A list represents a sequence of zero or more elements (which may be any Lisp objects). The important difference between lists and vectors is that two or more lists can share part of their structure; in addition, you can insert or delete elements in a list without copying the whole list.
Lists in Lisp are not a primitive data type; they are built up from cons cells. A cons cell is a data object that represents an ordered pair. It records two Lisp objects, one labeled as the CAR, and the other labeled as the CDR. These names are traditional; see section Cons Cell and List Types. CDR is pronounced "could-er."
A list is a series of cons cells chained together, one cons cell per
element of the list. By convention, the CARs of the cons cells are
the elements of the list, and the CDRs are used to chain the list:
the CDR of each cons cell is the following cons cell. The CDR
of the last cons cell is nil. This asymmetry between the
CAR and the CDR is entirely a matter of convention; at the
level of cons cells, the CAR and CDR slots have the same
characteristics.
Because most cons cells are used as part of lists, the phrase list structure has come to mean any structure made out of cons cells.
The symbol nil is considered a list as well as a symbol; it is
the list with no elements. For convenience, the symbol nil is
considered to have nil as its CDR (and also as its
CAR).
The CDR of any nonempty list l is a list containing all the elements of l except the first.
A cons cell can be illustrated as a pair of boxes. The first box
represents the CAR and the second box represents the CDR.
Here is an illustration of the two-element list, (tulip lily),
made from two cons cells:
--------------- --------------- | car | cdr | | car | cdr | | tulip | o---------->| lily | nil | | | | | | | --------------- ---------------
Each pair of boxes represents a cons cell. Each box "refers to",
"points to" or "contains" a Lisp object. (These terms are
synonymous.) The first box, which is the CAR of the first cons
cell, contains the symbol tulip. The arrow from the CDR of
the first cons cell to the second cons cell indicates that the CDR
of the first cons cell points to the second cons cell.
The same list can be illustrated in a different sort of box notation like this:
___ ___ ___ ___
|___|___|--> |___|___|--> nil
| |
| |
--> tulip --> lily
Here is a more complex illustration, showing the three-element list,
((pine needles) oak maple), the first element of which is a
two-element list:
___ ___ ___ ___ ___ ___
|___|___|--> |___|___|--> |___|___|--> nil
| | |
| | |
| --> oak --> maple
|
| ___ ___ ___ ___
--> |___|___|--> |___|___|--> nil
| |
| |
--> pine --> needles
The same list represented in the first box notation looks like this:
-------------- -------------- --------------
| car | cdr | | car | cdr | | car | cdr |
| o | o------->| oak | o------->| maple | nil |
| | | | | | | | | |
-- | --------- -------------- --------------
|
|
| -------------- ----------------
| | car | cdr | | car | cdr |
------>| pine | o------->| needles | nil |
| | | | | |
-------------- ----------------
See section Cons Cell and List Types, for the read and print syntax of cons cells and lists, and for more "box and arrow" illustrations of lists.
The following predicates test whether a Lisp object is an atom, is a
cons cell or is a list, or whether it is the distinguished object
nil. (Many of these predicates can be defined in terms of the
others, but they are used so often that it is worth having all of them.)
t if object is a cons cell, nil
otherwise. nil is not a cons cell, although it is a list.
t if object is an atom, nil
otherwise. All objects except cons cells are atoms. The symbol
nil is an atom and is also a list; it is the only Lisp object
that is both.
(atom object) == (not (consp object))
t if object is a cons cell or
nil. Otherwise, it returns nil.
(listp '(1))
=> t
(listp '())
=> t
listp: it returns t if
object is not a list. Otherwise, it returns nil.
(listp object) == (not (nlistp object))
t if object is nil, and
returns nil otherwise. This function is identical to not,
but as a matter of clarity we use null when object is
considered a list and not when it is considered a truth value
(see not in section Constructs for Combining Conditions).
(null '(1))
=> nil
(null '())
=> t
As a special case, if cons-cell is nil, then car
is defined to return nil; therefore, any list is a valid argument
for car. An error is signaled if the argument is not a cons cell
or nil.
