Our evaluator for Lisp will be implemented as a Lisp program. It may seem
circular to think about evaluating Lisp programs using an evaluator that is
itself implemented in Lisp. However, evaluation is a process, so it is
appropriate to describe the evaluation process using Lisp, which, after all, is
our tool for describing processes.
An evaluator that is written in the same language that it evaluates is said to
be
metacircular.
The metacircular evaluator is essentially a Scheme formulation of the
environment model of evaluation described in 3.2. Recall that
the model has two basic parts:
1. To evaluate a combination (a compound expression other than a special form),
evaluate the subexpressions and then apply the value of the operator
subexpression to the values of the operand subexpressions.2. To apply a compound procedure to a set of arguments, evaluate the body of the
procedure in a new environment. To construct this environment, extend the
environment part of the procedure object by a frame in which the formal
parameters of the procedure are bound to the arguments to which the procedure
is applied.
These two rules describe the essence of the evaluation process, a basic cycle
in which expressions to be evaluated in environments are reduced to procedures
to be applied to arguments, which in turn are reduced to new expressions to be
evaluated in new environments, and so on, until we get down to symbols, whose
values are looked up in the environment, and to primitive procedures, which are
applied directly (see Figure 4.1).
This evaluation cycle will be embodied by the interplay between the two
critical procedures in the evaluator, eval and apply, which are
described in 4.1.1 (see Figure 4.1).
Figure 4.1:Theeval-applycycle exposes the essence of a computer language.
The implementation of the evaluator will depend upon procedures that define the
syntax of the expressions to be evaluated. We will use data
abstraction to make the evaluator independent of the representation of the
language. For example, rather than committing to a choice that an assignment
is to be represented by a list beginning with the symbol set! we use an
abstract predicate assignment? to test for an assignment, and we use
abstract selectors assignment-variable and assignment-value to
access the parts of an assignment. Implementation of expressions will be
described in detail in 4.1.2. There are also operations,
described in 4.1.3, that specify the representation of procedures
and environments. For example, make-procedure constructs compound
procedures, lookup-variable-value accesses the values of variables, and
apply-primitive-procedure applies a primitive procedure to a given list
of arguments.
4.1.1The Core of the Evaluator
The evaluation process can be described as the interplay between two
procedures: eval and apply.
Eval
Eval takes as arguments an expression and an environment. It classifies
the expression and directs its evaluation. Eval is structured as a case
analysis of the syntactic type of the expression to be evaluated. In order to
keep the procedure general, we express the determination of the type of an
expression abstractly, making no commitment to any particular representation
for the various types of expressions. Each type of expression has a predicate
that tests for it and an abstract means for selecting its parts. This
abstract syntax makes it easy to see how we can change the syntax of
the language by using the same evaluator, but with a different collection of
syntax procedures.
Primitive expressions
- For self-evaluating expressions, such as numbers, eval returns the
expression itself.- Eval must look up variables in the environment to find their values.
Special forms
- For quoted expressions, eval returns the expression that was quoted.- An assignment to (or a definition of) a variable must recursively call
eval to compute the new value to be associated with the variable. The
environment must be modified to change (or create) the binding of the variable.- An if expression requires special processing of its parts, so as to
evaluate the consequent if the predicate is true, and otherwise to evaluate the
alternative.- A lambda expression must be transformed into an applicable procedure by
packaging together the parameters and body specified by the lambda
expression with the environment of the evaluation.- A begin expression requires evaluating its sequence of expressions in
the order in which they appear.- A case analysis (cond) is transformed into a nest of if
expressions and then evaluated.
Combinations
- For a procedure application, eval must recursively evaluate the operator
part and the operands of the combination. The resulting procedure and
arguments are passed to apply, which handles the actual procedure
application.
Here is the definition of eval:
For clarity, eval has been implemented as a case analysis using
cond. The disadvantage of this is that our procedure handles only a few
distinguishable types of expressions, and no new ones can be defined without
editing the definition of eval. In most Lisp implementations,
dispatching on the type of an expression is done in a data-directed style.
This allows a user to add new types of expressions that eval can
distinguish, without modifying the definition of eval itself. (See
Exercise 4.3.)
Apply
Apply takes two arguments, a procedure and a list of arguments to which
the procedure should be applied. Apply classifies procedures into two
kinds: It calls apply-primitive-procedure to apply primitives; it
applies compound procedures by sequentially evaluating the expressions that
make up the body of the procedure. The environment for the evaluation of the
body of a compound procedure is constructed by extending the base environment
carried by the procedure to include a frame that binds the parameters of the
procedure to the arguments to which the procedure is to be applied. Here is
the definition of apply:
def apply(procedure, arguments):
# Apply a procedure to a list of arguments
if is_primitive_procedure(procedure):
return apply_primitive_procedure(procedure, arguments)
elif is_compound_procedure(procedure):
return eval_sequence(
procedure_body(procedure),
extend_environment(
procedure_parameters(procedure),
arguments,
procedure_environment(procedure)
)
)
else:
return error("Unknown procedure type: APPLY", procedure)
When eval processes a procedure application, it uses
list-of-values to produce the list of arguments to which the procedure
is to be applied. List-of-values takes as an argument the operands of
the combination. It evaluates each operand and returns a list of the
corresponding values:
def list_of_values(exps, env):
# if there are no operands, return empty list
if not exps:
return []
# evaluate first operand and cons onto list of values of rest
return [eval(exps[0], env)] + list_of_values(exps[1:], env)
Eval-if evaluates the predicate part of an if expression in the
given environment. If the result is true, eval-if evaluates the
consequent, otherwise it evaluates the alternative:
def eval_if(exp, env):
if eval(if_predicate(exp), env) is not False:
return eval(if_consequent(exp), env)
else:
return eval(if_alternative(exp), env)
The use of true? in eval-if highlights the issue of the
connection between an implemented language and an implementation language. The
if-predicate is evaluated in the language being implemented and thus
yields a value in that language. The interpreter predicate true?
translates that value into a value that can be tested by the if in the
implementation language: The metacircular representation of truth might not be
the same as that of the underlying Scheme.
