Fundamentals of Computer Science I: Media Computing (CS151.02 2007F)
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Reference: [Scheme Report (R5RS)] [Scheme Reference] [DrScheme Manual]
Related Courses: [CSC151.01 2007F (Davis)] [CSC151 2007S (Rebelsky)] [CSCS151 2005S (Stone)]
Summary:
As you should know by now, cons
is one of the core Scheme procedures. Most
typically, cons
is applied to two arguments, a value and a list, and we
think of it as prepending the value to the front of the list. You also
know from experimentation that cons
can not take fewer than two arguments
nor more than two arguments. You have also found that
cons
can still be
called with a non-list as the second argument, and in this case the thing
built has a strange dot before that element. In this reading, we consider
what is happening behind the scenes when you call
cons
. We also use cons
to build structures other than lists.
Before we explore pairs, we'll take a quick detour into one of Scheme's other key data types, the symbol. Symbols look a lot like words (and perhaps even strings), except that they don't have quotation marks around them. Internally, Scheme uses symbols for all of the names that you write in your program. Many Scheme programmers find them useful when they just need a simple, readable, value to pass around.
You write symbols with a single quote mark ('
)
and the word you want to
serve as a symbol. For example, 'aardvark
is a symbol that corresponds to
the word “aardvark”
and 'zebra
is a symbol that corresponds to the word
“zebra”.
When Scheme prints out symbols, it does not give you the quote mark. Nonetheless, you must still write it. Hence,
>
'aardvark
aardvark
>
(list 'aardvark 'zebra)
(aardvark zebra)
>
aardvark
Error: Undefined value: aardvark
Symbol are atomic. Unlike strings (which symbols seem to resemble),
symbols do not support procedures, like string-ref
or substring
, that
extract some part of the symbol.
Hence, there are really only a few procedures applicable to symbols:
symbol?
checks whether a value is a symbol
and both eq?
and equal?
can be
used to compare symbols.
Why would we use symbols, if those are the only available procedures? Because they're simple and sometimes a bit more fun than numbers. Why do we introduce them now? Because they're also nice to use in diagrams, and we have a lot of diagrams in this reading.
As we have seen, Scheme uses cons
to build lists. As you may recall, cons
takes two arguments. Up to this point, the first element has been a value
and the second has been a list. When you call cons
,
Scheme actually builds
a structure in memory with two parts, one of which refers to the first
argument to cons
and the other of which refers to the second. This
structure is called a cons cell or a pair.
Let us now consider a graphical way to represent the result of a
cons
procedure. The basic idea is to use a rectangle, divided in half, to
represent the result of the cons
.
From the first half of the rectangle, we
draw an arrow to the first element of a list, its car; from the second
half of the rectangle, we draw an arrow to the rest of the list, its cdr.
When the cdr is null (the empty list), we draw a diagonal line through the
right half of the rectangle to indicate that the list stops at that point.
For instance, the value of the expression (cons 'a null)
would be represented in this notation as follows:
Since the value of the expression (cons 'a null)
is the list
(a)
, this
diagram represents (a)
as well.
Now consider the value of the expression
(cons 'b (cons 'a null))
,
or in other words, the list (b a)
.
Here, we draw another rectangle, where the
head points to b
and the tail points to the representation of (a)
that we
already have seen. The result is:
Similarly, the list (d c b a)
is the value of the expression
(cons 'd (cons 'c (cons 'b (cons 'a null))))
and would be drawn as follows:
A similar approach may be used for lists that have other lists as
elements. For example, consider the list ((a) b (c d) e)
.
This list contains four components, so at the top level we will need four
rectangles, just as in the previous example for the list
(d c b a)
. Here,
however, the first component designates the list (a)
, which itself
involves the box-and-pointer diagram already discussed. Similarly, the
list (c d)
has two boxes for its two components
(as in the diagram for (b a)
above). The resulting diagram is:
Throughout these diagrams, the empty list is represented by a null
pointer, a diagonal line. Thus, the list containing the empty list,
(())
-- that is, the value of the expression
(cons null null)
-- is represented
by a rectangle with lines through both halves:
While we consistently have discussed cons
in the context of lists, Scheme
allows cons
to be applied even when the second argument is not a list. For
example, (cons 'a 'b)
is a legal expression; its value is represented by
the following box-and-pointer diagram:
You may have noticed that some of your lists ended with a dot before the
last character.
In fact, whenever Scheme is asked to print out a sequence
of linked pairs that don't end with null, it uses dot notation, as in
(a . b)
.
Here, the dot indicates that cons has been applied, but the second
argument is not a list. Similarly, the value of (cons 1 'a)
is the pair (1 . a)
, and the value of
(cons "Henry" "Walker")
is ("Henry" . "Walker")
.
Using a box-and-pointer representation, this last result would be drawn as
follows:
The car
and cdr
procedures can be used to recover the halves of one of
these improper lists:
>
(car (cons 'a 'b))
a
>
(cdr (cons 'a 'b))
b
Note that the cdr
of such a structure is not a list.
