
So far all the programs we've written in this book have had no memory of the past history of the computation. We invoke a function with certain arguments, and we get back a value that depends only on those arguments. Compare this with the operation of Scheme itself:
> (foo 3) ERROR: FOO HAS NO VALUE > (define (foo x) (word x x)) > (foo 3) 33
Scheme remembers that you have defined foo
, so its response to
the very same expression is different the second time. Scheme maintains a
record of certain results of its past interaction with you; in particular,
Scheme remembers the global variables that you have defined. This record is
called its state.
Most of the programs that people use routinely are full of state; your text editor, for example, remembers all the characters in your file. In this chapter you will learn how to write programs with state.
The Indianapolis 500 is an annual 500mile automobile race, famous among people who like that sort of thing. It's held at the Indianapolis Motor Speedway, a racetrack in Indianapolis, Indiana. (Indiana is better known as the home of Dan Friedman, the coauthor of some good books about Scheme.) The racetrack is 2½ miles long, so, as you might imagine, the racers have to complete 200 laps in order to finish the race. This means that someone has to keep track of how many laps each car has completed so far.
Let's write a program to help this person keep count. Each car has a
number, and the count person will invoke the procedure lap
with that
number as argument every time a car completes a lap. The procedure will
return the number of laps that that car has completed altogether:
> (lap 87) 1 > (lap 64) 1 > (lap 17) 1 > (lap 64) 2 > (lap 64) 3
(Car 64 managed to complete three laps before the other cars
completed two because the others had flat tires.) Note that we typed
the expression (lap 64)
three times and got three different answers.
Lap
isn't a function! A function has to return the same answer
whenever it's invoked with the same arguments.
The point of this chapter is to show how procedures like lap
can be
written. To accomplish this, we're going to use a data structure called a
vector. (You may have seen something similar in other
programming languages under the name "array.")
A vector is, in effect, a row of boxes into which values can be put. Each vector has a fixed number of boxes; when you create a vector, you have to say how many boxes you want. Once a vector is created, there are two things you can do with it: You can put a new value into a box (replacing any old value that might have been there), or you can examine the value in a box. The boxes are numbered, starting with zero.
> (define v (makevector 5)) > (vectorset! v 0 'shoe) > (vectorset! v 3 'bread) > (vectorset! v 2 '(savoy truffle)) > (vectorref v 3) BREAD
There are several details to note here. When we invoke makevector
we
give it one argument, the number of boxes we want the vector to have. (In
this example, there are five boxes, numbered 0 through 4. There is no
box 5.) When we create the vector, there is nothing in any of the
boxes.[1]
We put things in boxes using the vectorset!
procedure. The
exclamation point in its name, indicates that this is a mutator—a procedure that changes the value of some previously
created data structure. The exclamation point is pronounced "bang," as in
"vector set bang." (Scheme actually has several such mutators, including
mutators for lists, but this is the only one we'll use in this book.
A procedure that modifies its argument is also called destructive.) The arguments to vectorset!
are the vector, the
number of the box (the index), and the desired new value. Like define
, vectorset!
returns an unspecified value.
We examine the contents of a box using vectorref
, which takes two
arguments, the vector and an index. Vectorref
is similar to listref
, except that it operates on vectors instead of lists.
We can change the contents of a box that already has something in it.
> (vectorset! v 3 'jewel) > (vectorref v 3) JEWEL
The old value of box 3, bread
, is no longer there. It's
been replaced by the new value.
> (vectorset! v 1 741) > (vectorset! v 4 #t) > v #(SHOE 741 (SAVOY TRUFFLE) JEWEL #T)
Once the vector is completely full, we can print its value. Scheme
prints vectors in a format like that of lists, except that there is a number
sign (#
) before the open parenthesis. If you ever have need for a
constant vector (one that you're not going to mutate), you can quote it
using the same notation:
> (vectorref '#(a b c d) 2) C
To implement our lap
procedure, we'll keep its state information, the
lap counts, in a vector. We'll use the car number as the index into the
vector. It's not enough to create the vector; we have to make sure that
each box has a zero as its initial value.