Sequences
Eval-sequence is used by apply to evaluate the sequence of
expressions in a procedure body and by eval to evaluate the sequence of
expressions in a begin expression. It takes as arguments a sequence of
expressions and an environment, and evaluates the expressions in the order in
which they occur. The value returned is the value of the final expression.
The following procedure handles assignments to variables. It calls eval
to find the value to be assigned and transmits the variable and the resulting
value to set-variable-value! to be installed in the designated
environment.
We have chosen here to return the symbol ok as the value of an
assignment or a definition.
Exercise 4.1: Notice that we cannot tell whether
the metacircular evaluator evaluates operands from left to right or from right
to left. Its evaluation order is inherited from the underlying Lisp: If the
arguments to cons in list-of-values are evaluated from left to
right, then list-of-values will evaluate operands from left to right;
and if the arguments to cons are evaluated from right to left, then
list-of-values will evaluate operands from right to left.
Write a version of list-of-values that evaluates operands from left to
right regardless of the order of evaluation in the underlying Lisp. Also write
a version of list-of-values that evaluates operands from right to left.
4.1.2Representing Expressions
The evaluator is reminiscent of the symbolic differentiation program discussed
in 2.3.2. Both programs operate on symbolic expressions. In
both programs, the result of operating on a compound expression is determined
by operating recursively on the pieces of the expression and combining the
results in a way that depends on the type of the expression. In both programs
we used data abstraction to decouple the general rules of operation from the
details of how expressions are represented. In the differentiation program
this meant that the same differentiation procedure could deal with algebraic
expressions in prefix form, in infix form, or in some other form. For the
evaluator, this means that the syntax of the language being evaluated is
determined solely by the procedures that classify and extract pieces of
expressions.
Here is the specification of the syntax of our language:
- The only self-evaluating items are numbers and strings:
(define (self-evaluating? exp)
(cond ((number? exp) true)
((string? exp) true)
(else false)))- Variables are represented by symbols:
(define (variable? exp) (symbol? exp))- Quotations have the form (quote ⟨text-of-quotation⟩):
(define (quoted? exp)
(tagged-list? exp ‘quote))
(define (text-of-quotation exp)
(cadr exp))
Quoted? is defined in terms of the procedure tagged-list?, which
identifies lists beginning with a designated symbol:
(define (tagged-list? exp tag)
(if (pair? exp)
(eq? (car exp) tag)
false))- Assignments have the form (set! ⟨var⟩ ⟨value⟩):
(define (definition-value exp)
(if (symbol? (cadr exp))
(caddr exp)
(make-lambda
(cdadr exp) ; formal parameters
(cddr exp)))) ; body- Lambda expressions are lists that begin with the symbol lambda:
(define (lambda? exp)
(tagged-list? exp ‘lambda))
(define (lambda-parameters exp) (cadr exp))
(define (lambda-body exp) (cddr exp))
We also provide a constructor for lambda expressions, which is used by
definition-value, above:
(define (make-lambda parameters body)
(cons ‘lambda (cons parameters body)))- Conditionals begin with if and have a predicate, a consequent, and an
(optional) alternative. If the expression has no alternative part, we provide
false as the alternative.
(define (if? exp) (tagged-list? exp ‘if))
(define (if-predicate exp) (cadr exp))
(define (if-consequent exp) (caddr exp))
(define (if-alternative exp)
(if (not (null? (cdddr exp)))
(cadddr exp)
‘false))
We also provide a constructor for if expressions, to be used by
cond->if to transform cond expressions into if
expressions:
(define (make-if predicate
consequent
alternative)
(list ‘if
predicate
consequent
alternative))- Begin packages a sequence of expressions into a single expression. We
include syntax operations on begin expressions to extract the actual
sequence from the begin expression, as well as selectors that return the
first expression and the rest of the expressions in the
sequence.
(define (begin? exp)
(tagged-list? exp ‘begin))
(define (begin-actions exp) (cdr exp))
(define (last-exp? seq) (null? (cdr seq)))
(define (first-exp seq) (car seq))
(define (rest-exps seq) (cdr seq))
We also include a constructor sequence->exp (for use by cond->if)
that transforms a sequence into a single expression, using begin if
necessary:
(define (make-begin seq) (cons ‘begin seq))- A procedure application is any compound expression that is not one of the above
expression types. The car of the expression is the operator, and the
cdr is the list of operands:
Some special forms in our language can be defined in terms of expressions
involving other special forms, rather than being implemented directly. One
example is cond, which can be implemented as a nest of if
expressions. For example, we can reduce the problem of evaluating the
expression
def cond_result(x):
if x > 0:
return x
elif x == 0:
print('zero')
return 0
else:
return -x
(cond ((> x 0) x)
((= x 0) (display 'zero) 0)
(else (- x)))
to the problem of evaluating the following expression involving if and
begin expressions:
def process(x):
if x > 0:
return x
elif x == 0:
print('zero')
return 0
else:
return -x
(if (> x 0)
x
(if (= x 0)
(begin (display 'zero) 0)
(- x)))
Implementing the evaluation of cond in this way simplifies the evaluator
because it reduces the number of special forms for which the evaluation process
must be explicitly specified.
We include syntax procedures that extract the parts of a cond
expression, and a procedure cond->if that transforms cond
expressions into if expressions. A case analysis begins with
cond and has a list of predicate-action clauses. A clause is an
else clause if its predicate is the symbol else.
Expressions (such as cond) that we choose to implement as syntactic
transformations are called
derived expressions. Let
expressions are also derived expressions (see
Exercise 4.6).
Exercise 4.2: Louis Reasoner plans to reorder the
cond clauses in eval so that the clause for procedure
applications appears before the clause for assignments. He argues that this
will make the interpreter more efficient: Since programs usually contain more
applications than assignments, definitions, and so on, his modified eval
will usually check fewer clauses than the original eval before
identifying the type of an expression.
