Summary: We consider letrec, a variant of
let that permits us to write recursive local procedures.
Contents:
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 problem with this technique is that 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
and let*). Can we do the same for procedures?
Yes. 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 recursively create the list of values from 10 to 1.
> (let ((count-down (lambda (n)
(if (zero? n)
null
(cons n (count-down (- n 1)))))))
(count-down 10))
reference to undefined identifier: count-down
The difficulty is that when the lambda-expression is
evaluated, the identifier count-down 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
identifier count-down 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.
Changing let to let* wouldn't help in this case,
since even under let* the lambda-expression would
be completely evaluated before the binding is established.
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 let is named
letrec:
> (letrec ((count-down (lambda (n)
(if (zero? n)
null
(cons n (count-down (- n 1)))))))
(count-down 10))
(10 9 8 7 6 5 4 3 2 1)
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
(+ (down-sum (cdr ls)) (car ls)))))
(down-sum
(lambda (ls)
(if (null? ls)
0
(- (up-sum (cdr ls)) (car 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:
;;; index
;;; Parameters:
;;; sought, a value.
;;; stuff, a list.
;;; Purpose:
;;; Find the position of a given value in a given list.
;;; Produces:
;;; pos
;;; Preconditions:
;;; None.
;;; Postconditions:
;;; pos is either (1) #f, if sought is not in stuff or (2) a
;;; nonnegative integer, if sought is in stuff.
;;; Affects neither sought nor stuff.
;;; If pos is #f, then sought is not in stuff.
;;; If pos is a nonnegative integer, then (list-ref stuff pos)
;;; is equal to sought.
(define index
(lambda (sought stuff)
(index-kernel sought stuff 0)))
;;; Kernel:
;;; index-kernel
;;; Parameters:
;;; sought, as above
;;; rest, a sublist of stuff
;;; bypassed, an integer that counts how many values have
;;; been bypassed in stuff
;;; Purpose:
;;; To keep looking for sought in part of the list.
(define index-kernel
(lambda (sought rest bypassed)
(cond ((null? rest) #f)
((equal? (car rest) sought) bypassed)
(else (index-kernel sought (cdr rest) (+ bypassed 1))))))
This 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 index
(lambda (sought stuff)
(letrec ((kernel (lambda (rest bypassed)
(cond ((null? rest) #f)
((equal? (car rest) sought) bypassed)
(else (kernel (cdr rest) (+ bypassed 1)))))))
(kernel stuff 0))))
Notice, too, that since the recursive kernel procedure is now entirely
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 sought as if it were a constant,
since its value doesn't change during any of the recursive calls.
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 greatest-of-list procedure from
the reading on preconditions
and postconditions:
(define greatest-of-list
(lambda (ls)
(letrec (
; all-real? checks if all the values in a list are real.
(all-real? (lambda (ls)
(or (null? ls)
(and (real? (car ls))
(all-real? (cdr ls))))))
; kernel finds the greatest in a list w/o verifying preconditions.
(kernel (lambda (rest)
(if (null? (cdr rest))
(car rest)
(max (car rest) (kernel (cdr rest)))))))
; Check all the preconditions and then do the work.
(cond ((not (list? ls))
(error "greatest-of-list"
"Argument must a list."))
((null? ls)
(error "greatest-of-list"
"Argument must be a non-empty list."))
((not (all-real? ls))
(error "greatest-of-list"
"Argument list must contain only numbers."))
(else (kernel ls))))))
Embedding the kernel inside the definition of greatest-of-list
rather than writing a separate greatest-of-list-kernel
procedure has another advantage: It is impossible for an incautious user to
invoke the 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 index and greatest-of-list
have a helper named kernel, we can safely write.
(index (greatest-of-list (list 18 6 14 7 2))
(list 18 6 14 7 2))
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 index procedure as follows:
(define 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 rest
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 identifier
kernel is bound to a procedure that has rest and
bypassed as parameters and the body of the named
let as its body. As we evaluate the
cond-expression, we may encounter a recursive call to the
kernel procedure -- in effect, we re-enter the body of the
named let, with rest now re-bound to the former
value of (cdr rest) and bypassed to the former
value of (+ bypassed 1).
As another example, here's a version of sum
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. My colleagues like to use it. I'll admit that
I first found it murky, but eventually came to like it. I hope that you
will, too.
;
Check with Bobby