often one needs to define a number of helper macros to get real work
done. Sometimes one wants to introduce a binding into the lexical context of the
generated code; this is impossible with @code{syntax-rules}. Relatedly, they
cannot programmatically generate identifiers.

The solution to all of these problems is to use @code{syntax-case} if you need
its features. But if for some reason you're stuck with @code{syntax-rules}, you
might enjoy Joe Marshall's
@uref{http://sites.google.com/site/evalapply/eccentric.txt,@code{syntax-rules}
Primer for the Merely Eccentric}.

@node Syntax Case
@subsection Support for the @code{syntax-case} System

@code{syntax-case} macros are procedural syntax transformers, with a power
worthy of Scheme.

@deffn {Syntax} syntax-case syntax literals (pattern [guard] exp)...
Match the syntax object @var{syntax} against the given patterns, in order. If a
@var{pattern} matches, return the result of evaluating the associated @var{exp}.
@end deffn

Compare the following definitions of @code{when}:

@example
(define-syntax when
  (syntax-rules ()
    ((_ test e e* ...)
     (if test (begin e e* ...)))))

(define-syntax when
  (lambda (x)
    (syntax-case x ()
      ((_ test e e* ...)
       #'(if test (begin e e* ...))))))
@end example

Clearly, the @code{syntax-case} definition is similar to its @code{syntax-rules}
counterpart, and equally clearly there are some differences. The
@code{syntax-case} definition is wrapped in a @code{lambda}, a function of one
argument; that argument is passed to the @code{syntax-case} invocation; and the
``return value'' of the macro has a @code{#'} prefix.

All of these differences stem from the fact that @code{syntax-case} does not
define a syntax transformer itself -- instead, @code{syntax-case} expressions
provide a way to destructure a @dfn{syntax object}, and to rebuild syntax
objects as output.

So the @code{lambda} wrapper is simply a leaky implementation detail, that
syntax transformers are just functions that transform syntax to syntax. This
should not be surprising, given that we have already described macros as
``programs that write programs''. @code{syntax-case} is simply a way to take
apart and put together program text, and to be a valid syntax transformer it
needs to be wrapped in a procedure.

Unlike traditional Lisp macros (@pxref{Defmacros}), @code{syntax-case} macros
transform syntax objects, not raw Scheme forms. Recall the naive expansion of
@code{my-or} given in the previous section:

@example
(let ((t #t))
  (my-or #f t))
;; naive expansion:
(let ((t #t))
  (let ((t #f))
    (if t t t)))
@end example

Raw Scheme forms simply don't have enough information to distinguish the first
two @code{t} instances in @code{(if t t t)} from the third @code{t}. So instead
of representing identifiers as symbols, the syntax expander represents
identifiers as annotated syntax objects, attaching such information to those
syntax objects as is needed to maintain referential transparency.

@deffn {Syntax} syntax form
Create a syntax object wrapping @var{form} within the current lexical context.
@end deffn

Syntax objects are typically created internally to the process of expansion, but
it is possible to create them outside of syntax expansion:

@example
(syntax (foo bar baz))
@result{} #<some representation of that syntax>
@end example

@noindent
However it is more common, and useful, to create syntax objects when building
output from a @code{syntax-case} expression.

@example
(define-syntax add1
  (lambda (x)
    (syntax-case x ()
      ((_ exp)
       (syntax (+ exp 1))))))
@end example

It is not strictly necessary for a @code{syntax-case} expression to return a
syntax object, because @code{syntax-case} expressions can be used in helper
functions, or otherwise used outside of syntax expansion itself. However a
syntax transformer procedure must return a syntax object, so most uses of
@code{syntax-case} do end up returning syntax objects.

Here in this case, the form that built the return value was @code{(syntax (+ exp
1))}. The interesting thing about this is that within a @code{syntax}
expression, any appearance of a pattern variable is substituted into the
resulting syntax object, carrying with it all relevant metadata from the source
expression, such as lexical identity and source location.

