((_ #((var val) ...) exp exp* ...)
(let ((var val) ...) exp exp* ...))))
(letv #((foo 'bar)) foo)
-@result{} foo
+@result{} bar
@end example
Literals are used to match specific datums in an expression, like the use of
@code{syntax-case} it is easy:
@deffn {Scheme Procedure} identifier? syntax-object
-Returns @code{#t} iff @var{syntax-object} is an identifier.
+Returns @code{#t} if @var{syntax-object} is an identifier, or @code{#f}
+otherwise.
@end deffn
@example
type.
@deffn {Scheme Procedure} bound-identifier=? a b
-Return @code{#t} iff the syntax objects @var{a} and @var{b} refer to the
-same lexically-bound identifier.
+Return @code{#t} if the syntax objects @var{a} and @var{b} refer to the
+same lexically-bound identifier, or @code{#f} otherwise.
@end deffn
@deffn {Scheme Procedure} free-identifier=? a b
-Return @code{#t} iff the syntax objects @var{a} and @var{b} refer to the
-same free identifier.
+Return @code{#t} if the syntax objects @var{a} and @var{b} refer to the
+same free identifier, or @code{#f} otherwise.
@end deffn
@deffn {Scheme Procedure} generate-temporaries ls
(set! foo @var{val})
;; expands via
(foo-transformer #'(set! foo @var{val}))
-;; iff foo-transformer is a "variable transformer"
+;; if foo-transformer is a "variable transformer"
@end example
As the example notes, the transformer procedure must be explicitly
@deffn {Scheme Procedure} macro? obj
@deffnx {C Function} scm_macro_p (obj)
-Return @code{#t} iff @var{obj} is a syntax transformer.
+Return @code{#t} if @var{obj} is a syntax transformer, or @code{#f}
+otherwise.
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