Summary: We extend our understanding of the algorithm and structures that the Scheme interpreter users to evaluate Scheme expressions. In particular, we see how the introduction of user-defined procedures affects the operation of the interpreter.
As you may recall, we left our original exploration of the way Scheme evaluates with a high-level overview of an algorithm the interpreter might use and a data structure that it relies on. The data structure is a dictionary that maps names to values. It keeps track of the values we've defined. The algorithm was something like the following:
Look at the next non-space character If the next non-space character is an open parenthesis Look at the next thing after the open parenthesis. If the next thing is the keyword define Read the next thing (a name to define) Read the next thing (an expression) Evaluate the expression, giving a value Add the [name:value] entry to the dictionary Otherwise the next thing must be a function Read and evaluate each argument Apply the function to the evaluated arguments Otherwise, see if the next non-space character is a digit (or + or -) Read all the parts of a number The number is the result Otherwise, see if the next non-space character is a letter Read everything up to the next space (or close paren). Look up the thing just read in the dictionary. If it's not there, crash and burn Otherwise, the result is the value found in the dictionary Otherwise, we've encountered an unknown kind of value
We've just introduced user-defined procedures, which have the following form.
(define procedure-name (lambda (parameter-list) expression-1 ... expression-n))
What effects does this introduction have? Surprisingly many. We need to consider what happens in a procedure definition and what happens when we apply a procedure.
Suppose we have the following procedure definition.
(define square (lambda (x) (* x x)))
What does the interpreter do with this? Well, we know that when
it sees an open paren, it looks at the next thing. If the next
thing is the keyword
define, it assumes it has a
definition of the form
expression), evaluates the expression and then
associates it with name. So, how do we evaluate
the “expression”? Here's an interesting trick: In
Scheme, you don't evaluate lambda expressions;
you leave them as is.
Hence, the effect of the previous definition is to extend the dictionary as follows
|square||(lambda (x) (* x x))|
To accommodate this approach, we need to change the interpretation
algorithm a bit. In particular, if the first thing after an open
parenthesis is the keyword
lambda, we simply return
the whole thing.
If the next non-space character is an open parenthesis Look at the next thing. If the next thing is the keyword define ... Otherwise, if the next thing is the keyword lambda Read until the corresponding close parenthesis. The whole expression is the result. Otherwise, the thing after the open paren is a function ...
But if the lambda is never evaluated, what good is it? It turns out the lambda is evaluated, just not until you use the defined procedure.
So, what happens when we apply a user defined procedure, which
we've just learned is stored as an unevaluated lambda expression?
Once again, we evaluate all of the arguments. It's then time to apply
the procedure. In effect, Scheme begins by updating the dictionary so
that each formal parameter (the things that start the lambda expression)
gets associated with the corresponding argument. That is, in applying
Scheme updates the dictionary to associate
x. Similarly, in applying
(lambda (first second) (- (max first second) (min first
the interpreter associates the name
first with the
7 and the name
second with the value
Once Scheme updates the dictionary, it evaluates the expressions in the body of the procedure, using the updated dictionary. When it reaches the end of the body, it uses the value of the last expression as the value of the procedure application.
Is the interpreter done once it's computed the value of the last expression? No. It also needs to clean up the dictionary by removing any definitions that it added at the beginning.
1. Look at the next non-space character 2. If the next non-space character is an open parenthesis 2.1 Look at the next thing after the open paren. 2.2 If the next thing is the keyword define 2.2.1. Read the next thing after the keyword (a name to define) 2.2.2. Read the next thing after the name (an expression) 2.2.3. Evaluate the expression, giving a value 2.2.4. Add the [name:value] entry to the dictionary 2.3. Otherwise, if the next thing is the keyword lambda 2.3.1. Read until the corresponding close parenthesis. 2.3.2. The whole expression is the result. 2.4. Otherwise the next thing must be a function 2.4.1. Look up the function in the dictionary 2.4.2. Read and evaluate each argument 2.4.3. If the function is a built-in function, apply it directly 2.4.4. Otherwise, the function is a user-defined function of the form (lambda (param1 ... paramm) body1 ... bodyn) 188.8.131.52. Make sure that the number of arguments equals the number of parameters (m). If not, report an error. 184.108.40.206. For each i, from 1 to m, 220.127.116.11.1 Update the dictionary to associate parameter i with argument i, 18.104.22.168. Evaluate expressions body1 ... bodyn. 22.214.171.124. Undo the updates made to the dictionary. 126.96.36.199. The value is the value of the bodyn. 3. Otherwise, see if the next non-space character is a digit (or + or -) 3.1. Read all the parts of a number 3.2. The number is the result 4. Otherwise, see if the next non-space character is a double quotation mark ("). 4.1. Read until the corresponding double quotation mark is found. 4.2. The result is a string. 5. Otherwise, see if the next non-space character is a letter 5.1. Read everything up to the next space (or close paren). 5.2. Look up the thing just read in the dictionary. 5.3. If it's not there, crash and burn 5.4. Otherwise, the result is the value found in the dictionary 6. Otherwise, we've encountered an unknown kind of value
Let's consider an example. Suppose we have three things to evaluate: A
definition of the procedure
square, a call to that procedure
on the value 5, and a call to that procedure on the sum of 3 and 4.
(define square (lambda (x (* x x))))
We'll begin with the
define. What happens?
(lambda (x) (* x x)).
(lambda (x) (* x x)).
(lambda (x) (* x x))] to the dictionary.
As things stand, the dictionary now contains only one entry.
Now, we're ready to evaluate the expression
(lambda (x) (* x x))>
(lambda (x) (* x x))
(* x x).
(* x x).
xin the dictionary and find 5.
xin the dictionary and find 5.
That's a lot of steps to compute the 25. Fortunately, computers are fast!
In the discussion of applying lambda expressions to values, we suggested that the Scheme interpreter updates the dictionary and then evaluates the body of the expression. However, Scheme does a bit more trickery to limit access to the parameters of a procedure. Hence, the claim that the interpreter simply updates the dictionary is a simplification, albeit a useful one. For most of the code you write in this class (and elsewhere), the simplification is acceptable. However, there are some cases in which the simplification breaks down. We'll mention them when they occur.
For those who like esoteric terminology, the model we've given you is dynamically scoped. However, the Scheme language is actually statically scoped. For those of you who don't like esoteric terminology, ignore the preceding two sentences. (We thought about telling you to ignore the whole paragraph, but that means you would have to ignore the instruction telling you to ignore the paragraph, which could lead to some confusion.)
Copyright (c) 2007-9 Janet Davis, Matthew Kluber, Samuel A. Rebelsky, and Jerod Weinman. (Selected materials copyright by John David Stone and Henry Walker and used by permission.)
This material is based upon work partially supported by the National Science Foundation under Grant No. CCLI-0633090. Any opinions, findings, and conclusions or recommendations expressed in this material are those of the author(s) and do not necessarily reflect the views of the National Science Foundation.
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