Indeed, a pattern variable may only be referenced from inside a @code{syntax}
form. The syntax expander would raise an error when defining @code{add1} if it
found @var{exp} referenced outside a @code{syntax} form.

Since @code{syntax} appears frequently in macro-heavy code, it has a special
reader macro: @code{#'}. @code{#'foo} is transformed by the reader into
@code{(syntax foo)}, just as @code{'foo} is transformed into @code{(quote foo)}.

The pattern language used by @code{syntax-case} is conveniently the same
language used by @code{syntax-rules}. Given this, Guile actually defines
@code{syntax-rules} in terms of @code{syntax-case}:

@example
(define-syntax syntax-rules
  (lambda (x)
    (syntax-case x ()
      ((_ (k ...) ((keyword . pattern) template) ...)
       #'(lambda (x)
           (syntax-case x (k ...)
             ((dummy . pattern) #'template)
             ...))))))
@end example

And that's that.

@subsubsection Why @code{syntax-case}?

The examples we have shown thus far could just as well have been expressed with
@code{syntax-rules}, and have just shown that @code{syntax-case} is more
verbose, which is true. But there is a difference: @code{syntax-case} creates
@emph{procedural} macros, giving the full power of Scheme to the macro expander.
This has many practical applications.

A common desire is to be able to match a form only if it is an identifier. This
is impossible with @code{syntax-rules}, given the datum matching forms. But with
@code{syntax-case} it is easy:

@deffn {Scheme Procedure} identifier? syntax-object
Returns @code{#t} iff @var{syntax-object} is an identifier.
@end deffn

@example
;; relying on previous add1 definition
(define-syntax add1!
  (lambda (x)
    (syntax-case x ()
      ((_ var) (identifier? #'var)
       #'(set! var (add1 var))))))

(define foo 0)
(add1! foo)
foo @result{} 1
(add1! "not-an-identifier") @result{} error
@end example

With @code{syntax-rules}, the error for @code{(add1! "not-an-identifier")} would
be something like ``invalid @code{set!}''. With @code{syntax-case}, it will say
something like ``invalid @code{add1!}'', because we attach the @dfn{guard
clause} to the pattern: @code{(identifier? #'var)}. This becomes more important
with more complicated macros. It is necessary to use @code{identifier?}, because
to the expander, an identifier is more than a bare symbol.

Note that even in the guard clause, we reference the @var{var} pattern variable
within a @code{syntax} form, via @code{#'var}.

Another common desire is to introduce bindings into the lexical context of the
output expression. One example would be in the so-called ``anaphoric macros'',
like @code{aif}. Anaphoric macros bind some expression to a well-known
identifier, often @code{it}, within their bodies. For example, in @code{(aif
(foo) (bar it))}, @code{it} would be bound to the result of @code{(foo)}.

To begin with, we should mention a solution that doesn't work:

@example
;; doesn't work
(define-syntax aif
  (lambda (x)
    (syntax-case x ()
      ((_ test then else)
       #'(let ((it test))
           (if it then else))))))
@end example

The reason that this doesn't work is that, by default, the expander will
preserve referential transparency; the @var{then} and @var{else} expressions
won't have access to the binding of @code{it}.

But they can, if we explicitly introduce a binding via @code{datum->syntax}.

@deffn {Scheme Procedure} datum->syntax for-syntax datum
Create a syntax object that wraps @var{datum}, within the lexical context
corresponding to the syntax object @var{for-syntax}.
@end deffn

For completeness, we should mention that it is possible to strip the metadata
from a syntax object, returning a raw Scheme datum:

@deffn {Scheme Procedure} syntax->datum syntax-object
Strip the metadata from @var{syntax-object}, returning its contents as a raw
Scheme datum.
@end deffn

In this case we want to introduce @code{it} in the context of the whole
expression, so we can create a syntax object as @code{(datum->syntax x 'it)},
where @code{x} is the whole expression, as passed to the transformer procedure.