(define *lapvector* (makevector 100)) (define (initializelapvector index) (if (< index 0) 'done (begin (vectorset! *lapvector* index 0) (initializelapvector ( index 1))))) > (initializelapvector 99) DONE
We've created a global variable whose value is the vector. We used a recursive procedure to put a zero into each box of the vector.[2] Note that the vector is of length 100, but its largest index is 99. Also, the base case of the recursion is that the index is less than zero, not equal to zero as in many earlier examples. That's because zero is a valid index.
Now that we have the vector, we can write lap
.
(define (lap carnumber) (vectorset! *lapvector* carnumber (+ (vectorref *lapvector* carnumber) 1)) (vectorref *lapvector* carnumber))
Remember that a procedure body can include more than one expression. When the procedure is invoked, the expressions will be evaluated in order. The value returned by the procedure is the value of the last expression (in this case, the second one).
Lap
has both a return value and a side effect. The job of the first
expression is to carry out that side effect, that is, to add 1 to the lap
count for the specified car. The second expression looks at the value we
just put in a box to determine the return value.
We remarked earlier that lap
isn't a function because invoking it
twice with the same argument doesn't return the same value both
times.[3]
It's not a coincidence that lap
also violates functional programming
by maintaining state information. Any procedure whose return value is not a
function of its arguments (that is, whose return value is not always the
same for any particular arguments) must depend on knowledge of what has
happened in the past. After all, computers don't pull results out of the
air; if the result of a computation doesn't depend entirely on the arguments
we give, then it must depend on some other information available to the
program.
Suppose somebody asks you, "Car 54 has just completed a lap; how many has it completed in all?" You can't answer that question with only the information in the question itself; you have to remember earlier events in the race. By contrast, if someone asks you, "What's the plural of `book'?" what has happened in the past doesn't matter at all.
The connection between nonfunctional procedures and state also applies to
nonfunctional Scheme primitives. The read
procedure, for example,
returns different results when you invoke it repeatedly with the same
argument because it remembers how far it's gotten in the file. That's why
the argument is a port instead of a file name: A port is an abstract data
type that includes, among other things, this piece of state. (If you're
reading from the keyboard, the state is in the person doing the typing.)
A more surprising example is the random
procedure that you met in
Chapter 2. Random
isn't a function because it doesn't always
return the same value when called with the same argument. How does random
compute its result? Some versions of random
compute a number
that's based on the current time (in tiny units like milliseconds so you
don't get the same answer from two calls in quick succession). How does
your computer know the time? Every so often some procedure (or some
hardware device) adds 1 to a remembered value, the number of milliseconds
since midnight. That's state, and random
relies on it.
The most commonly used algorithm for random numbers is a little trickier;
each time you invoke random
, the result is a function of the
result from the last time you invoked it. (The procedure is pretty
complicated; typically the old number is multiplied by some large, carefully
chosen constant, and only the middle digits of the product are kept.) Each
time you invoke random
, the returned value is stashed away somehow so
that the next invocation can remember it. That's state too.
Just because a procedure remembers something doesn't necessarily make it stateful. Every procedure remembers the arguments with which it was invoked, while it's running. Otherwise the arguments wouldn't be able to affect the computation. A procedure whose result depends only on its arguments (the ones used in the current invocation) is functional. The procedure is nonfunctional if it depends on something outside of its current arguments. It's that sort of "longterm" memory that we consider to be state.
In particular, a procedure that uses let
isn't stateful merely because
the body of the let
remembers the values of the variables created by the
let
. Once let
returns a value, the variables that it created no
longer exist. You couldn't use let
, for example, to carry out the
kind of remembering that random
needs. Let
doesn't remember a
value between invocations, just during a single invocation.