1. What is wrong with Louis’s plan? (Hint: What will Louis’s evaluator do with
the expression (define x 3)?)2. Louis is upset that his plan didn’t work. He is willing to go to any lengths
to make his evaluator recognize procedure applications before it checks for
most other kinds of expressions. Help him by changing the syntax of the
evaluated language so that procedure applications start with call. For
example, instead of (factorial 3) we will now have to write (call
factorial 3) and instead of (+ 1 2) we will have to write (call +
1 2).
Exercise 4.3: Rewrite eval so that the
dispatch is done in data-directed style. Compare this with the data-directed
differentiation procedure of Exercise 2.73. (You may use the car
of a compound expression as the type of the expression, as is appropriate for
the syntax implemented in this section.)
Exercise 4.4: Recall the definitions of the
special forms and and or from Chapter 1:
- and: The expressions are evaluated from left to right. If any
expression evaluates to false, false is returned; any remaining expressions are
not evaluated. If all the expressions evaluate to true values, the value of
the last expression is returned. If there are no expressions then true is
returned.- or: The expressions are evaluated from left to right. If any expression
evaluates to a true value, that value is returned; any remaining expressions
are not evaluated. If all expressions evaluate to false, or if there are no
expressions, then false is returned.
Install and and or as new special forms for the evaluator by
defining appropriate syntax procedures and evaluation procedures
eval-and and eval-or. Alternatively, show how to implement
and and or as derived expressions.
Exercise 4.5: Scheme allows an additional syntax
for cond clauses, (⟨test⟩ => ⟨recipient⟩). If
⟨test⟩ evaluates to a true value, then ⟨recipient⟩ is evaluated.
Its value must be a procedure of one argument; this procedure is then invoked
on the value of the ⟨test⟩, and the result is returned as the value of
the cond expression. For example
def assoc(key, alist):
for pair in alist:
if pair[0] == key:
return pair
return None
def cadr(pair):
return pair[1]
>>> v = assoc('b', [('a', 1), ('b', 2)])
>>> cadr(v) if v else False
2
Implement a syntactic transformation let->combination that reduces
evaluating let expressions to evaluating combinations of the type shown
above, and add the appropriate clause to eval to handle let
expressions.
Exercise 4.7:Let* is similar to
let, except that the bindings of the let* variables are performed
sequentially from left to right, and each binding is made in an environment in
which all of the preceding bindings are visible. For example
>>> x = 3
>>> y = x + 2
>>> z = x + y + 5
>>> x * z
39
> (let* ((x 3)
> (y (+ x 2))
> (z (+ x y 5)))
> (* x z))
>
returns 39. Explain how a let* expression can be rewritten as a set of
nested let expressions, and write a procedure let*->nested-lets
that performs this transformation. If we have already implemented let
(Exercise 4.6) and we want to extend the evaluator to handle let*,
is it sufficient to add a clause to eval whose action is
>>> eval(let_star_to_nested_lets(exp), env)
> (eval (let*->nested-lets exp) env)
>
or must we explicitly expand let* in terms of non-derived expressions?
Exercise 4.8: “Named let” is a variant
of let that has the form
# (let ⟨var⟩ ⟨bindings⟩ ⟨body⟩)
def _let():
var = ... # binding(s) from ⟨bindings⟩
# ... body using var ...
return ...
_let()
> (let ⟨var⟩ ⟨bindings⟩ ⟨body⟩)
>
The ⟨bindings⟩ and ⟨body⟩ are just as in ordinary let,
except that ⟨var⟩ is bound within ⟨body⟩ to a procedure whose body
is ⟨body⟩ and whose parameters are the variables in the ⟨bindings⟩.
Thus, one can repeatedly execute the ⟨body⟩ by invoking the procedure
named ⟨var⟩. For example, the iterative Fibonacci procedure
(1.2.2) can be rewritten using named let as follows:
def fib(n):
# translated from the named let fib-iter ((a 1) (b 0) (count n))
a, b, count = 1, 0, n
while count != 0:
a, b, count = a + b, a, count - 1
return b
> (define (fib n)
> (let fib-iter ((a 1) (b 0) (count n))
> (if (= count 0)
> b
> (fib-iter (+ a b)
> a
> (- count 1)))))
>
Modify let->combination of Exercise 4.6 to also support named
let.
Exercise 4.9: Many languages support a variety of
iteration constructs, such as do, for, while, and
until. In Scheme, iterative processes can be expressed in terms of
ordinary procedure calls, so special iteration constructs provide no essential
gain in computational power. On the other hand, such constructs are often
convenient. Design some iteration constructs, give examples of their use, and
show how to implement them as derived expressions.
Exercise 4.10: By using data abstraction, we
were able to write an eval procedure that is independent of the
particular syntax of the language to be evaluated. To illustrate this, design
and implement a new syntax for Scheme by modifying the procedures in this
section, without changing eval or apply.
4.1.3Evaluator Data Structures
In addition to defining the external syntax of expressions, the evaluator
implementation must also define the data structures that the evaluator
manipulates internally, as part of the execution of a program, such as the
representation of procedures and environments and the representation of true
and false.
Testing of predicates
For conditionals, we accept anything to be true that is not the explicit
false object.
def is_true(x):
# (define (true? x) (not (eq? x false)))
return not (x is False)
def is_false(x):
# (define (false? x) (eq? x false))
return x is False
(define (true? x)
(not (eq? x false)))
(define (false? x)
(eq? x false))
Representing procedures
To handle primitives, we assume that we have available the following
procedures:
- (apply-primitive-procedure ⟨proc⟩ ⟨args⟩)
applies the given primitive procedure to the argument values in the list
⟨args⟩ and returns the result of the application.- (primitive-procedure? ⟨proc⟩)
tests whether ⟨proc⟩ is a primitive procedure.
These mechanisms for handling primitives are further described in
4.1.4.
Compound procedures are constructed from parameters, procedure bodies, and
environments using the constructor make-procedure:
The evaluator needs operations for manipulating environments. As explained in
3.2, an environment is a sequence of frames, where each frame is
a table of bindings that associate variables with their corresponding values.
