Fundamentals of Computer Science I: Media Computing (CS151.01 2008S)
letrec, a variant of
let that permits us to write recursive local procedures.
As you have probably noted, we often find it useful to write helper procedures that accompany our main procedures. For example, we might use a helper to recurse on part of a larger structure or to act as the kernel of a husk-and-kernel procedure.
One issue with this technique is that it often makes little sense for
other procedures to use the helper, so we should restrict access to the
helper procedure. In particular, only the procedure that uses the helper
(unless it's a very generic helper) should be able to access the helper.
We know how to restrict access to variables (using
let*). Can we do the same for procedures?
Yes. As the reading on local bindings suggested, it is possible for a let expression to bind an identifier to a non-recursive procedure:
(let ((square (lambda (n) (* n n)))) (square 12))144
Like any other binding that is introduced in a let expression, this binding is local. Within the body of the let expression, it supersedes any previous binding of the same identifier, but as soon as the value of the let expression has been computed, the local binding evaporates.
However, it is not possible to bind an identifier to a recursively defined procedure in this way. For example, consider the following expression which is intended to find the largest value in a list
(let ((largest (lambda (lst) (if (null? (cdr lst)) (car lst) (max (car lst) (largest (cdr lst))))))) (largest (list 2 1 4 -5 6 13 3)))Error: eval: unbound variable: largest
The difficulty is that when the lambda expression is
evaluated, the identifier
largest has not yet been bound,
so the value of the lambda expression is a procedure that
includes an unbound identifier. Binding this procedure value to the
largest creates a new environment, but does not
affect the behavior of procedures that were constructed in the old
environment. So, when the body of the let expression invokes
this procedure, we get the unbound-identifier error.
wouldn't help in this case, since even under
lambda expression would be completely evaluated before the binding
What we need is some variant of let that binds the
identifier to some kind of a place-holder and adds the binding
to the environment first, then computes the value of the
lambda expression in the new environment, and then finally
substitutes that value for the place-holder. This will work in Scheme,
so long as the procedure is not actually invoked until we get into the
body of the expression.
Fortunately, the designers of Scheme decided to let us write local
procedures using that technique. The keyword associated with this
“recursive binding” variant of
(letrec ((largest (lambda (lst) (if (null? (cdr lst)) (car lst) (max (car lst) (largest (cdr lst))))))) (largest (list 2 1 4 -5 6 13 3)))
A letrec expression constructs all of its place-holder bindings simultaneously (in effect), then evaluates all of the lambda expressions simultaneously, and finally replaces all of the place-holders simultaneously. This makes it possible to include binding specifications for mutually recursive procedures (which invoke each other) in the same binding list. Here's a particularly silly example, which takes a list of numbers and alternately adds and subtracts them.
(letrec ((up-sum (lambda (ls) (if (null? ls) 0 (+ (car ls) (down-sum (cdr ls)))))) (down-sum (lambda (ls) (if (null? ls) 0 (+ (- (car ls)) (up-sum (cdr ls))))))) (up-sum (list 1 23 6 12 7)))
-21; which is 1 - 23 + 6 - 12 + 7.
We can use letrec expressions to separate the husk and the kernel of a recursive procedure without having to define two procedures.
;;; Procedure: ;;; list-index ;;; Parameters: ;;; stuff, a list. ;;; sought, a value. ;;; Purpose: ;;; Find the position of a given value in a given list. ;;; Produces: ;;; pos, a Scheme value ;;; Preconditions: ;;; [No additional.] ;;; Postconditions: ;;; If sought is not in stuff, pos is #f. ;;; If sought is an element of stuff, then ;;; pos is a nonnegative integer. ;;; (list-ref stuff pos) is equal to sought. (define list-index (lambda (stuff sought) (list-index-kernel stuff sought 0))) i;;; Kernel: ;;; list-index-kernel ;;; Parameters: ;;; rest, a list (a sublist of stuff, described above) ;;; sought, a value ;;; bypassed, an integer that counts how many values have ;;; been bypassed in stuff (that is, how many times we've ;;; called cdr since the outside call. ;;; Purpose: ;;; To keep looking for sought in part of the list. (define list-index-kernel (lambda (rest sought bypassed) (cond ((null? rest) #f) ((equal? (car rest) sought) bypassed) (else (index-kernel (cdr rest) sought (+ bypassed 1))))))
This technique works, but it's more stylish to construct the kernel procedure inside a letrec expression, so that the extra identifier can be bound to it locally:
(define list-index (lambda (stuff sought) (letrec ((kernel (lambda (rest sought bypassed) (cond ((null? rest) #f) ((equal? (car rest) sought) bypassed) (else (kernel (cdr rest) sought (+ bypassed 1))))))) (kernel stuff 0))))
Of course, now that the recursive kernel procedure is now inside the
body of the
index procedure, it is not necessary
to pass the value of
sought to the kernel as a
parameter. Instead, the kernel can treat
as if it were a constant, since its value doesn't change during any
of the recursive calls.
