The substitution model of evaluation
Now that you know how to create and run a LISP function, it's time to become
familiar with how LISP functions are evaluated by a LISP interpreter. We
won't go into all the details here; we'll leave some of the details for the
weeks to come. But this little discussion should give you enough
understanding to make you dangerous.
Let's say you've created a file with all the LISP code for evaluating the
quadratic formula that I gave you in the previous lecture notes, and that
you've corrected the little detail about the number of arguments to the
denominator function. Then let's say that you've also added your
implementation of the function for computing the discriminant. Now you tell
the LISP interpreter to evaluate all those defun functions, as you were shown
in your lab session. You're ready to compute stuff. If you type
(pos-root 1 0 -4) at the LISP evaluator, here's what will happen:
? (pos-root 1 0 -4)
2
?
You can follow that a little bit more closely by telling the evaluator that
you want to "trace" some of the functions that are being called:
? (trace pos-root pos-numerator discriminant denominator)
NIL
?
Then when you type (pos-root 1 0 -4), you'll see something like this:
? (pos-root 1 0 -4)
Calling (POS-ROOT 1 0 -4)
Calling (POS-NUMERATOR 1 0 -4)
Calling (DISCRIMINANT 1 0 -4)
DISCRIMINANT returned 4
POS-NUMERATOR returned 4
Calling (DENOMINATOR 1)
DENOMINATOR returned 2
POS-ROOT returned 2
2
What's really happening is something like this: When I type the list
(pos-root 1 0 -4) at the interpreter, LISP assumes I want to invoke the
function "pos-root". It evaluates the arguments, 1, 0, and -4, which in
this case conveniently evaluate to themselves. LISP then looks up the
definition of "pos-root", extracts that definition, and then applies that
definition to the evaluated arguments. This can be viewed as a
substitution, where:
(pos-root 1 0 -4)
becomes
(/ (pos-numerator 1 0 -4) (denominator 1 0 -4))
This expression is sent to the evaluator again. LISP evaluates the arguments,
but this time the arguments aren't numbers but are instead lists. So these
expressions are handed to the evaluator, and LISP treats them as function
calls.
The order in which these function calls, or any arguments in this example,
are evaluated won't be important. Why? Because of the functional programming
constraints we've followed, there are no side-effects. For example, computing
pos-numerator in no way affects the computation of denominator and vice-versa,
so it doesn't matter which one is computed first. (This will be a very nice
feature in the parallel-processing world of the future.) If, on the other
hand, one of these functions updated some global variable that was accessed
by the other, we would most definitely worry about which one was computed
first, resulting in a harder programming job, more complex software, and so on.
Still, something like Macintosh Common LISP has to evaluate one of these
arguments before the other, and the way it works (as do most LISP systems) is
to evaluate the arguments left to right. So LISP evaluates the expression
(pos-numerator 1 0 -4) and does the appropriate substitution:
(/ (pos-numerator 1 0 -4) (denominator 1 0 -4))
becomes
(/ (+ (- 0) (discriminant 1 0 -4)) (denominator 1 0 -4))
This "substitution" process continues until LISP arrives at a solution
(which requires that all the functions you defined are eventually replaced by
functions that were already defined in LISP). In short, the LISP function
evaluation process can be described like this:
1. Look up and retrieve the function definition.
2. Evaluate the arguments to the function (by passing the
arguments themselves to the LISP evaluation function---
numbers evaluate to themselves, symbols evaluate to
the objects they're bound to, while lists are treated
as function calls and evaluated accordingly).
3. Apply the function definition to the evaluated arguments
(i.e., replace the argument place-holders in the new
function definition with the corresponding evaluated
arguments).
4. Substitute the result of all that for the original
function call from step 1.
5. Go back to step 1 with this new expression.
Warning: The is a very sketchy and hand-wavy description of the evaluation
process. It should suffice for now, but as time goes by, we'll beef this
description up with details about how various types of things get evaluated.
But rather than dump all the details on you at once, we'll add things
incrementally as we need them.
LISP's favorite data structure
LISP comes with its own "abstract data types" or ADTs. As you undoubtedly
remember from previous courses, an ADT is (1) the logical data structure
itself (an abstraction, not the detailed implementation), combined with (2) a
set of operations which work on the data structure.