One of the advantages of the vector data structure is that it allows elements to be rearranged. As an example, we'll create and shuffle a deck of cards.
We'll start with a procedure cardlist
that returns a list of all the
cards, in standard order:
(define (cardlist) (reduce append (map (lambda (suit) (map (lambda (rank) (word suit rank)) '(a 2 3 4 5 6 7 8 9 10 j q k))) '(h s d c)))) > (cardlist) (HA H2 H3 H4 H5 H6 H7 H8 H9 H10 HJ HQ HK SA S2 S3 S4 S5 S6 S7 S8 S9 S10 SJ SQ SK DA D2 D3 D4 D5 D6 D7 D8 D9 D10 DJ DQ DK CA C2 C3 C4 C5 C6 C7 C8 C9 C10 CJ CQ CK)
In writing cardlist
, we need reduce append
because the result
from the outer invocation of map
is a list of lists: ((HA H2 …) (SA …) …)
.[4]
Each time we want a new deck of cards, we start with this list of 52 cards,
copy the list into a vector, and shuffle that vector. We'll use the Scheme
primitive
, which takes a list as argument and
returns a vector of the same length, with the boxes initialized to the
corresponding elements of the list. (There is also a procedure list>vector
that does the reverse. The characters vector>list
>
in
these function names are meant to look like an arrow
(→); this is a Scheme convention for functions
that convert information from one data type to another.)
(define (makedeck) (shuffle! (list>vector (cardlist)) 51)) (define (shuffle! deck index) (if (< index 0) deck (begin (vectorswap! deck index (random (+ index 1))) (shuffle! deck ( index 1))))) (define (vectorswap! vector index1 index2) (let ((temp (vectorref vector index1))) (vectorset! vector index1 (vectorref vector index2)) (vectorset! vector index2 temp)))
Now, each time we call makedeck
, we get a randomly shuffled
vector of cards:
> (makedeck) #(C4 SA C7 DA S4 D9 SQ H4 C10 D5 H9 S10 D6 S9 CA C9 S2 H7 S5 H6 D7 HK S7 C3 C2 C6 HJ SK CQ CJ D4 SJ D8 S8 HA C5 DK D3 HQ D10 H8 DJ C8 H2 H5 H3 CK S3 DQ S6 D2 H10) > (makedeck) #(CQ H7 D10 D5 S8 C7 H10 SQ H4 H3 D8 C9 S7 SK DK S6 DA D4 C6 HQ D6 S2 H5 CA H2 HJ CK D7 H6 HA CJ C4 SJ HK SA C2 D2 S4 DQ S5 C10 H9 D9 C5 D3 DJ C3 S9 S3 C8 S10 H8)
How does the shuffling algorithm work? Conceptually it's not complicated, but there are some implementation details that make the actual procedures a little tricky. The general idea is this: We want all the cards shuffled into a random order. So we choose any card at random, and make it the first card. We're then left with a onecardsmaller deck to shuffle, and we do that by recursion. (This algorithm is similar to selection sort from Chapter 15, except that we select a random card each time instead of selecting the smallest value.)
The details that complicate this algorithm have to do with the fact that we're using a vector, in which it's easy to change the value in one particular position, but it's not easy to do what would otherwise be the most natural thing: If you had a handful of actual cards and wanted to move one of them to the front, you'd slide the other cards over to make room. There's no "sliding over" in a vector. Instead we use a trick; we happen to have an empty slot, the one from which we removed the randomly chosen card, so instead of moving several cards, we just move the one card that was originally at the front into that slot. In other words, we exchange two cards, the randomly chosen one and the one that used to be in front.
Second, there's nothing comparable to cdr
to provide a
onecardsmaller vector to the recursive invocation. Instead, we must use
the entire vector and also provide an additional index
argument, a
number that keeps track of how many cards remain to be shuffled. It's
simplest if each recursive invocation is responsible for the range of cards
from position 0
to position index
of the vector, and therefore
the program actually moves each randomly selected card to the end of
the remaining portion of the deck.