We use the following operations for manipulating environments:
- (lookup-variable-value ⟨var⟩ ⟨env⟩)
returns the value that is bound to the symbol ⟨var⟩ in the environment
⟨env⟩, or signals an error if the variable is unbound.- (extend-environment ⟨variables⟩ ⟨values⟩ ⟨base-env⟩)
returns a new environment, consisting of a new frame in which the symbols in
the list ⟨variables⟩ are bound to the corresponding elements in the list
⟨values⟩, where the enclosing environment is the environment
⟨base-env⟩.- (define-variable! ⟨var⟩ ⟨value⟩ ⟨env⟩)
adds to the first frame in the environment ⟨env⟩ a new binding that
associates the variable ⟨var⟩ with the value ⟨value⟩.- (set-variable-value! ⟨var⟩ ⟨value⟩ ⟨env⟩)
changes the binding of the variable ⟨var⟩ in the environment ⟨env⟩
so that the variable is now bound to the value ⟨value⟩, or signals an
error if the variable is unbound.
To implement these operations we represent an environment as a list of frames.
The enclosing environment of an environment is the cdr of the list. The
empty environment is simply the empty list.
Each frame of an environment is represented as a pair of lists: a list of the
variables bound in that frame and a list of the associated
values.
def make_frame(variables, values):
return [variables, values]
def frame_variables(frame):
return frame[0]
def frame_values(frame):
return frame[1]
def add_binding_to_frame(var, val, frame):
# mutate the frame by adding var and val to the front of variables and values
frame[0] = [var] + frame[0]
frame[1] = [val] + frame[1]
(define (make-frame variables values)
(cons variables values))
(define (frame-variables frame) (car frame))
(define (frame-values frame) (cdr frame))
(define (add-binding-to-frame! var val frame)
(set-car! frame (cons var (car frame)))
(set-cdr! frame (cons val (cdr frame))))
To extend an environment by a new frame that associates variables with values,
we make a frame consisting of the list of variables and the list of values, and
we adjoin this to the environment. We signal an error if the number of
variables does not match the number of values.
def extend_environment(vars, vals, base_env):
# (define (extend-environment vars vals base-env)
# (if (= (length vars) (length vals))
# (cons (make-frame vars vals) base-env)
# (if (< (length vars) (length vals))
# (error "Too many arguments supplied"
# vars
# vals)
# (error "Too few arguments supplied"
# vars
# vals))))
if len(vars) == len(vals):
return [make_frame(vars, vals)] + base_env
if len(vars) < len(vals):
raise Exception("Too many arguments supplied", vars, vals)
raise Exception("Too few arguments supplied", vars, vals)
(define (extend-environment vars vals base-env)
(if (= (length vars) (length vals))
(cons (make-frame vars vals) base-env)
(if (< (length vars) (length vals))
(error "Too many arguments supplied"
vars
vals)
(error "Too few arguments supplied"
vars
vals))))
To look up a variable in an environment, we scan the list of variables in the
first frame. If we find the desired variable, we return the corresponding
element in the list of values. If we do not find the variable in the current
frame, we search the enclosing environment, and so on. If we reach the empty
environment, we signal an “unbound variable” error.
def lookup_variable_value(var, env):
def env_loop(env):
def scan(vars, vals):
if not vars:
return env_loop(enclosing_environment(env))
elif var == vars[0]:
return vals[0]
else:
return scan(vars[1:], vals[1:])
if env is the_empty_environment:
raise NameError(f"Unbound variable: {var}")
else:
frame = first_frame(env)
return scan(frame_variables(frame), frame_values(frame))
return env_loop(env)
To set a variable to a new value in a specified environment, we scan for the
variable, just as in lookup-variable-value, and change the corresponding
value when we find it.
def set_variable_value(var, val, env):
def env_loop(env):
if env is the_empty_environment:
raise Exception(f"Unbound variable: SET! {var}")
frame = first_frame(env)
vars = frame_variables(frame)
vals = frame_values(frame)
for i in range(len(vars)):
if vars[i] == var:
vals[i] = val
return
return env_loop(enclosing_environment(env))
return env_loop(env)
To define a variable, we search the first frame for a binding for the variable,
and change the binding if it exists (just as in set-variable-value!).
If no such binding exists, we adjoin one to the first frame.
def define_variable_bang(var, val, env):
# Translate of (define (define-variable! var val env) ...)
frame = first_frame(env)
vars = frame_variables(frame)
vals = frame_values(frame)
for i, v in enumerate(vars):
# Scheme uses eq? for symbols; use == for Python objects here
if v == var:
vals[i] = val
return
add_binding_to_frame_bang(var, val, frame)
(define (define-variable! var val env)
(let ((frame (first-frame env)))
(define (scan vars vals)
(cond ((null? vars)
(add-binding-to-frame!
var val frame))
((eq? var (car vars))
(set-car! vals val))
(else (scan (cdr vars)
(cdr vals)))))
(scan (frame-variables frame)
(frame-values frame))))
The method described here is only one of many plausible ways to represent
environments. Since we used data abstraction to isolate the rest of the
evaluator from the detailed choice of representation, we could change the
environment representation if we wanted to. (See Exercise 4.11.) In a
production-quality Lisp system, the speed of the evaluator’s environment
operations—especially that of variable lookup—has a major impact on the
performance of the system. The representation described here, although
conceptually simple, is not efficient and would not ordinarily be used in a
production system.
Exercise 4.11: Instead of representing a frame
as a pair of lists, we can represent a frame as a list of bindings, where each
binding is a name-value pair. Rewrite the environment operations to use this
alternative representation.
Exercise 4.12: The procedures
define-variable!, set-variable-value! and
lookup-variable-value can be expressed in terms of more abstract
procedures for traversing the environment structure. Define abstractions that
capture the common patterns and redefine the three procedures in terms of these
abstractions.