The two basic data types in LISP are "atoms" and "lists". An atom is a
non-divisible thing, like symbols (foo, bar, +) and numbers (42, 3.15).
They're not exactly interesting, nor are the operations associated with them.
The more interesting data structure is the list. A list is an ordered set of
atoms or lists (a recursive definition, no?). That ordered set must begin
with a "(" and end with a ")", and it may contain any number of atoms or lists
in any order, and therefore includes the empty list. The following are all
examples of valid lists:
()
(a)
(a b c)
(a (b) c)
(a (b (c)))
(alex ate the green meat)
(defun pos-root (a b c) (/ (pos-numerator a b c)(denominator a b c)))
(Both atoms and lists are often called symbolic expressions, or even just
expressions, which is why we talked about evaluating expressions occasionally
in the discussion a few paragraphs above.)
Simple operations on lists
There are three fundamental operations on lists: two of them are used for
decomposing lists and getting at their components, while the other operation
is used for composing lists.
The first list operator is called "first". Given an argument which is a list,
"first" returns the first element of that list. For example, if the symbol A
has been bound to the list (X Y Z 4), then (first A) will return X. Note
that the original list will not be altered in any way; A is still bound to
the list (X Y Z 4).
The second list operator is called "rest". Given an argument which is a list,
"rest" will return the list consisting of all the original elements of the
list except the first element. Thus, assuming A is still bound to (X Y Z 4),
(rest A) will return (Y Z 4).
The third operator is called "cons", which you can think of as being short for
"construct". Given two arguments, where the first is anything and the second
is a list, "cons" returns the list that you get by inserting the first
argument as the first element of the list that's the second argument.
Got it? So, (cons (first A)(rest A)) returns (X Y Z 4).
Let's go over it again:
What does (first (X Y Z 4)) return?
Time's up. If you said it returns X, then you're dead wrong. Why?
Because you forgot the evaluation rules:
(first A) where A is bound to the list (X Y Z 4) is not the same as
(first (X Y Z 4))!
In the first case, the argument A evaluates to the list (X Y Z 4), and then
when the definition of "first" is applied to that, the first element of that
list, X, is returned. Great. In the second case, however, LISP looks at the
argument (X Y Z 4) and tries to evaluate that as a call to the function named
X with the arguments Y, Z, and 4. If you don't have a previously-defined
function named X, or if Y or Z aren't bound to something, you'll see LISP
grind to a screeching halt.
In short, anything you throw at LISP, LISP will try to evaluate. Give it a
symbol and LISP will try to find what the symbol is bound to. Give it a list,
and LISP will try to evaluate it as a function call...
Preventing evaluation
...unless you explicitly tell LISP not to evaluate what you're giving it!
How do you do that? With another function, called "quote", which is merely a
function that takes an argument, doesn't evaluate it, and returns that
argument. For example, (quote (X Y Z 4)) returns (X Y Z 4). So
(first (quote (X Y Z 4))) returns X, not an error.
The "quote" function is used so often that it gets an abbreviation: the
apostrophe or single quote mark. So (quote (X Y Z 4)) is the same thing
as '(X Y Z 4), and
? (first '(X Y Z 4))
X
? (rest '(X Y Z 4))
(Y Z 4)
? (cons 'X '(Y Z 4))
(X Y Z 4)
?
A dangerous piece of knowledge
Lots of LISP programming errors result from using "quote" when not necessary,
or not using it when you should, so you might want to sit down for a while in
front of your favorite LISP system and play with this stuff for awhile.
And in doing this kind of practice, it might be helpful to know how to bind
symbols to values using an assignment operator. So now I'm going to show you
how to do that, but for now you can only use assignment to help you get a feel
for quoting and evaluation and stuff like that. DO NOT USE THIS ASSIGNMENT
OPERATOR IN ANYTHING YOU SUBMIT FOR GRADING UNTIL WE TELL YOU TO AS IT
VIOLATES FUNCTIONAL PROGRAMMING CONSTRAINTS. Someday in the weeks ahead,
we'll let you use assignment in a responsible fashion, but for now just use
it in practice. Otherwise, you'll find yourself getting amazingly low
grades while doing extra work.