(define list-index (lambda (stuff sought) (letrec ((kernel (lambda (rest bypassed) (cond ((null? rest) #f) ((equal? (car rest) sought) bypassed) (else (kernel (cdr rest) (+ bypassed 1))))))) (kernel stuff 0))))
The same approach can be used to perform precondition tests efficiently,
by placing them with the husk in the body of a letrec expression and
omitting them from the kernel. For instance, here's how to introduce
precondition tests into the
procedure from the
reading on verifying preconditions:
(define spots-leftmost (let ((spot-leftmost (lambda (spot1 spot2) (if (< (spot-col spot1) (spot-col spot2)) spot1 spot2)))) (lambda (spots) (letrec ( ; all-spots? checks to see if all the values in a list are spots (all-spots? (lambda (ls) (or (null? ls) (and (spot? (car ls)) (all-spots? (cdr ls)))))) ; kernel finds the leftmost spot w/o verifying preconditions (kernel (lambda (spots) (if (null? (cdr spots)) (car spots) (spot-leftmost (car spots) (spots-leftmost (cdr spots))))))) (cond ((null? spots) (throw "spots-leftmost: Argument must be a *non-empty* list of spots")) ((not (list? spots)) (throw "spots-leftmost: Argument must be a *list* of spots; received" spots)) ((not (all-spots? spots)) (throw "spots-leftmost: Argument list must contain only spots ; received" spots)) (else (kernel spots)))))))
Embedding the kernel inside the definition of
spots-leftmost rather than writing a separate
spots-leftmost-kernel procedure has another
advantage: It is impossible for an incautious user to invoke
kernel procedure directly, bypassing the
precondition tests. The only way to get at the
recursive procedure to which
kernel is bound is
to invoke the procedure within which the binding is established.
We've recycled the name
kernel in this example to drive
home the point that local bindings in separate procedures don't
interfere with one another. Even if both procedures were active at the
same time, the correct
kernel procedure would be used in
each case because the correct local binding would supersede all others.
Hence, even though both
spots-leftmost have a helper named
kernel, we can safely write
(list-index figure (spots-leftmost figure))
to find the index of the leftmost spot in
Many programmers use letrec expressions in writing most of these husk-and-kernel procedures. When there is only one recursive procedure to bind, however, a contemporary Scheme programmer might well use yet another variation of the let expression -- the “named let”.
The named let has the same syntax as a regular let expression, except that there is an identifier between the keyword let and the binding list. The named let binds this extra identifier to a kernel procedure whose parameters are the same as the variables in the binding list and whose body is the same as the body of the let expression. Here's the basic form
(let name ((param1 val1) (param2 val2) ... (paramn valn)) body)
You can think of this as a more elegant (and, eventually, more readable) shorthand for
(letrec ((name (lambda (param1 ... paramn) body))) (name val1 ... valn))
So, for example, one might write the
(define list-index (lambda (sought ls) (let kernel ((rest ls) (bypassed 0)) (cond ((null? rest) #f) ((equal? (car rest) sought) bypassed) (else (kernel (cdr rest) (+ bypassed 1)))))))
When we enter the named let, the identifier
is bound to the value of
ls and the identifier
bypassed is bound to 0, just as if we were
entering an ordinary let expression. In addition, however, the
kernel is bound to a procedure that
as parameters and the body of the named let as its body. As we
evaluate the cond expression, we may encounter a recursive call
kernel procedure -- in effect, we
re-enter the body of the named let, with
now re-bound to the former value of
(cdr rest) and
bypassed to the former value of
As another example, here's a version of
that uses a named let:
(define sum (lambda (ls) (let kernel ((rest ls) (running-total 0)) (if (null? rest) running-total (kernel (cdr rest) (+ (car rest) running-total))))))
Scheme programmers seem to be mixed in their reaction to the named let. Some find it clear and elegant, others find it murky and too special-purpose. Our colleagues like to use it. We'll admit that we first found it murky, but eventually came to like it. We hope that you will, too.
Copyright (c) 2007-8 Janet Davis, Matthew Kluber, and Samuel A. Rebelsky. (Selected materials copyright by John David Stone and Henry Walker and used by permission.)
This material is based upon work partially supported by the National Science Foundation under Grant No. CCLI-0633090. Any opinions, findings, and conclusions or recommendations expressed in this material are those of the author(s) and do not necessarily reflect the views of the National Science Foundation.
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