(car '(a b c))
=> a
(car '())
=> nil
As a special case, if cons-cell is nil, then cdr
is defined to return nil; therefore, any list is a valid argument
for cdr. An error is signaled if the argument is not a cons cell
or nil.
(cdr '(a b c))
=> (b c)
(cdr '())
=> nil
nil otherwise. This is in contrast
to car, which signals an error if object is not a list.
(car-safe object)
==
(let ((x object))
(if (consp x)
(car x)
nil))
nil otherwise.
This is in contrast to cdr, which signals an error if
object is not a list.
(cdr-safe object)
==
(let ((x object))
(if (consp x)
(cdr x)
nil))
nil.
If n is negative, nth returns the first element of
list.
(nth 2 '(1 2 3 4))
=> 3
(nth 10 '(1 2 3 4))
=> nil
(nth -3 '(1 2 3 4))
=> 1
(nth n x) == (car (nthcdr n x))
If n is zero or negative, nthcdr returns all of
list. If the length of list is n or less,
nthcdr returns nil.
(nthcdr 1 '(1 2 3 4))
=> (2 3 4)
(nthcdr 10 '(1 2 3 4))
=> nil
(nthcdr -3 '(1 2 3 4))
=> (1 2 3 4)
Many functions build lists, as lists reside at the very heart of Lisp.
cons is the fundamental list-building function; however, it is
interesting to note that list is used more times in the source
code for Emacs than cons.
(cons 1 '(2))
=> (1 2)
(cons 1 '())
=> (1)
(cons 1 2)
=> (1 . 2)
cons is often used to add a single element to the front of a
list. This is called consing the element onto the list. For
example:
(setq list (cons newelt list))
Note that there is no conflict between the variable named list
used in this example and the function named list described below;
any symbol can serve both purposes.
nil-terminated. If no objects
are given, the empty list is returned.
(list 1 2 3 4 5)
=> (1 2 3 4 5)
(list 1 2 '(3 4 5) 'foo)
=> (1 2 (3 4 5) foo)
(list)
=> nil
make-list with make-string (see section Creating Strings).
(make-list 3 'pigs)
=> (pigs pigs pigs)
(make-list 0 'pigs)
=> nil
More generally, the final argument to append may be any Lisp
object. The final argument is not copied or converted; it becomes the
CDR of the last cons cell in the new list. If the final argument
is itself a list, then its elements become in effect elements of the
result list. If the final element is not a list, the result is a
"dotted list" since its final CDR is not nil as required
in a true list.
See nconc in section Functions that Rearrange Lists, for a way to join lists with no
copying.
Here is an example of using append:
(setq trees '(pine oak))
=> (pine oak)
(setq more-trees (append '(maple birch) trees))
=> (maple birch pine oak)
trees
=> (pine oak)
more-trees
=> (maple birch pine oak)
(eq trees (cdr (cdr more-trees)))
=> t
You can see how append works by looking at a box diagram. The
variable trees is set to the list (pine oak) and then the
variable more-trees is set to the list (maple birch pine
oak). However, the variable trees continues to refer to the
original list:
more-trees trees
| |
| ___ ___ ___ ___ -> ___ ___ ___ ___
--> |___|___|--> |___|___|--> |___|___|--> |___|___|--> nil
| | | |
| | | |
--> maple -->birch --> pine --> oak
An empty sequence contributes nothing to the value returned by
append. As a consequence of this, a final nil argument
forces a copy of the previous argument.
trees
=> (pine oak)
(setq wood (append trees ()))
=> (pine oak)
wood
=> (pine oak)
(eq wood trees)
=> nil
This once was the usual way to copy a list, before the function
copy-sequence was invented. See section Sequences, Arrays, and Vectors.
With the help of apply, we can append all the lists in a list of
lists:
(apply 'append '((a b c) nil (x y z) nil))
=> (a b c x y z)
If no sequences are given, nil is returned:
(append)
=> nil
Here are some examples where the final argument is not a list:
(append '(x y) 'z)
=> (x y z)
(append '(x y) [z])
=> (x y [z])
The second example shows that when the final argument is a sequence but not a list, the sequence's elements do not become elements of the resulting list. Instead, the sequence becomes the final CDR, like any other non-list final argument.