Exercise 4.13: Scheme allows us to create new
bindings for variables by means of define, but provides no way to get
rid of bindings. Implement for the evaluator a special form
make-unbound! that removes the binding of a given symbol from the
environment in which the make-unbound! expression is evaluated. This
problem is not completely specified. For example, should we remove only the
binding in the first frame of the environment? Complete the specification and
justify any choices you make.
4.1.4Running the Evaluator as a Program
Given the evaluator, we have in our hands a description (expressed in Lisp) of
the process by which Lisp expressions are evaluated. One advantage of
expressing the evaluator as a program is that we can run the program. This
gives us, running within Lisp, a working model of how Lisp itself evaluates
expressions. This can serve as a framework for experimenting with evaluation
rules, as we shall do later in this chapter.
Our evaluator program reduces expressions ultimately to the application of
primitive procedures. Therefore, all that we need to run the evaluator is to
create a mechanism that calls on the underlying Lisp system to model the
application of primitive procedures.
There must be a binding for each primitive procedure name, so that when
eval evaluates the operator of an application of a primitive, it will
find an object to pass to apply. We thus set up a global environment
that associates unique objects with the names of the primitive procedures that
can appear in the expressions we will be evaluating. The global environment
also includes bindings for the symbols true and false, so that
they can be used as variables in expressions to be evaluated.
It does not matter how we represent the primitive procedure objects, so long as
apply can identify and apply them by using the procedures
primitive-procedure? and apply-primitive-procedure. We have
chosen to represent a primitive procedure as a list beginning with the symbol
primitive and containing a procedure in the underlying Lisp that
implements that primitive.
Setup-environment will get the primitive names and implementation
procedures from a list:
def car(pair):
return pair[0]
def cdr(pair):
return pair[1:]
def cons(x, y):
# If y is a list, cons x onto it; otherwise create a pair-like list
if isinstance(y, list):
return [x] + y
else:
return [x, y]
def nullp(x):
return x == [] or x is None
primitive_procedures = [
['car', car],
['cdr', cdr],
['cons', cons],
['null?', nullp],
...
]
def primitive_procedure_names():
return [proc[0] for proc in primitive_procedures]
def primitive_procedure_objects():
return [['primitive', proc[1]] for proc in primitive_procedures]
For convenience in running the metacircular evaluator, we provide a
driver loop that models the read-eval-print loop of the underlying
Lisp system. It prints a
prompt, reads an input expression,
evaluates this expression in the global environment, and prints the result. We
precede each printed result by an
output prompt so as to distinguish
the value of the expression from other output that may be printed.
# M-Eval prompts
input_prompt = ";;; M-Eval input:"
output_prompt = ";;; M-Eval value:"
def driver_loop():
prompt_for_input(input_prompt)
inp = read()
output = scheme_eval(inp, the_global_environment)
announce_output(output_prompt)
user_print(output)
driver_loop()
def prompt_for_input(string):
print()
print()
print(string)
def announce_output(string):
print()
print(string)
print()
# Placeholders for the rest of the evaluator's components
def read():
... # Reads the next expression (to be implemented)
def scheme_eval(exp, env):
... # Evaluates exp in env (to be implemented)
the_global_environment = ... # The global environment (to be provided)
def user_print(val):
print(val) # Prints the value (can be replaced with a Scheme-style printer)
We use a special printing procedure, user-print, to avoid printing the
environment part of a compound procedure, which may be a very long list (or may
even contain cycles).
def user_print(obj):
if is_compound_procedure(obj):
# display the compound-procedure representation
print(['compound-procedure',
procedure_parameters(obj),
procedure_body(obj),
'<procedure-env>'])
else:
print(obj)
Now all we need to do to run the evaluator is to initialize the global
environment and start the driver loop. Here is a sample interaction:
the_global_environment = setup_environment()
driver_loop()
#;;; M-Eval input:
>>> def append(x, y):
... # x and y are Python lists
... if not x:
... return y
... return [x[0]] + append(x[1:], y)
ok
#;;; M-Eval input:
>>> append(['a', 'b', 'c'], ['d', 'e', 'f'])
['a', 'b', 'c', 'd', 'e', 'f']
(define the-global-environment
(setup-environment))
(driver-loop)
;;; M-Eval input:
(define (append x y)
(if (null? x)
y
(cons (car x) (append (cdr x) y))))
;;; M-Eval value:
ok
;;; M-Eval input:
(append '(a b c) '(d e f))
;;; M-Eval value:
(a b c d e f)
Exercise 4.14: Eva Lu Ator and Louis Reasoner
are each experimenting with the metacircular evaluator. Eva types in the
definition of map, and runs some test programs that use it. They work
fine. Louis, in contrast, has installed the system version of map as a
primitive for the metacircular evaluator. When he tries it, things go terribly
wrong. Explain why Louis’s map fails even though Eva’s works.
4.1.5Data as Programs
In thinking about a Lisp program that evaluates Lisp expressions, an analogy
might be helpful. One operational view of the meaning of a program is that a
program is a description of an abstract (perhaps infinitely large) machine.
For example, consider the familiar program to compute factorials:
def factorial(n):
if n == 1:
return 1
else:
return factorial(n - 1) * n
(define (factorial n)
(if (= n 1)
1
(* (factorial (- n 1)) n)))
We may regard this program as the description of a machine containing parts
that decrement, multiply, and test for equality, together with a two-position
switch and another factorial machine. (The factorial machine is infinite
because it contains another factorial machine within it.) Figure 4.2 is
a flow diagram for the factorial machine, showing how the parts are wired
together.
Figure 4.2:The factorial program, viewed as an abstract machine.
In a similar way, we can regard the evaluator as a very special machine that
takes as input a description of a machine. Given this input, the evaluator
configures itself to emulate the machine described. For example, if we feed
our evaluator the definition of factorial, as shown in Figure 4.3,
the evaluator will be able to compute factorials.
Figure 4.3:The evaluator emulating a factorial machine.
From this perspective, our evaluator is seen to be a
universal machine.
