ANSI Common Lisp 2 Syntax 2.4 Standard Macro Characters
The backquote introduces a template of a data structure to be built. For example, writing
`(cond ((numberp ,x) ,@y) (t (print ,x) ,@y))is roughly equivalent to writing
(list 'cond (cons (list 'numberp x) y) (list* 't (list 'print x) y))Where a comma occurs in the template, the expression following the comma is to be evaluated to produce an object to be inserted at that point. Assume b has the value 3, for example, then evaluating the form denoted by `(a b ,b ,(+ b 1) b) produces the result (a b 3 4 b).
If a comma is immediately followed by an at-sign, then the form following the at-sign is evaluated to produce a list of objects. These objects are then "spliced" into place in the template. For example, if x has the value (a b c), then
`(x ,x ,@x foo ,(cadr x) bar ,(cdr x) baz ,@(cdr x)) (x (a b c) a b c foo b bar (b c) baz b c)The backquote syntax can be summarized formally as follows.
Anywhere ",@" may be used, the syntax ",." may be used instead to indicate that it is permissible to operate destructively on the list structure produced by the form following the ",." (in effect, to use nconc instead of append).
If the backquote syntax is nested, the innermost backquoted form should be expanded first. This means that if several commas occur in a row, the leftmost one belongs to the innermost backquote.
An implementation is free to interpret a backquoted form F1 as any form F2 that, when evaluated, will produce a result that is the same under equal as the result implied by the above definition, provided that the side-effect behavior of the substitute form F2 is also consistent with the description given above. The constructed copy of the template might or might not share list structure with the template itself. As an example, the above definition implies that
`((,a b) ,c ,@d)will be interpreted as if it were
(append (list (append (list a) (list 'b))) (list c) d) (append (list (append (list a) '(b))) (list c) d) (list* (cons a '(b)) c d) (list* (cons a (list 'b)) c d) (append (list (cons a '(b))) (list c) d) (list* (cons a '(b)) c (copy-list d))