letrec and “named let”, two variants of
let that permit 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 and
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)))
largest: undefined; cannot reference an identifier before its definition
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 identifier 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.
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.
letrecWhat 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 ([largest (lambda (lst)
(if (null? (cdr lst))
(car lst)
(max (car lst) (largest (cdr lst)))))])
(largest (list 2 1 4 -5 6 13 3)))
13
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)))
;;; 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.
;;; Produces:
;;; index, an integer (or #f
(define list-index-kernel
(lambda (rest sought bypassed)
(cond
[(null? rest)
#f]
[(equal? (car rest) sought)
bypassed]
[else
(list-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 sought 0))))
Of course, now that the recursive kernel procedure is inside the body
of the list-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.
(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 one way to introduce
precondition tests into a recursively defined version of a version of
the furthest-from-zero procedure from the reading on patterns of list
recursion.
;;; Procedure:
;;; furthest-from-zero
;;; Parameters:
;;; numbers, a nonempty list of real numbers
;;; Purpose:
;;; Find the number in the list with the largest absolute value.
;;; Produces:
;;; furthest, a real number
;;; Preconditions:
;;; [No additional]
;;; Postconditions:
;;; furthest is an element of numbers
;;; For any number, n, in numbers
;;; (>= (abs furthest) (abs n))
(define furthest-from-zero
(lambda (numbers)
(letrec ([kernel (lambda (furthest-so-far remaining)
(cond
[(null? remaining)
furthest-so-far]
[(>= (abs furthest-so-far) (abs (car remaining)))
(kernel furthest-so-far (cdr remaining))]
[else
(kernel (car remaining) (cdr remaining))]))])
(cond
[(null? numbers)
(error "furthest-from-zero: Argument must be a non-empty list of real numbers, given the empty list")]
[(not (list? numbers))
(error "furthest-from-zero: Argument must be a non-empty list of real numbers, given a non-list" numbers)]
[(not (all-real? numbers))
(error "furthest-from-zero: Argument must be a non-empty list of real numbers, given a list containing other values" numbers)]
[else
(kernel (car numbers) (cdr numbers))]))))
Embedding the kernel inside the definition of furthest-from rather
than writing a separate furthest-from-zero-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 list-index and furthest-from-zero have a
helper named kernel, we can safely write
(index-of (furthest-from-zero values) values)
to find the index of the value in values that is furthest from zero.
Note that furthest-from effectively has three local procedures
that it relies upon. Not only does it use kernel (formerly called
furthest-from-kernel), but also all-real?. We might even write
a local helper to choose the further from zero of two values.
While we have used letrec to define a local kernel, we can also use it
to define other local recursive procedures, such as the all-real?
predicate. As we’ve seen in the past, we can also use let or let*
to define non-recursive procedures. Hence, if we want to define all
three helpers within furthest-from, we might write something like
the following.
(define furthest-from-zero
(letrec (; finds which of two numbers is further from zero
[further
(lambda (val1 val2)
(if (> (abs val1) (abs val2))
val1
val2))]
; Determines if its list parameter contains only real #s.
[all-real?
(lambda (vals)
(or (null? vals)
(and (real? (car vals))
(all-real? (cdr vals)))))])
(lambda (numbers)
(letrec ([kernel (lambda (furthest-so-far remaining)
(if (null? remaining)
furthest-so-far
(kernel (further furthest-so-far (car remaining))
(cdr remaining))))])
(cond
[(null? numbers)
(error "furthest-from-zero: Argument must be a non-empty list of real numbers, given the empty list")]
[(not (list? numbers))
(error "furthest-from-zero: Argument must be a non-empty list of real numbers, given a non-list" numbers)]
[(not (all-real? numbers))
(error "furthest-from-zero: Argument must be a non-empty list of real numbers, given a list containing other values" numbers)]
[else
(kernel (car numbers) (cdr numbers))])))))
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)
What does this do? In essence, it defines a procedure named name with
parameters param1 through paramn. Then, it calls that procedure on
arguments val1 through valn. You can think of this as a more elegant
(and, eventually, more readable) shorthand for
(letrec ([name (lambda (param1 ... paramn)
body)])
(name val1 ... valn))
Alternately, you can think of this as first binding each of param1
through paramn to the corresponding value. It also creates a procedure
with the same parameters. It then executes the body. When the body
calls the procedure name, those bindings get updated, and the body
starts again.
So, for example, one might write the list-index procedure as
(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 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 ... 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) ... 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 (numbers)
(let kernel ([sum-so-far 0]
[remaining numbers])
(if (null? remaining)
sum-so-far
(kernel (+ sum-so-far (car remaining))
(cdr remaining))))))
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 prefer it. We’ll admit that we first found it murky, but eventually came to prefer it. We hope that you will, too. (That is, we hope that you will come to prefer it, not that we hope you find it murky.)
letrec?This question is review: Explain why you cannot name a recursive procedure
locally with let.
differenceYou now have seen a version of list-index that uses a letrec to define
its helper locally. Rewrite newer-difference so that it similarly uses a
letrec to define its helper locally.
(define newer-difference
(lambda (lst)
(new-difference-helper (car lst) (cdr lst))))
(define new-difference-helper
(lambda (difference-so-far remaining)
(if (null? remaining)
difference-so-far
(new-difference-helper (- difference-so-far (car remaining))
(cdr remaining)))))
let us name a local differenceThis one might be slightly more difficult, but the best way to engage
the topics is to try them out. You have seen a version of sum that
uses a named let to define its helper locally:
(define new-sum
(lambda (numbers)
(new-sum-helper 0 numbers)))
(define new-sum-helper
(lambda (sum-so-far remaining)
(if (null? remaining)
sum-so-far
(new-sum-helper (+ sum-so-far (car remaining))
(cdr remaining)))))
(define sum
(lambda (numbers)
(let kernel ([sum-so-far 0]
[remaining numbers])
(if (null? remaining)
sum-so-far
(kernel (+ sum-so-far (car remaining))
(cdr remaining))))))
Rewrite newer-difference one more time to similarly use a named let
to define its helper locally.