Here's another solution that doesn't work:

@example
;; doesn't work either
(define-syntax aif
  (lambda (x)
    (syntax-case x ()
      ((_ test then else)
       (let ((it (datum->syntax x 'it)))
         #'(let ((it test))
             (if it then else)))))))
@end example

The reason that this one doesn't work is that there are really two
environments at work here -- the environment of pattern variables, as
bound by @code{syntax-case}, and the environment of lexical variables,
as bound by normal Scheme. The outer let form establishes a binding in
the environment of lexical variables, but the inner let form is inside a
syntax form, where only pattern variables will be substituted. Here we
need to introduce a piece of the lexical environment into the pattern
variable environment, and we can do so using @code{syntax-case} itself:

@example
;; works, but is obtuse
(define-syntax aif
  (lambda (x)
    (syntax-case x ()
      ((_ test then else)
       ;; invoking syntax-case on the generated
       ;; syntax object to expose it to `syntax'
       (syntax-case (datum->syntax x 'it) ()
         (it
           #'(let ((it test))
               (if it then else))))))))

(aif (getuid) (display it) (display "none")) (newline)
@print{} 500
@end example

However there are easier ways to write this. @code{with-syntax} is often
convenient:

@deffn {Syntax} with-syntax ((pat val)...) exp...
Bind patterns @var{pat} from their corresponding values @var{val}, within the
lexical context of @var{exp...}.

@example
;; better
(define-syntax aif
  (lambda (x)
    (syntax-case x ()
      ((_ test then else)
       (with-syntax ((it (datum->syntax x 'it)))
         #'(let ((it test))
             (if it then else)))))))
@end example
@end deffn

As you might imagine, @code{with-syntax} is defined in terms of
@code{syntax-case}. But even that might be off-putting to you if you are an old
Lisp macro hacker, used to building macro output with @code{quasiquote}. The
issue is that @code{with-syntax} creates a separation between the point of
definition of a value and its point of substitution.

@pindex quasisyntax
@pindex unsyntax
@pindex unsyntax-splicing
So for cases in which a @code{quasiquote} style makes more sense,
@code{syntax-case} also defines @code{quasisyntax}, and the related
@code{unsyntax} and @code{unsyntax-splicing}, abbreviated by the reader as
@code{#`}, @code{#,}, and @code{#,@@}, respectively.

For example, to define a macro that inserts a compile-time timestamp into a
source file, one may write:

@example
(define-syntax display-compile-timestamp
  (lambda (x)
    (syntax-case x ()
      ((_)
       #`(begin
          (display "The compile timestamp was: ")
          (display #,(current-time))
          (newline))))))
@end example

Finally, we should mention the following helper procedures defined by the core
of @code{syntax-case}:

@deffn {Scheme Procedure} bound-identifier=? a b
Returns @code{#t} iff the syntax objects @var{a} and @var{b} refer to the same
lexically-bound identifier.
@end deffn

@deffn {Scheme Procedure} free-identifier=? a b
Returns @code{#t} iff the syntax objects @var{a} and @var{b} refer to the same
free identifier.
@end deffn

@deffn {Scheme Procedure} generate-temporaries ls
Return a list of temporary identifiers as long as @var{ls} is long.
@end deffn

Readers interested in further information on @code{syntax-case} macros should
see R. Kent Dybvig's excellent @cite{The Scheme Programming Language}, either
edition 3 or 4, in the chapter on syntax. Dybvig was the primary author of the
@code{syntax-case} system. The book itself is available online at
@uref{http://scheme.com/tspl4/}.

@node Defmacros
@subsection Lisp-style Macro Definitions

The traditional way to define macros in Lisp is very similar to procedure
definitions. The key differences are that the macro definition body should
return a list that describes the transformed expression, and that the definition
is marked as a macro definition (rather than a procedure definition) by the use
of a different definition keyword: in Lisp, @code{defmacro} rather than
@code{defun}, and in Scheme, @code{define-macro} rather than @code{define}.