If you want to make a vector with only a few boxes, and you know in advance
what values you want in those boxes, you can use the constructor
vector
. Like list
, it takes any number of arguments and
returns a vector containing those arguments as elements:
> (define beatles (vector 'john 'paul 'george 'pete)) > (vectorset! beatles 3 'ringo) > beatles #(JOHN PAUL GEORGE RINGO)
The procedure vectorlength
takes a vector as argument and returns the
number of boxes in the vector.
> (vectorlength beatles) 4
The predicate equal?
, which we've used with words and lists, also
accepts vectors as arguments. Two vectors are equal if they are the same
size and all their corresponding elements are equal. (A list and a vector
are never equal, even if their elements are equal.)
Finally, the predicate vector?
takes anything as argument and returns
#t
if and only if its argument is a vector.
Here are two procedures that you've seen earlier in this chapter, which do something to each element of a vector:
(define (initializelapvector index) (if (< index 0) 'done (begin (vectorset! *lapvector* index 0) (initializelapvector ( index 1))))) (define (shuffle! deck index) (if (< index 0) deck (begin (vectorswap! deck index (random (+ index 1))) (shuffle! deck ( index 1)))))
These procedures have a similar structure, like the similarities we found in other recursive patterns. Both of these procedures take an index as an argument, and both have
(< index 0)
as their base case. Also, both have, as their recursive case, a
begin
in which the first action does something to the vector element
selected by the current index, and the second action is a recursive call
with the index decreased by one. These procedures are initially called with
the largest possible index value.
In some cases it's more convenient to count the index upward from zero:
(define (list>vector lst) (l>vhelper (makevector (length lst)) lst 0)) (define (l>vhelper vec lst index) (if (= index (vectorlength vec)) vec (begin (vectorset! vec index (car lst)) (l>vhelper vec (cdr lst) (+ index 1)))))
Since lists are naturally processed from left to right (using
car
and cdr
), this program must process the vector from left to
right also.
Since we introduced vectors to provide mutability, you may have the impression that mutability is the main difference between vectors and lists. Actually, lists are mutable too, although the issues are more complicated; that's why we haven't used list mutation in this book.
The most important difference between lists and vectors is that each kind of aggregate lends itself to a different style of programming, because some operations are faster than others in each. List programming is characterized by two operations: dividing a list into its first element and all the rest, and sticking one new element onto the front of a list. Vector programming is characterized by selecting elements in any order, from a collection whose size is set permanently when the vector is created.
To make these rather vague descriptions more concrete, here are two procedures, one of which squares every number in a list, and the other of which squares every number in a vector:
(define (listsquare numbers) (if (null? numbers) '() (cons (square (car numbers)) (listsquare (cdr numbers))))) (define (vectorsquare numbers) (vecsqhelper (makevector (vectorlength numbers)) numbers ( (vectorlength numbers) 1))) (define (vecsqhelper new old index) (if (< index 0) new (begin (vectorset! new index (square (vectorref old index))) (vecsqhelper new old ( index 1)))))
In the list version, the intermediate stages of the algorithm deal with lists that are smaller than the original argument. Each recursive invocation "strips off" one element of its argument and "glues on" one extra element in its return value. In the vector version, the returned vector is created, at full size, as the first step in the algorithm; its component parts are filled in as the program proceeds.
This example can plausibly be done with either vectors or lists, so we've used it to compare the two techniques. But some algorithms fit most naturally with one kind of aggregate and would be awkward and slow using the other kind. The swapping of pairs of elements in the shuffling algorithm would be much harder using lists, while mergesort would be harder using vectors.