It mimics other machines when these are described as Lisp
programs. This is striking. Try to imagine an analogous evaluator
for electrical circuits. This would be a circuit that takes as input a signal
encoding the plans for some other circuit, such as a filter. Given this input,
the circuit evaluator would then behave like a filter with the same
description. Such a universal electrical circuit is almost unimaginably
complex. It is remarkable that the program evaluator is a rather simple
program.
Another striking aspect of the evaluator is that it acts as a bridge between
the data objects that are manipulated by our programming language and the
programming language itself. Imagine that the evaluator program (implemented
in Lisp) is running, and that a user is typing expressions to the evaluator and
observing the results. From the perspective of the user, an input expression
such as (* x x) is an expression in the programming language, which the
evaluator should execute. From the perspective of the evaluator, however, the
expression is simply a list (in this case, a list of three symbols: *,
x, and x) that is to be manipulated according to a well-defined
set of rules.
That the user’s programs are the evaluator’s data need not be a source of
confusion. In fact, it is sometimes convenient to ignore this distinction, and
to give the user the ability to explicitly evaluate a data object as a Lisp
expression, by making eval available for use in programs. Many Lisp
dialects provide a primitive eval procedure that takes as arguments an
expression and an environment and evaluates the expression relative to the
environment. Thus,
Exercise 4.15: Given a one-argument procedure
p and an object a, p is said to “halt” on a if
evaluating the expression (p a) returns a value (as opposed to
terminating with an error message or running forever). Show that it is
impossible to write a procedure halts? that correctly determines whether
p halts on a for any procedure p and object a. Use
the following reasoning: If you had such a procedure halts?, you could
implement the following program:
Now consider evaluating the expression (try try) and show that any
possible outcome (either halting or running forever) violates the intended
behavior of halts?.
4.1.6Internal Definitions
Our environment model of evaluation and our metacircular evaluator execute
definitions in sequence, extending the environment frame one definition at a
time. This is particularly convenient for interactive program development, in
which the programmer needs to freely mix the application of procedures with the
definition of new procedures. However, if we think carefully about the
internal definitions used to implement block structure (introduced in
1.1.8), we will find that name-by-name extension of the environment may
not be the best way to define local variables.
Consider a procedure with internal definitions, such as
def f(x):
def even_(n):
if n == 0:
return True
return odd_(n - 1)
def odd_(n):
if n == 0:
return False
return even_(n - 1)
...
(define (f x)
(define (even? n)
(if (= n 0)
true
(odd? (- n 1))))
(define (odd? n)
(if (= n 0)
false
(even? (- n 1))))
⟨rest of body of f⟩)
Our intention here is that the name odd? in the body of the procedure
even? should refer to the procedure odd? that is defined after
even?. The scope of the name odd? is the entire body of
f, not just the portion of the body of f starting at the point
where the define for odd? occurs. Indeed, when we consider that
odd? is itself defined in terms of even?—so that even?
and odd? are mutually recursive procedures—we see that the only
satisfactory interpretation of the two defines is to regard them as if
the names even? and odd? were being added to the environment
simultaneously. More generally, in block structure, the scope of a local name
is the entire procedure body in which the define is evaluated.
As it happens, our interpreter will evaluate calls to f correctly, but
for an “accidental” reason: Since the definitions of the internal procedures
come first, no calls to these procedures will be evaluated until all of them
have been defined. Hence, odd? will have been defined by the time
even? is executed. In fact, our sequential evaluation mechanism will
give the same result as a mechanism that directly implements simultaneous
definition for any procedure in which the internal definitions come first in a
body and evaluation of the value expressions for the defined variables doesn’t
actually use any of the defined variables. (For an example of a procedure that
doesn’t obey these restrictions, so that sequential definition isn’t equivalent
to simultaneous definition, see Exercise 4.19.)
There is, however, a simple way to treat definitions so that internally defined
names have truly simultaneous scope—just create all local variables that will
be in the current environment before evaluating any of the value expressions.
One way to do this is by a syntax transformation on lambda expressions.
Before evaluating the body of a lambda expression, we “scan out” and
eliminate all the internal definitions in the body. The internally defined
variables will be created with a let and then set to their values by
assignment. For example, the procedure
def lambda_fn(vars):
u = ...
v = ...
return ...
(lambda ⟨vars⟩
(define u ⟨e1⟩)
(define v ⟨e2⟩)
⟨e3⟩)
would be transformed into
def proc(...):
u = '*unassigned*'
v = '*unassigned*'
u = ...
v = ...
return ...
(lambda ⟨vars⟩
(let ((u '*unassigned*)
(v '*unassigned*))
(set! u ⟨e1⟩)
(set! v ⟨e2⟩)
⟨e3⟩))
where *unassigned* is a special symbol that causes looking up a variable
to signal an error if an attempt is made to use the value of the
not-yet-assigned variable.
An alternative strategy for scanning out internal definitions is shown in
Exercise 4.18. Unlike the transformation shown above, this enforces the
restriction that the defined variables’ values can be evaluated without using
any of the variables’ values.
Exercise 4.16: In this exercise we implement the
method just described for interpreting internal definitions. We assume that
the evaluator supports let (see Exercise 4.6).
1. Change lookup-variable-value (4.1.3) to signal an error if
the value it finds is the symbol *unassigned*.2. Write a procedure scan-out-defines that takes a procedure body and
returns an equivalent one that has no internal definitions, by making the
transformation described above.3. Install scan-out-defines in the interpreter, either in
make-procedure or in procedure-body (see 4.1.3).
Which place is better? Why?
Exercise 4.17: Draw diagrams of the environment
in effect when evaluating the expression ⟨e3⟩ in the procedure in the
text, comparing how this will be structured when definitions are interpreted
sequentially with how it will be structured if definitions are scanned out as
described. Why is there an extra frame in the transformed program? Explain
why this difference in environment structure can never make a difference in the
behavior of a correct program. Design a way to make the interpreter implement
the “simultaneous” scope rule for internal definitions without constructing
the extra frame.
Exercise 4.18: Consider an alternative strategy
for scanning out definitions that translates the example in the text to
def proc(...):
u = '*unassigned*'
v = '*unassigned*'
a = ...
b = ...
u = a
v = b
return ...