@fnindex defmacro
@fnindex define-macro
Guile supports this style of macro definition using both @code{defmacro}
and @code{define-macro}.  The only difference between them is how the
macro name and arguments are grouped together in the definition:

@lisp
(defmacro @var{name} (@var{args} @dots{}) @var{body} @dots{})
@end lisp

@noindent
is the same as

@lisp
(define-macro (@var{name} @var{args} @dots{}) @var{body} @dots{})
@end lisp

@noindent
The difference is analogous to the corresponding difference between
Lisp's @code{defun} and Scheme's @code{define}.

Having read the previous section on @code{syntax-case}, it's probably clear that
Guile actually implements defmacros in terms of @code{syntax-case}, applying the
transformer on the expression between invocations of @code{syntax->datum} and
@code{datum->syntax}. This realization leads us to the problem with defmacros,
that they do not preserve referential transparency. One can be careful to not
introduce bindings into expanded code, via liberal use of @code{gensym}, but
there is no getting around the lack of referential transparency for free
bindings in the macro itself.

Even a macro as simple as our @code{when} from before is difficult to get right:

@example
(define-macro (when cond exp . rest)
  `(if ,cond
       (begin ,exp . ,rest)))

(when #f (display "Launching missiles!\n"))
@result{} #f

(let ((if list))
  (when #f (display "Launching missiles!\n")))
@print{} Launching missiles!
@result{} (#f #<unspecified>)
@end example

Guile's perspective is that defmacros have had a good run, but that modern
macros should be written with @code{syntax-rules} or @code{syntax-case}. There
are still many uses of defmacros within Guile itself, but we will be phasing
them out over time. Of course we won't take away @code{defmacro} or
@code{define-macro} themselves, as there is lots of code out there that uses
them.


@node Identifier Macros
@subsection Identifier Macros

When the syntax expander sees a form in which the first element is a macro, the
whole form gets passed to the macro's syntax transformer. One may visualize this
as:

@example
(define-syntax foo foo-transformer)
(foo @var{arg}...)
;; expands via
(foo-transformer #'(foo @var{arg}...))
@end example

If, on the other hand, a macro is referenced in some other part of a form, the
syntax transformer is invoked with only the macro reference, not the whole form.

@example
(define-syntax foo foo-transformer)
foo
;; expands via
(foo-transformer #'foo)
@end example

This allows bare identifier references to be replaced programmatically via a
macro. @code{syntax-rules} provides some syntax to effect this transformation
more easily.

@deffn {Syntax} identifier-syntax exp
Returns a macro transformer that will replace occurrences of the macro with
@var{exp}.
@end deffn

For example, if you are importing external code written in terms of @code{fx+},
the fixnum addition operator, but Guile doesn't have @code{fx+}, you may use the
following to replace @code{fx+} with @code{+}:

@example
(define-syntax fx+ (identifier-syntax +))
@end example

There is also special support for recognizing identifiers on the
left-hand side of a @code{set!} expression, as in the following:

@example
(define-syntax foo foo-transformer)
(set! foo @var{val})
;; expands via
(foo-transformer #'(set! foo @var{val}))
;; iff foo-transformer is a "variable transformer"
@end example

As the example notes, the transformer procedure must be explicitly
marked as being a ``variable transformer'', as most macros aren't
written to discriminate on the form in the operator position.

@deffn {Scheme Procedure} make-variable-transformer transformer
Mark the @var{transformer} procedure as being a ``variable
transformer''. In practice this means that, when bound to a syntactic
keyword, it may detect references to that keyword on the left-hand-side
of a @code{set!}.