When Scheme tries to print out a pair structure, it uses what we might call an optimistic assumption. If the next thing is null or a pair, it assumes that it's a list, and therefore uses a space before the next object. When it hits the end and finds no null, it inserts the dot there, but not earlier.
The pair?
predicate returns #t
when it is given any structure that is
printed as a dotted pair, or indeed any structure that
cons
can possibly
return as its value. (Basically, pair?
determines whether the object it is
given is one of those two-box rectangles.)
Just as lists can be nested within lists, so pairs can be nested within pairs, as deeply as you like. For instance, here is a pair structure that contains the first eight natural numbers:
To build this structure in Scheme, we can use repeated calls to
cons
, thus:
(cons (cons (cons 0 1) (cons 2 3)) (cons (cons 4 5) (cons 6 7)))
or we can use the dotted-pair notation inside a literal constant beginning with a quote:
'(((0 . 1) . (2 . 3)) . ((4 . 5) . (6 . 7)))
(As we've said previously, we'd prefer that you use
cons
rather than quote
to build structures.)
If we have a pair structure that is constructed by repeated invocations of
cons
,
starting from constituents of some simple type such as numbers or
strings, we call such a structure a tree. (We often prefix the word tree
with the type from which the tree is built, such as number tree;
alternately we suffix the word tree with of and then the type, as in tree
of numbers.) We will look at trees in some more depth in the reading on
deep recursion. For now, let's consider a basic approach.
In particular, when we are dealing with a tree, we can use
pair recursion,
which adapts the shape of the computation to the shape of the particular
pair structure on which we operate. In pair recursion, the base cases are
the values that are not pairs, and must therefore be operated on directly.
For the non-base cases -- those that are pairs -- we invoke the procedure
recursively twice (once for the car
,
once for the cdr
) and combine the
values of the recursive calls to get the final result of the operation.
For instance, here is how we'd find the sum of the numbers in a pair structure like the one diagrammed above.
;;; Procedure: ;;; sum-of-number-tree ;;; Parameters: ;;; ntree, a number tree ;;; Purpose: ;;; Sums all the numbers in ntree. ;;; Produces: ;;; sum, a number ;;; Preconditions: ;;; ntree is a number tree. That is, it consists only of numbers ;;; and cons cells. ;;; Postconditions: ;;; sum is the sum of all numbers in ntree. (define sum-of-number-tree (lambda (ntree) (if (pair? ntree) (+ (sum-of-number-tree (car ntree)) (sum-of-number-tree (cdr ntree))) ntree)))
>
(sum-of-number-tree (cons (cons (cons 0 1) (cons 2 3)) (cons (cons 4 5) (cons 6 7))))
28
When this procedure is applied to a base case -- that is, just a number rather than a collection of numbers fitted into a pair structure -- it returns the number unchanged:
>
(sum-of-number-tree 19)
19
There is no such thing as an empty pair analogous to an empty list. Every
pair has exactly two components, and it is always valid to take the
car
and the cdr
of a pair. So the base case for a pair recursion is just any
value that is not itself a pair.
You may be wondering why we ask you to pay attention to these pair things. (You'll probably be wondering why we ask you to pay so much attention after doing the lab). There are a few reasons. First, we find that students better understand lists (and related structures) if they have an understanding of what's going on behind the scenes. Second, there are many instances in which we are better off building trees (like those above) than lists. Third, pair structures provide an additional mechanism for thinking about recursion.
The pair structures also reveal a bit about Scheme terminology. In the
first computers on which LISP (the forerunner of Scheme) was implemented,
there was an underlying memory structure that had two cells, which made it
a convenient way to implement pairs. On that computer, the operations used
to remove values from the structure were car
(shorthand for “contents of address register”)
and cdr
(shorthand for “contents of decrement register”,
even though some people mistakenly claim it stands for
“contents of data register”).
Primary: [Front Door] [Glance] - [Academic Honesty] [Instructions]
Current: [Outline] [EBoard] [Reading] [Lab] [Assignment]
Groupings: [Assignments] [EBoards] [Examples] [Exams] [Handouts] [Labs] [Outlines] [Projects] [Readings] [Reference]
Reference: [Scheme Report (R5RS)] [Scheme Reference] [DrScheme Manual]
Related Courses: [CSC151.01 2007F (Davis)] [CSC151 2007S (Rebelsky)] [CSCS151 2005S (Stone)]
Copyright © 2007 Janet Davis, Matthew Kluber, and Samuel A. Rebelsky. (Selected materials copyright by John David Stone and Henry Walker and used by permission.)
This material is based upon work partially supported by the National Science Foundation under Grant No. CCLI-0633090. Any opinions, findings, and conclusions or recommendations expressed in this material are those of the author(s) and do not necessarily reflect the views of the National Science Foundation.
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