The best way to understand these differences in style is to know the
operations that are most efficient for each kind of aggregate. In each
case, there are certain operations that can be done in one small unit of
time, regardless of the number of elements in the aggregate, while other
operations take more time for more elements. The constant time
operations for lists are cons
, car
, cdr
, and null?
;
the ones for vectors are vectorref
, vectorset!
, and vectorlength
.[5] And if you reread the
squaring programs, you'll find that these are precisely the operations
they use.
We might have used listref
in the list version, but we didn't, and
Scheme programmers usually don't, because we know that it would be slower.
Similarly, we could implement something like cdr
for vectors, but that
would be slow, too, since it would have to make a onesmaller vector and
copy the elements one at a time. There are two possible morals to this
story, and they're both true: First, programmers invent and learn the
algorithms that make sense for whatever data structure is available. Thus
we have wellknown programming patterns, such as the filter
pattern,
appropriate for lists, and different patterns appropriate for vectors.
Second, programmers choose which data structure to use depending on what
algorithms they need. If you want to shuffle cards, use a vector, but if
you want to split the deck into a bunch of variablesize piles, lists might
be more appropriate. In general, vectors are good at selecting elements in
arbitrary order from a fixedsize collection; lists are good only at
selecting elements strictly from left to right, but they can vary in size.
In this book, despite what we're saying here about efficiency, we've generally tried to present algorithms in the way that's easiest to understand, even when we know that there's a faster way. For example, we've shown several recursive procedures in which the base case test was
(= (count sent) 1)
If we were writing the program for practical use, rather than for a book, we would have written
(empty? (butfirst sent))
because we know that empty?
and butfirst
are both
constant time operations (because for sentences they're implemented as
null?
and cdr
), while count
takes a long time for large
sentences. But the version using count
makes the intent
clearer.[6]
Effects, sequence, and state are three sides of the same coin.[7]
In Chapter 20 we explained the connection between effect (printing something on the screen) and sequence: It matters what you print first. We also noted that there's no benefit to a sequence of expressions unless those expressions produce an effect, since the values returned by all but the last expression are discarded.
In this chapter we've seen another connection. The way our vector programs
maintain state information is by carrying out effects, namely, vectorset!
invocations. Actually, every effect changes some kind
of state; if not in Scheme's memory, then on the computer screen or in a
file.
The final connection to be made is between state and sequence. Once a
program maintains state, it matters whether some computation is carried out
before or after another computation that changes the state. The example at
the beginning of this chapter in which an expression had different results
before and after defining a variable illustrates this point. As another
example, if we evaluate (lap 1)
200 times and (lap 2)
200 times,
the program's determination of the winner of the race depends on whether the
last evaluation of (lap 1)
comes before or after the last invocation
of (lap 2)
.
Because these three ideas are so closely connected, the names sequential programming (emphasizing sequence) and imperative programming (emphasizing effect) are both used to refer to a style of programming that uses all three. This style is in contrast with functional programming, which, as you know, uses none of them.
Although functional and sequential programming are, in a sense, opposites, it's perfectly possible to use both styles within one program, as we pointed out in the tictactoe program of Chapter 20. We'll show more such hybrid programs in the following chapters.
Don't forget that the first element of a vector is number zero, and there is no element whose index number is equal to the length of the vector. (Although these points are equally true for lists, it doesn't often matter, because we rarely select the elements of a list by number.) In particular, in a vector recursion, if zero is the base case, then there's probably still one element left to process.
Try the following experiment:
> (define dessert (vector 'chocolate 'sundae)) > (define twodesserts (list dessert dessert)) > (vectorset! (car twodesserts) 1 'shake) > twodesserts (#(CHOCOLATE SHAKE) #(CHOCOLATE SHAKE))
You might have expected that after asking to change one word in
twodesserts
, the result would be
(#(CHOCOLATE SHAKE) #(CHOCOLATE SUNDAE))
However, because of the way we created twodesserts
, both of
its elements are the same vector. If you think of a list as a
collection of things, it's strange to imagine the very same thing in two
different places, but that's the situation. If you want to have two separate
vectors that happen to have the same values in their elements, but are
individually mutable, you'd have to say
> (define twodesserts (list (vector 'chocolate 'sundae) (vector 'chocolate 'sundae))) > (vectorset! (car twodesserts) 1 'shake) > twodesserts (#(CHOCOLATE SHAKE) #(CHOCOLATE SUNDAE))
Each invocation of vector
or makevector
creates a
new, independent vector.