Integers are also allowed as arguments to append. They are
converted to strings of digits making up the decimal print
representation of the integer, and these strings are then appended.
Here's what happens:
(setq trees '(pine oak))
=> (pine oak)
(char-to-string 54)
=> "6"
(setq longer-list (append trees 6 '(spruce)))
=> (pine oak 54 spruce)
(setq x-list (append trees 6 6))
=> (pine oak 54 . 6)
This special case exists for compatibility with Mocklisp, and we don't
recommend you take advantage of it. If you want to convert an integer
in this way, use format (see section Formatting Strings) or
number-to-string (see section Conversion of Characters and Strings).
(setq x '(1 2 3 4))
=> (1 2 3 4)
(reverse x)
=> (4 3 2 1)
x
=> (1 2 3 4)
You can modify the CAR and CDR contents of a cons cell with the
primitives setcar and setcdr.
Common Lisp note: Common Lisp uses functions
rplacaandrplacdto alter list structure; they change structure the same way assetcarandsetcdr, but the Common Lisp functions return the cons cell whilesetcarandsetcdrreturn the new CAR or CDR.
setcar
Changing the CAR of a cons cell is done with setcar. When
used on a list, setcar replaces one element of a list with a
different element.
(setq x '(1 2))
=> (1 2)
(setcar x 4)
=> 4
x
=> (4 2)
When a cons cell is part of the shared structure of several lists, storing a new CAR into the cons changes one element of each of these lists. Here is an example:
;; Create two lists that are partly shared.
(setq x1 '(a b c))
=> (a b c)
(setq x2 (cons 'z (cdr x1)))
=> (z b c)
;; Replace the CAR of a shared link.
(setcar (cdr x1) 'foo)
=> foo
x1 ; Both lists are changed.
=> (a foo c)
x2
=> (z foo c)
;; Replace the CAR of a link that is not shared.
(setcar x1 'baz)
=> baz
x1 ; Only one list is changed.
=> (baz foo c)
x2
=> (z foo c)
Here is a graphical depiction of the shared structure of the two lists
in the variables x1 and x2, showing why replacing b
changes them both:
___ ___ ___ ___ ___ ___
x1---> |___|___|----> |___|___|--> |___|___|--> nil
| --> | |
| | | |
--> a | --> b --> c
|
___ ___ |
x2--> |___|___|--
|
|
--> z
Here is an alternative form of box diagram, showing the same relationship:
x1:
-------------- -------------- --------------
| car | cdr | | car | cdr | | car | cdr |
| a | o------->| b | o------->| c | nil |
| | | -->| | | | | |
-------------- | -------------- --------------
|
x2: |
-------------- |
| car | cdr | |
| z | o----
| | |
--------------
The lowest-level primitive for modifying a CDR is setcdr:
Here is an example of replacing the CDR of a list with a different list. All but the first element of the list are removed in favor of a different sequence of elements. The first element is unchanged, because it resides in the CAR of the list, and is not reached via the CDR.
(setq x '(1 2 3))
=> (1 2 3)
(setcdr x '(4))
=> (4)
x
=> (1 4)
You can delete elements from the middle of a list by altering the
CDRs of the cons cells in the list. For example, here we delete
the second element, b, from the list (a b c), by changing
the CDR of the first cell:
(setq x1 '(a b c))
=> (a b c)
(setcdr x1 (cdr (cdr x1)))
=> (c)
x1
=> (a c)
Here is the result in box notation:
--------------------
| |
-------------- | -------------- | --------------
| car | cdr | | | car | cdr | -->| car | cdr |
| a | o----- | b | o-------->| c | nil |
| | | | | | | | |
-------------- -------------- --------------
The second cons cell, which previously held the element b, still
exists and its CAR is still b, but it no longer forms part
of this list.