The generic assignment operator is "setf", and takes two arguments. The first
argument is a symbol (e.g., a variable name), and the second argument is the
thing you want the symbol bound to. The first argument is not evaluated, but
the second argument is, which should tell you that "setf" is not an ordinary
LISP function. The "setf" function evaluates the second argument, binds it
to the first argument, and returns the evaluated second argument:
? (setf A '(X Y Z 4))
(X Y Z 4)
? (first A)
X
?
? (setf A (X Y Z 4))
> Error: Unbound variable: Y
> While executing: SYMBOL-VALUE
> Type Command-/ to continue, Command-. to abort.
> If continued: Retry getting the value of Y.
See the Restarts... menu item for further choices.
1 >
A variation of "setf" (actually, it's the predecessor to "setf") is "setq".
They're different, but it's not important for you to know how they're
different just yet. We'll deal with that later. For now, assume that
they're interchangeable.
Box-and-pointer notation
Now you're comfortable, I hope, with the notion that (X Y Z 4) is a list, but
unlike Pascal and other languages, LISP doesn't require you to worry about
the pointer details. But sometimes, LISP folks like to see the pointers to
help understand what's going on. So here's a more tangible (i.e., less
abstract) picture of what our sample list looks like:
A
|
|
| _______ _______ _______ _______
+-->| | | | | | | | | | | /|
| | | --+---->| | | --+---->| | | --+---->| | | / |
|_|_|___| |_|_|___| |_|_|___| |_|_|/__|
| | | |
X Y Z 4
In this lovely figure, each element of the list is represented as a pair of
words in memory. The first or left word contains a pointer to the symbol
that "is" the first element, and the second or right word contains a pointer
to the next element. It's now easy to see that, given a pointer to this list,
the function "first" (e.g., (first A)) looks at the pair of words at the end
of that pointer, follows the pointer stored in the left word, and returns
what's at the end of that pointer. In this case, it's the symbol X.
The "rest" function (e.g., (rest A)) looks at the pair of words pointed at by
A, follows the pointer stored in the right word, and returns what's at the
end of that pointer, which in this case is the list (Y Z 4). Note again that
these operations didn't change the structure of the list, they merely followed
pointers and returned what they pointed to.
Each of these two-word pairs are called "cons cells", because it's what you
get when you "cons" two things together. (This is where the dynamic memory
allocation mechanism comes into play.)
Earlier, I said that the second argument to the "cons" function should be a
list, but I lied for the sake of simplicity. You can in fact have something
that's not a list as the second argument to a "cons" function, but the result
is something slightly weird, called a "dotted pair". For example:
? (cons 'A 'B)
(A . B)
?
The box-and-pointer notation for this dotted pair looks like this:
_______
| | |
| | | --+---->B
|_|_|___|
|
A
You used to see dotted pairs used a lot in LISP programming, but they're not
used as frequently any more (except in some circumstances, which we might get
to in this course). I personally interpret these as meaning that I "cons'ed"
two things together that I didn't mean to.
Something for nothing
What's the technical difference between the list and the dotted pair? It's
how they are terminated. A "well-formed" list always ends with something
called "nil", which we represented in the box-and-pointer diagram up there
with a big slash through the right word of the last element. A dotted pair,
on the other hand, ends with something other than "nil".
So what's a nil? A nil is a very special thing in LISP. First, it has the
unusual property of being both an atom and a list at the same time. What
kind of list? It's the empty list, represented by "()". Consequently, "nil"
and "()" are the same thing. It's also the symbol meaning Boolean "false" in
LISP (Boolean "true" is the predefined "T" or anything non-nil). And, as we
noted just above, it's always the end of a "well-formed" list, which allows
LISP to maintain the following sorts of consistencies:
The first element of the empty list is, naturally, nothing, which is nil,
which is the empty list:
? (first nil)
NIL
?
The rest of the empty list, which is just the empty list with the first
element (i.e., nothing) removed, is also the empty list:
? (rest nil)
NIL
?
But if we try to put (first nil) and (rest nil) back together using "cons",
we should get the original thing we started with, which was nil, no? No:
? (cons (first nil) (rest nil))
(NIL)
?
It's a result of the unique nature of nil in LISP. Think of it as one of
LISP's endearing idiosyncracies. And since ending with nil is a sign of
goodness, that's all for now. Go practice.
Copyright 1996 by Kurt Eiselt. All rights reserved.
Last revised: April 3, 1996