Do not solve any of the following exercises by converting a vector to a list, using list procedures, and then converting the result back to a vector.
23.1 Write a procedure sumvector
that takes a vector full of
numbers as its argument and returns the sum of all the numbers:
> (sumvector '#(6 7 8)) 21
23.2 Some versions of Scheme provide a procedure vectorfill!
that takes a
vector and anything as its two arguments. It replaces every element of the
vector with the second argument, like this:
> (define vec (vector 'one 'two 'three 'four)) > vec #(one two three four) > (vectorfill! vec 'yeah) > vec #(yeah yeah yeah yeah)
Write vectorfill!
. (It doesn't matter what value it
returns.)
23.3 Write a function vectorappend
that works just like regular append
, but for vectors:
> (vectorappend '#(not a) '#(second time)) #(not a second time)
23.4 Write vector>list
.
23.5 Write a procedure vectormap
that takes two arguments, a
function and a vector, and returns a new vector in which each box contains
the result of applying the function to the corresponding element of the
argument vector.
23.6 Write a procedure vectormap!
that takes two arguments, a
function and a vector, and modifies the argument vector by replacing each
element with the result of applying the function to that element. Your
procedure should return the same vector.
23.7 Could you write vectorfilter
? How about vectorfilter!
?
Explain the issues involved.
23.8 Modify the lap
procedure to print "Car 34 wins!" when car 34
completes its 200th lap. (A harder but more correct modification is
to print the message only if no other car has completed 200 laps.)
23.9
Write a procedure leader
that says which car is in the lead
right now.
23.10 Why doesn't this solution to Exercise 23.9 work?
(define (leader) (leaderhelper 0 1)) (define (leaderhelper leader index) (cond ((= index 100) leader) ((> (lap index) (lap leader)) (leaderhelper index (+ index 1))) (else (leaderhelper leader (+ index 1)))))
23.11 In some restaurants, the servers use computer terminals to keep track of what each table has ordered. Every time you order more food, the server enters your order into the computer. When you're ready for the check, the computer prints your bill.
You're going to write two procedures, order
and bill.
Order
takes a table number and an item as arguments and
adds the cost of that item to that table's bill. Bill
takes a table
number as its argument, returns the amount owed by that table, and resets
the table for the next customers. (Your order
procedure can examine a
global variable *menu*
to find the price of each item.)
> (order 3 'potstickers) > (order 3 'worwonton) > (order 5 'eggrolls) > (order 3 'shinshinspecialprawns) > (bill 3) 13.85 > (bill 5) 2.75
23.12 Rewrite selection sort (from Chapter 15) to sort a vector. This can
be done in a way similar to the procedure for shuffling a deck: Find
the smallest element of the vector and exchange it (using vectorswap!
) with the value in the first box. Then find the smallest
element not including the first box, and exchange that with the second box,
and so on. For example, suppose we have a vector of numbers:
#(23 4 18 7 95 60)
Your program should transform the vector through these intermediate stages:
#(4 23 18 7 95 60) ; exchange 4 with 23 #(4 7 18 23 95 60) ; exchange 7 with 23 #(4 7 18 23 95 60) ; exchange 18 with itself #(4 7 18 23 95 60) ; exchange 23 with itself #(4 7 18 23 60 95) ; exchange 60 with 95