It is equally easy to insert a new element by changing CDRs:
(setq x1 '(a b c))
=> (a b c)
(setcdr x1 (cons 'd (cdr x1)))
=> (d b c)
x1
=> (a d b c)
Here is this result in box notation:
-------------- ------------- -------------
| car | cdr | | car | cdr | | car | cdr |
| a | o | -->| b | o------->| c | nil |
| | | | | | | | | | |
--------- | -- | ------------- -------------
| |
----- --------
| |
| --------------- |
| | car | cdr | |
-->| d | o------
| | |
---------------
Here are some functions that rearrange lists "destructively" by modifying the CDRs of their component cons cells. We call these functions "destructive" because they chew up the original lists passed to them as arguments, to produce a new list that is the returned value.
The function delq in the following section is another example
of destructive list manipulation.
append (see section Building Cons Cells and Lists), the lists are
not copied. Instead, the last CDR of each of the
lists is changed to refer to the following list. The last of the
lists is not altered. For example:
(setq x '(1 2 3))
=> (1 2 3)
(nconc x '(4 5))
=> (1 2 3 4 5)
x
=> (1 2 3 4 5)
Since the last argument of nconc is not itself modified, it is
reasonable to use a constant list, such as '(4 5), as in the
above example. For the same reason, the last argument need not be a
list:
(setq x '(1 2 3))
=> (1 2 3)
(nconc x 'z)
=> (1 2 3 . z)
x
=> (1 2 3 . z)
A common pitfall is to use a quoted constant list as a non-last
argument to nconc. If you do this, your program will change
each time you run it! Here is what happens:
(defun add-foo (x) ; We want this function to add
(nconc '(foo) x)) ; foo to the front of its arg.
(symbol-function 'add-foo)
=> (lambda (x) (nconc (quote (foo)) x))
(setq xx (add-foo '(1 2))) ; It seems to work.
=> (foo 1 2)
(setq xy (add-foo '(3 4))) ; What happened?
=> (foo 1 2 3 4)
(eq xx xy)
=> t
(symbol-function 'add-foo)
=> (lambda (x) (nconc (quote (foo 1 2 3 4) x)))
reverse, nreverse alters its argument by reversing
the CDRs in the cons cells forming the list. The cons cell that
used to be the last one in list becomes the first cell of the
value.
For example:
(setq x '(1 2 3 4))
=> (1 2 3 4)
x
=> (1 2 3 4)
(nreverse x)
=> (4 3 2 1)
;; The cell that was first is now last.
x
=> (1)
To avoid confusion, we usually store the result of nreverse
back in the same variable which held the original list:
(setq x (nreverse x))
Here is the nreverse of our favorite example, (a b c),
presented graphically:
Original list head: Reversed list:
------------- ------------- ------------
| car | cdr | | car | cdr | | car | cdr |
| a | nil |<-- | b | o |<-- | c | o |
| | | | | | | | | | | | |
------------- | --------- | - | -------- | -
| | | |
------------- ------------
The argument predicate must be a function that accepts two
arguments. It is called with two elements of list. To get an
increasing order sort, the predicate should return t if the
first element is "less than" the second, or nil if not.
The destructive aspect of sort is that it rearranges the cons
cells forming list by changing CDRs. A nondestructive sort
function would create new cons cells to store the elements in their
sorted order. If you wish to make a sorted copy without destroying the
original, copy it first with copy-sequence and then sort.
Sorting does not change the CARs of the cons cells in list; each cons cell in the result contains the same element that it contained before. The result differs from the argument list because the cells themselves have been reordered.
Sorting does not change the CARs of the cons cells in list;
the cons cell that originally contained the element a in
list still has a in its CAR after sorting, but it now
appears in a different position in the list due to the change of
CDRs. For example:
(setq nums '(1 3 2 6 5 4 0))
=> (1 3 2 6 5 4 0)
(sort nums '<)
=> (0 1 2 3 4 5 6)
nums
=> (1 2 3 4 5 6)
Note that the list in nums no longer contains 0; this is the same
cons cell that it was before, but it is no longer the first one in the
list. Don't assume a variable that formerly held the argument now holds
the entire sorted list! Instead, save the result of sort and use
that. Most often we store the result back into the variable that held
the original list:
(setq nums (sort nums '<))
See section Sorting Text, for more functions that perform sorting.