> (lambda ⟨vars⟩
> (let ((u '*unassigned*)
> (v '*unassigned*))
> (let ((a ⟨e1⟩)
> (b ⟨e2⟩))
> (set! u a)
> (set! v b))
> ⟨e3⟩))
>
Here a and b are meant to represent new variable names, created
by the interpreter, that do not appear in the user’s program. Consider the
solve procedure from 3.5.4:
def solve(f, y0, dt):
# (define (solve f y0 dt)
# (define y (integral (delay dy) y0 dt))
# (define dy (stream-map f y))
# y)
y = integral(delay(lambda: dy), y0, dt)
dy = stream_map(f, y)
return y
> (define (solve f y0 dt)
> (define y (integral (delay dy) y0 dt))
> (define dy (stream-map f y))
> y)
>
Will this procedure work if internal definitions are scanned out as shown in
this exercise? What if they are scanned out as shown in the text? Explain.
Exercise 4.19: Ben Bitdiddle, Alyssa P. Hacker,
and Eva Lu Ator are arguing about the desired result of evaluating the
expression
> (let ((a 1))
> (define (f x)
> (define b (+ a x))
> (define a 5)
> (+ a b))
> (f 10))
>
Ben asserts that the result should be obtained using the sequential rule for
define: b is defined to be 11, then a is defined to be 5,
so the result is 16. Alyssa objects that mutual recursion requires the
simultaneous scope rule for internal procedure definitions, and that it is
unreasonable to treat procedure names differently from other names. Thus, she
argues for the mechanism implemented in Exercise 4.16. This would lead
to a being unassigned at the time that the value for b is to be
computed. Hence, in Alyssa’s view the procedure should produce an error. Eva
has a third opinion. She says that if the definitions of a and b
are truly meant to be simultaneous, then the value 5 for a should be
used in evaluating b. Hence, in Eva’s view a should be 5,
b should be 15, and the result should be 20. Which (if any) of these
viewpoints do you support? Can you devise a way to implement internal
definitions so that they behave as Eva prefers?
Exercise 4.20: Because internal definitions look
sequential but are actually simultaneous, some people prefer to avoid them
entirely, and use the special form letrec instead. Letrec looks
like let, so it is not surprising that the variables it binds are bound
simultaneously and have the same scope as each other. The sample procedure
f above can be written without internal definitions, but with exactly
the same meaning, as
def f(x):
# letrec: mutually recursive definitions of even and odd
def even(n):
if n == 0:
return True
else:
return odd(n - 1)
def odd(n):
if n == 0:
return False
else:
return even(n - 1)
return ...
> (define (f x)
> (letrec
> ((even?
> (lambda (n)
> (if (= n 0)
> true
> (odd? (- n 1)))))
> (odd?
> (lambda (n)
> (if (= n 0)
> false
> (even? (- n 1))))))
> ⟨rest of body of f⟩))
>
Letrec expressions, which have the form
# Generic template for translating:
# (letrec ((var1 exp1)
# (var2 exp2))
# body)
def letrec_template():
# placeholders for mutually recursive bindings
var1 = None
var2 = None
# exp1 and exp2 are defined as normal Python functions;
# they may refer to var1 and var2 (which will be assigned below).
def _exp1(x):
# example behavior: call var2 (which will be bound later)
return var2(x - 1) if x > 0 else "base1"
def _exp2(x):
return var1(x - 1) if x > 0 else "base2"
# complete the letrec by assigning the names to the expressions
var1 = _exp1
var2 = _exp2
# body that uses var1 and var2
return var1(3)
# Concrete example: mutual recursion for even?/odd? from SICP
def is_even(n):
# Simulates:
# (letrec ((even? (lambda (n) (if (= n 0) #t (odd? (- n 1)))))
# (odd? (lambda (n) (if (= n 0) #f (even? (- n 1))))))
# (even? n))
def even(k):
return True if k == 0 else odd(k - 1)
def odd(k):
return False if k == 0 else even(k - 1)
return even(n)
if __name__ == "__main__":
# Example usages
print(letrec_template()) # -> "base1"
print(is_even(88)) # -> True
are a variation on let in which the expressions
$\langle \;exp_{k} \rangle$ that provide the initial values for the
variables $\langle \;var_{k} \rangle$ are evaluated in an environment
that includes all the letrec bindings. This permits recursion in the
bindings, such as the mutual recursion of even? and odd? in the
example above, or the evaluation of 10 factorial with
>>> def fact(n):
... if n == 1:
... return 1
... return n * fact(n - 1)
>>> fact(10)
3628800
> (letrec
> ((fact
> (lambda (n)
> (if (= n 1)
> 1
> (* n (fact (- n 1)))))))
> (fact 10))
>
Implement letrec as a derived expression, by transforming a
letrec expression into a let expression as shown in the text
above or in Exercise 4.18. That is, the letrec variables should
be created with a let and then be assigned their values with
set!.2. Louis Reasoner is confused by all this fuss about internal definitions. The
way he sees it, if you don’t like to use define inside a procedure, you
can just use let. Illustrate what is loose about his reasoning by
drawing an environment diagram that shows the environment in which the
⟨rest of body of f⟩ is evaluated during evaluation of the
expression (f 5), with f defined as in this exercise. Draw an
environment diagram for the same evaluation, but with let in place of
letrec in the definition of f.
Exercise 4.21: Amazingly, Louis’s intuition in
Exercise 4.20 is correct. It is indeed possible to specify recursive
procedures without using letrec (or even define), although the
method for accomplishing this is much more subtle than Louis imagined. The
following expression computes 10 factorial by applying a recursive factorial
procedure:
>>> (lambda n: (lambda fact: fact(fact, n))(lambda ft, k: 1 if k == 1 else k * ft(ft, k - 1)))(10)
3628800
> ((lambda (n)
> ((lambda (fact) (fact fact n))
> (lambda (ft k)
> (if (= k 1)
> 1
> (* k (ft ft (- k 1)))))))
> 10)
>
Check (by evaluating the expression) that this really does compute factorials.