23.13 Why doesn't this work?
(define (vectorswap! vector index1 index2) (vectorset! vector index1 (vectorref vector index2)) (vectorset! vector index2 (vectorref vector index1)))
23.14 Implement a twodimensional version of vectors. (We'll call one of these structures a matrix.) The implementation will use a vector of vectors. For example, a threebyfive matrix will be a threeelement vector, in which each of the elements is a fiveelement vector. Here's how it should work:
> (define m (makematrix 3 5)) > (matrixset! m 2 1 '(her majesty)) > (matrixref m 2 1) (HER MAJESTY)
23.15 Generalize Exercise 23.14 by implementing an array structure that can have any number of dimensions. Instead of taking two numbers as index arguments, as the matrix procedures do, the array procedures will take one argument, a list of numbers. The number of numbers is the number of dimensions, and it will be constant for any particular array. For example, here is a threedimensional array (4×5×6):
> (define a1 (makearray '(4 5 6))) > (arrayset! a1 '(3 2 3) '(the end))
23.16 We want to reimplement sentences as vectors instead of lists.
(a) Write versions of sentence
, empty?
, first
,
butfirst
, last
, and butlast
that use vectors. Your
selectors need only work for sentences, not for words.
> (sentence 'a 'b 'c) #(A B C) > (butfirst (sentence 'a 'b 'c)) #(B C)
(You don't have to make these procedures work on lists as well as vectors!)
(b) Does the following program still work with the new implementation of sentences? If not, fix the program.
(define (praise stuff) (sentence stuff '(is good)))
(c) Does the following program still work with the new implementation of sentences? If not, fix the program.
(define (praise stuff) (sentence stuff 'rules!))
(d) Does the following program still work with the new implementation of sentences? If not, fix the program. If so, is there some optional rewriting that would improve its performance?
(define (item n sent) (if (= n 1) (first sent) (item ( n 1) (butfirst sent))))
(e) Does the following program still work with the new implementation of sentences? If not, fix the program. If so, is there some optional rewriting that would improve its performance?
(define (every fn sent) (if (empty? sent) sent (sentence (fn (first sent)) (every fn (butfirst sent)))))
(f) In what ways does using vectors to implement sentences affect the speed of the selectors and constructor? Why do you think we chose to use lists?
[2] In some versions of Scheme, makevector
can take an
optional argument specifying an initial value to put in every box. In those
versions, we could just say
(define *lapvector* (makevector 100 0))
without having to use the initialization procedure.
[3] That's what we mean by "nonfunctional," not that it doesn't work!
[4] We could get around this problem in a different way:
(define (cardlist) (every (lambda (suit) (every (lambda (rank) (word suit rank)) '(a 2 3 4 5 6 7 8 9 10 j q k))) '(h s d c)))
In this version, we're taking advantage of the fact that our
sentence data type was defined in a way that prevents the creation of
sublists. A sentence of cards is a good representation for the deck.
However, with this approach we are mixing up the list and sentence data
types, because later we're going to invoke list>vector
with this deck
of cards as its argument. If we use sentence tools such as every
to
create the deck, then the procedure cardlist
should really be called
cardsentence
.
What difference does it make? The every
version works fine, as long
as sentences are implemented as lists, so that list>vector
can be
applied to a sentence. But the point about abstract data types such as
sentences is to avoid making assumptions about their implementation. If
for some reason we decided to change the internal representation of
sentences, then list>vector
could no longer be applied to a sentence.
Strictly speaking, if we're going to use this trick, we need a separate
conversion procedure sentence>vector
.
Of course, if you don't mind a little typing, you can avoid this whole issue
by having a quoted list of all 52 cards built into the definition of cardlist
.
[5] Where did this information come from? Just take our word for it. In later courses you'll study how vectors and lists are implemented, and then there will be reasons.
[6] For words, it turns out, the count
version is faster,
because words behave more like vectors than like lists.
[7] … to coin a phrase.
Brian Harvey,
bh@cs.berkeley.edu