See documentation in section Access to Documentation Strings, for a
useful example of sort.
A list can represent an unordered mathematical set--simply consider a
value an element of a set if it appears in the list, and ignore the
order of the list. To form the union of two sets, use append (as
long as you don't mind having duplicate elements). Other useful
functions for sets include memq and delq, and their
equal versions, member and delete.
Common Lisp note: Common Lisp has functions
union(which avoids duplicate elements) andintersectionfor set operations, but GNU Emacs Lisp does not have them. You can write them in Lisp if you wish.
memq returns a list starting with the
first occurrence of object. Otherwise, it returns nil.
The letter `q' in memq says that it uses eq to
compare object against the elements of the list. For example:
(memq 'b '(a b c b a))
=> (b c b a)
(memq '(2) '((1) (2))) ; (2) and (2) are not eq.
=> nil
eq to
object from list. The letter `q' in delq says
that it uses eq to compare object against the elements of
the list, like memq.
When delq deletes elements from the front of the list, it does so
simply by advancing down the list and returning a sublist that starts
after those elements:
(delq 'a '(a b c)) == (cdr '(a b c))
When an element to be deleted appears in the middle of the list, removing it involves changing the CDRs (see section Altering the CDR of a List).
(setq sample-list '(a b c (4)))
=> (a b c (4))
(delq 'a sample-list)
=> (b c (4))
sample-list
=> (a b c (4))
(delq 'c sample-list)
=> (a c (4))
sample-list
=> (a c (4))
Note that (delq 'b sample-list) modifies sample-list to
splice out the second element, but (delq 'a sample-list) does not
splice anything--it just returns a shorter list. Don't assume that a
variable which formerly held the argument list now has fewer
elements, or that it still holds the original list! Instead, save the
result of delq and use that. Most often we store the result back
into the variable that held the original list:
(setq flowers (delq 'rose flowers))
In the following example, the (4) that delq attempts to match
and the (4) in the sample-list are not eq:
(delq '(4) sample-list)
=> (a c (4))
The following two functions are like memq and delq but use
equal rather than eq to compare elements. They are new in
Emacs 19.
member tests to see whether object is a member
of list, comparing members with object using equal.
If object is a member, member returns a list starting with
its first occurrence in list. Otherwise, it returns nil.
Compare this with memq:
(member '(2) '((1) (2))) ;(2)and(2)areequal. => ((2)) (memq '(2) '((1) (2))) ;(2)and(2)are noteq. => nil ;; Two strings with the same contents areequal. (member "foo" '("foo" "bar")) => ("foo" "bar")
equal to
object from list. It is to delq as member is
to memq: it uses equal to compare elements with
object, like member; when it finds an element that matches,
it removes the element just as delq would. For example:
(delete '(2) '((2) (1) (2)))
=> '((1))
Common Lisp note: The functions
memberanddeletein GNU Emacs Lisp are derived from Maclisp, not Common Lisp. The Common Lisp versions do not useequalto compare elements.
An association list, or alist for short, records a mapping from keys to values. It is a list of cons cells called associations: the CAR of each cell is the key, and the CDR is the associated value.(1)
Here is an example of an alist. The key pine is associated with
the value cones; the key oak is associated with
acorns; and the key maple is associated with seeds.
'((pine . cones) (oak . acorns) (maple . seeds))
The associated values in an alist may be any Lisp objects; so may the
keys. For example, in the following alist, the symbol a is
associated with the number 1, and the string "b" is
associated with the list (2 3), which is the CDR of
the alist element:
((a . 1) ("b" 2 3))
Sometimes it is better to design an alist to store the associated value in the CAR of the CDR of the element. Here is an example:
'((rose red) (lily white) (buttercup yellow))
Here we regard red as the value associated with rose. One
advantage of this method is that you can store other related
information--even a list of other items--in the CDR of the
CDR. One disadvantage is that you cannot use rassq (see
below) to find the element containing a given value. When neither of
these considerations is important, the choice is a matter of taste, as
long as you are consistent about it for any given alist.