Devise an analogous expression for computing Fibonacci numbers.2. Consider the following procedure, which includes mutually recursive internal
definitions:
(define (f x)
(define (even? n)
(if (= n 0)
true
(odd? (- n 1))))
(define (odd? n)
(if (= n 0)
false
(even? (- n 1))))
(even? x))
Fill in the missing expressions to complete an alternative definition of
f, which uses neither internal definitions nor letrec:
The evaluator implemented above is simple, but it is very inefficient, because
the syntactic analysis of expressions is interleaved with their execution.
Thus if a program is executed many times, its syntax is analyzed many times.
Consider, for example, evaluating (factorial 4) using the following
definition of factorial:
def factorial(n):
if n == 1:
return 1
else:
return factorial(n - 1) * n
(define (factorial n)
(if (= n 1)
1
(* (factorial (- n 1)) n)))
Each time factorial is called, the evaluator must determine that the
body is an if expression and extract the predicate. Only then can it
evaluate the predicate and dispatch on its value. Each time it evaluates the
expression (* (factorial (- n 1)) n), or the subexpressions
(factorial (- n 1)) and (- n 1), the evaluator must perform the
case analysis in eval to determine that the expression is an
application, and must extract its operator and operands. This analysis is
expensive. Performing it repeatedly is wasteful.
We can transform the evaluator to be significantly more efficient by arranging
things so that syntactic analysis is performed only once. We split eval, which takes an expression and an
environment, into two parts. The procedure analyze takes only the
expression. It performs the syntactic analysis and returns a new procedure,
the
execution procedure, that encapsulates the work to be done in
executing the analyzed expression. The execution procedure takes an
environment as its argument and completes the evaluation. This saves work
because analyze will be called only once on an expression, while the
execution procedure may be called many times.
With the separation into analysis and execution, eval now becomes
def eval(exp, env):
return analyze(exp)(env)
(define (eval exp env) ((analyze exp) env))
The result of calling analyze is the execution procedure to be applied
to the environment. The analyze procedure is the same case analysis as
performed by the original eval of 4.1.1, except that the
procedures to which we dispatch perform only analysis, not full evaluation:
Here is the simplest syntactic analysis procedure, which handles
self-evaluating expressions. It returns an execution procedure that ignores
its environment argument and just returns the expression:
For a quoted expression, we can gain a little efficiency by extracting the text
of the quotation only once, in the analysis phase, rather than in the execution
phase.
Analyze-assignment also must defer actually setting the variable until
the execution, when the environment has been supplied. However, the fact that
the assignment-value expression can be analyzed (recursively) during
analysis is a major gain in efficiency, because the assignment-value
expression will now be analyzed only once. The same holds true for
definitions.
Analyzing a lambda expression also achieves a major gain in efficiency:
We analyze the lambda body only once, even though procedures resulting
from evaluation of the lambda may be applied many times.
Analysis of a sequence of expressions (as in a begin or the body of a
lambda expression) is more involved. Each expression in the
sequence is analyzed, yielding an execution procedure. These execution
procedures are combined to produce an execution procedure that takes an
environment as argument and sequentially calls each individual execution
procedure with the environment as argument.
def analyze_sequence(exps):
def sequentially(proc1, proc2):
def seq(env):
proc1(env)
return proc2(env)
return seq
def loop(first_proc, rest_procs):
if not rest_procs:
return first_proc
return loop(sequentially(first_proc, rest_procs[0]), rest_procs[1:])
procs = list(map(analyze, exps))
if not procs:
raise ValueError("Empty sequence: ANALYZE")
return loop(procs[0], procs[1:])
To analyze an application, we analyze the operator and operands and construct
an execution procedure that calls the operator execution procedure (to obtain
the actual procedure to be applied) and the operand execution procedures (to
obtain the actual arguments). We then pass these to
execute-application, which is the analog of apply in
4.1.1. Execute-application differs from apply in that the
procedure body for a compound procedure has already been analyzed, so there is
no need to do further analysis. Instead, we just call the execution procedure
for the body on the extended environment.
Our new evaluator uses the same data structures, syntax procedures, and
run-time support procedures as in 4.1.2, 4.1.3, and
4.1.4.
Exercise 4.22: Extend the evaluator in this
section to support the special form let. (See Exercise 4.6.)
Exercise 4.23: Alyssa P. Hacker doesn’t
understand why analyze-sequence needs to be so complicated. All the
other analysis procedures are straightforward transformations of the
corresponding evaluation procedures (or eval clauses) in
4.1.1. She expected analyze-sequence to look like this:
def analyze_sequence(exps):
# Execute a sequence of procedure-producing expressions in order.
def execute_sequence(procs, env):
if len(procs) == 1:
return procs[0](env)
else:
procs[0](env)
return execute_sequence(procs[1:], env)
procs = [analyze(exp) for exp in exps]
if not procs:
raise Exception("Empty sequence: ANALYZE")
return lambda env: execute_sequence(procs, env)
Eva Lu Ator explains to Alyssa that the version in the text does more of the
work of evaluating a sequence at analysis time. Alyssa’s sequence-execution
procedure, rather than having the calls to the individual execution procedures
built in, loops through the procedures in order to call them: In effect,
although the individual expressions in the sequence have been analyzed, the
sequence itself has not been.
Compare the two versions of analyze-sequence. For example, consider the
common case (typical of procedure bodies) where the sequence has just one
expression. What work will the execution procedure produced by Alyssa’s
program do? What about the execution procedure produced by the program in the
text above? How do the two versions compare for a sequence with two
expressions?
Exercise 4.24: Design and carry out some
experiments to compare the speed of the original metacircular evaluator with
the version in this section. Use your results to estimate the fraction of time
that is spent in analysis versus execution for various procedures.
Adapted from Structure and Interpretation of Computer Programs
by Harold Abelson and Gerald Jay Sussman (MIT Press, 1996).
Original Scheme examples translated to Python.