Assume you want to replace all occurences of
‘quux’, and so on. You want to cycle through the replacements:
(let ((master '("bar" "baz" "quux")) (items)) (while (search-forward "foo" nil t) (replace-match (or (car items) (car (setq items master)))) (setq items (cdr items))))
The code above has a master list of replacements and a working copy called
‘items’. It will use the first item in the list of items, or get a copy of the master list and use that. Then it shortens the list by one item and loops.
Or we can use a circular list:
(let ((items '("bar" "baz" "quux"))) (nconc items items) (while (search-forward "foo" nil t) (replace-match (car items)) (setq items (cdr items))))
‘nconc’ function modifies the ListStructure and makes it circular. The list of items basically holds
‘quux’, ad infinitum. See ListModification for more information about
With a circular list, we can now use the first item of it, and shorten it by one, and we’ll never run out of items.
To represent shared or circular structures in Lisp code, use the reader constructs
n is some integer. Use
#n= before an object to label it, then use
#n# to refer to the same object in another place.
For example, here is how to make a list in which the first element recurs as the third element:
(setq x '(#1=(a) b #1#)) (eq (nth 0 x) (nth 2 x)) => t
You can also use the same syntax to make a circular structure, which appears as an “element” within itself:
(setq x '#1=(a #1#)) (eq x (cadr x)) => t
The Lisp printer can produce this syntax if you bind the variable
print-circle to a non-
nil value. For more details, see Circular Objects.