Note that the same alist shown above could be regarded as having the
associated value in the CDR of the element; the value associated
with rose would be the list (red).
Association lists are often used to record information that you might otherwise keep on a stack, since new associations may be added easily to the front of the list. When searching an association list for an association with a given key, the first one found is returned, if there is more than one.
In Emacs Lisp, it is not an error if an element of an association list is not a cons cell. The alist search functions simply ignore such elements. Many other versions of Lisp signal errors in such cases.
Note that property lists are similar to association lists in several respects. A property list behaves like an association list in which each key can occur only once. See section Property Lists, for a comparison of property lists and association lists.
equal (see section Equality Predicates). It returns nil if no
association in alist has a CAR equal to key.
For example:
(setq trees '((pine . cones) (oak . acorns) (maple . seeds)))
=> ((pine . cones) (oak . acorns) (maple . seeds))
(assoc 'oak trees)
=> (oak . acorns)
(cdr (assoc 'oak trees))
=> acorns
(assoc 'birch trees)
=> nil
Here is another example, in which the keys and values are not symbols:
(setq needles-per-cluster
'((2 "Austrian Pine" "Red Pine")
(3 "Pitch Pine")
(5 "White Pine")))
(cdr (assoc 3 needles-per-cluster))
=> ("Pitch Pine")
(cdr (assoc 2 needles-per-cluster))
=> ("Austrian Pine" "Red Pine")
assoc in that it returns the first
association for key in alist, but it makes the comparison
using eq instead of equal. assq returns nil
if no association in alist has a CAR eq to key.
This function is used more often than assoc, since eq is
faster than equal and most alists use symbols as keys.
See section Equality Predicates.
(setq trees '((pine . cones) (oak . acorns) (maple . seeds)))
=> ((pine . cones) (oak . acorns) (maple . seeds))
(assq 'pine trees)
=> (pine . cones)
On the other hand, assq is not usually useful in alists where the
keys may not be symbols:
(setq leaves
'(("simple leaves" . oak)
("compound leaves" . horsechestnut)))
(assq "simple leaves" leaves)
=> nil
(assoc "simple leaves" leaves)
=> ("simple leaves" . oak)
nil if no association in alist has
a CDR eq to value.
rassq is like assq except that it compares the CDR of
each alist association instead of the CAR. You can think of
this as "reverse assq", finding the key for a given value.
For example:
(setq trees '((pine . cones) (oak . acorns) (maple . seeds)))
(rassq 'acorns trees)
=> (oak . acorns)
(rassq 'spores trees)
=> nil
Note that rassq cannot search for a value stored in the CAR
of the CDR of an element:
(setq colors '((rose red) (lily white) (buttercup yellow)))
(rassq 'white colors)
=> nil
In this case, the CDR of the association (lily white) is not
the symbol white, but rather the list (white). This
becomes clearer if the association is written in dotted pair notation:
(lily white) == (lily . (white))
(setq needles-per-cluster
'((2 . ("Austrian Pine" "Red Pine"))
(3 . ("Pitch Pine"))
(5 . ("White Pine"))))
=>
((2 "Austrian Pine" "Red Pine")
(3 "Pitch Pine")
(5 "White Pine"))
(setq copy (copy-alist needles-per-cluster))
=>
((2 "Austrian Pine" "Red Pine")
(3 "Pitch Pine")
(5 "White Pine"))
(eq needles-per-cluster copy)
=> nil
(equal needles-per-cluster copy)
=> t
(eq (car needles-per-cluster) (car copy))
=> nil
(cdr (car (cdr needles-per-cluster)))
=> ("Pitch Pine")
(eq (cdr (car (cdr needles-per-cluster)))
(cdr (car (cdr copy))))
=> t
This example shows how copy-alist makes it possible to change
the associations of one copy without affecting the other:
(setcdr (assq 3 needles-per-cluster)
'("Martian Vacuum Pine"))
(cdr (assq 3 needles-per-cluster))
=> ("Pitch Pine")