@example
(define bar 10)
(define-syntax bar-alias
  (make-variable-transformer
   (lambda (x)
     (syntax-case x (set!)
       ((set! var val) #'(set! bar val))
       ((var arg ...) #'(bar arg ...))
       (var (identifier? #'var) #'bar)))))

bar-alias @result{} 10
(set! bar-alias 20)
bar @result{} 20
(set! bar 30)
bar-alias @result{} 30
@end example
@end deffn

There is an extension to identifier-syntax which allows it to handle the
@code{set!} case as well:

@deffn {Syntax} identifier-syntax (var exp1) ((set! var val) exp2)
Create a variable transformer. The first clause is used for references
to the variable in operator or operand position, and the second for
appearances of the variable on the left-hand-side of an assignment.

For example, the previous @code{bar-alias} example could be expressed
more succinctly like this:

@example
(define-syntax bar-alias
  (identifier-syntax
    (var bar)
    ((set! var val) (set! bar val))))
@end example

@noindent
As before, the templates in @code{identifier-syntax} forms do not need
wrapping in @code{#'} syntax forms.
@end deffn


@node Eval When
@subsection Eval-when

As @code{syntax-case} macros have the whole power of Scheme available to them,
they present a problem regarding time: when a macro runs, what parts of the
program are available for the macro to use?

The default answer to this question is that when you import a module (via
@code{define-module} or @code{use-modules}), that module will be loaded up at
expansion-time, as well as at run-time. Additionally, top-level syntactic
definitions within one compilation unit made by @code{define-syntax} are also
evaluated at expansion time, in the order that they appear in the compilation
unit (file).

But if a syntactic definition needs to call out to a normal procedure at
expansion-time, it might well need need special declarations to indicate that
the procedure should be made available at expansion-time.

For example, the following code will work at a REPL, but not in a file:

@example
;; incorrect
(use-modules (srfi srfi-19))
(define (date) (date->string (current-date)))
(define-syntax %date (identifier-syntax (date)))
(define *compilation-date* %date)
@end example

It works at a REPL because the expressions are evaluated one-by-one, in order,
but if placed in a file, the expressions are expanded one-by-one, but not
evaluated until the compiled file is loaded.

The fix is to use @code{eval-when}.

@example
;; correct: using eval-when
(use-modules (srfi srfi-19))
(eval-when (compile load eval)
  (define (date) (date->string (current-date))))
(define-syntax %date (identifier-syntax (date)))
(define *compilation-date* %date)
@end example

@deffn {Syntax} eval-when conditions exp...
Evaluate @var{exp...} under the given @var{conditions}. Valid conditions include
@code{eval}, @code{load}, and @code{compile}. If you need to use
@code{eval-when}, use it with all three conditions, as in the above example.
Other uses of @code{eval-when} may void your warranty or poison your cat.
@end deffn

@node Internal Macros
@subsection Internal Macros

@deffn {Scheme Procedure} make-syntax-transformer name type binding
Construct a syntax transformer object. This is part of Guile's low-level support
for syntax-case.
@end deffn

@deffn {Scheme Procedure} macro? obj
@deffnx {C Function} scm_macro_p (obj)
Return @code{#t} iff @var{obj} is a syntax transformer.

Note that it's a bit difficult to actually get a macro as a first-class object;
simply naming it (like @code{case}) will produce a syntax error. But it is
possible to get these objects using @code{module-ref}:

@example
(macro? (module-ref (current-module) 'case))
@result{} #t
@end example
@end deffn

@deffn {Scheme Procedure} macro-type m
@deffnx {C Function} scm_macro_type (m)
Return the @var{type} that was given when @var{m} was constructed, via
@code{make-syntax-transformer}.
@end deffn

@deffn {Scheme Procedure} macro-name m
@deffnx {C Function} scm_macro_name (m)
Return the name of the macro @var{m}.
@end deffn

@deffn {Scheme Procedure} macro-binding m
@deffnx {C Function} scm_macro_binding (m)
Return the binding of the macro @var{m}.
@end deffn

@deffn {Scheme Procedure} macro-transformer m
@deffnx {C Function} scm_macro_transformer (m)
Return the transformer of the macro @var{m}. This will return a procedure, for
which one may ask the docstring. That's the whole reason this section is
documented. Actually a part of the result of @code{macro-binding}.
@end deffn


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