Racket: Why wont this compile? - racket

I'm attempting to program a simple function that adds integers to a list descending from a range of "high" and "low", incremented by "step"
For example,
if the input is (3 12 3), the expected output is '(12 9 6 3)
Below is the following code:
(define (downSeries step high low [(define ret '())])
(if (< high low)
ret
(cons ret (- high step))
(downSeries (step (- high step) low))))
I'm pretty new to racket, but I'm really not sure why this isn't compiling. Any tips? Thank you.

Since only racket is tagged and no special languages are describes it is expeted the first line in the definition window is #lang racket. Answer will be different for student languages.
1 The last argument is nested in two parentheses and is illegal syntax. Default arguments only have one set:
(define (test mandatory (optional '()))
(list mandatory optional))
(test 1) ; ==> (1 ())
(test 1 2) ; ==> (1 2)
2 You have 4 operands in your if form. It takes maximum 3!
(if prediate-expression
then-expression
else-expression)
Looking at the code you should have the cons expression in the position of ret argument. Having it before the recursion makes it dead code. ret will always be (). Eg this loks similar to a typical fold implementation:
(define (fold-1 combine init lst)
(if (null? lst)
init ; fully grown init returned
(fold-1 combine
(combine (car lst) init) ; init grows
(cdr lst))))

Related

Implementing an infinite list of consecutive integers in Lisp for lazy evaluation

Prelude
In Raku there's a notion called infinite list AKA lazy list which is defined and used like:
my #inf = (1,2,3 ... Inf);
for #inf { say $_;
exit if $_ == 7 }
# => OUTPUT
1
2
3
4
5
6
7
I'd like to implement this sort of thing in Common Lisp, specifically an infinite list of consecutive integers like:
(defun inf (n)
("the implementation"))
such that
(inf 5)
=> (5 6 7 8 9 10 .... infinity)
;; hypothetical output just for the demo purposes. It won't be used in reality
Then I'll use it for lazy evaluation like this:
(defun try () ;; catch and dolist
(catch 'foo ;; are just for demo purposes
(dolist (n (inf 1) 'done)
(format t "~A~%" n)
(when (= n 7)
(throw 'foo x)))))
CL-USER> (try)
1
2
3
4
5
6
7
; Evaluation aborted.
How can I implement such an infinite list in CL in the most practical way?
A good pedagogical approach to this is to define things which are sometimes called 'streams'. The single best introduction to doing this that I know of is in Structure and Interpretation of Computer Programs. Streams are introduced in section 3.5, but don't just read that: read the book, seriously: it is a book everyone interested in programming should read.
SICP uses Scheme, and this sort of thing is more natural in Scheme. But it can be done in CL reasonably easily. What I've written below is rather 'Schemy' CL: in particular I just assume tail calls are optimised. That's not a safe assumption in CL, but it's good enough to see how you can build these concepts into a language which does not already have them, if your language is competent.
First of all we need a construct which supports lazy evaluation: we need to be able to 'delay' something to create a 'promise' which will be evaluated only when it needs to be. Well, what functions do is evaluate their body only when they are asked to, so we'll use them:
(defmacro delay (form)
(let ((stashn (make-symbol "STASH"))
(forcedn (make-symbol "FORCED")))
`(let ((,stashn nil)
(,forcedn nil))
(lambda ()
(if ,forcedn
,stashn
(setf ,forcedn t
,stashn ,form))))))
(defun force (thing)
(funcall thing))
delay is mildly fiddly, it wants to make sure that a promise is forced only once, and it also wants to make sure that the form being delayed doesn't get infected by the state it uses to do that. You can trace the expansion of delay to see what it makes:
(delay (print 1))
-> (let ((#:stash nil) (#:forced nil))
(lambda ()
(if #:forced #:stash (setf #:forced t #:stash (print 1)))))
This is fine.
So now, we'll invent streams: streams are like conses (they are conses!) but their cdrs are delayed:
(defmacro cons-stream (car cdr)
`(cons ,car (delay ,cdr)))
(defun stream-car (s)
(car s))
(defun stream-cdr (s)
(force (cdr s)))
OK, let's write a function to get the nth element of a stream:
(defun stream-nth (n s)
(cond ((null s)
nil)
((= n 0) (stream-car s))
(t
(stream-nth (1- n) (stream-cdr s)))))
And we can test this:
> (stream-nth 2
(cons-stream 0 (cons-stream 1 (cons-stream 2 nil))))
2
And now we can write a function to enumerate an interval in the naturals, which by default will be an half-infinite interval:
(defun stream-enumerate-interval (low &optional (high nil))
(if (and high (> low high))
nil
(cons-stream
low
(stream-enumerate-interval (1+ low) high))))
And now:
> (stream-nth 1000 (stream-enumerate-interval 0))
1000
And so on.
Well, we'd like some kind of macro which lets us traverse a stream: something like dolist, but for streams. Well we can do this by first writing a function which will call a function for each element in the stream (this is not the way I'd do this in production CL code, but it's fine here):
(defun call/stream-elements (f s)
;; Call f on the elements of s, returning NIL
(if (null s)
nil
(progn
(funcall f (stream-car s))
(call/stream-elements f (stream-cdr s)))))
And now
(defmacro do-stream ((e s &optional (r 'nil)) &body forms)
`(progn
(call/stream-elements (lambda (,e)
,#forms)
,s)
,r))
And now, for instance
(defun look-for (v s)
;; look for an element of S which is EQL to V
(do-stream (e s (values nil nil))
(when (eql e v)
(return-from look-for (values e t)))))
And we can then say
> (look-for 100 (stream-enumerate-interval 0))
100
t
Well, there is a lot more mechanism you need to make streams really useful: you need to be able to combine them, append them and so on. SICP has many of these functions, and they're generally easy to turn into CL, but too long here.
For practical purposes it would be wise to use existing libraries, but since the question is about how to implemented lazy lists, we will do it from scratch.
Closures
Lazy iteration is a matter of producing an object that can generate the new value of a lazy sequence each time it is asked to do so.
A simple approach for this is to return a closure, i.e. a function that closes over variables, which produces values while updating its state by side-effect.
If you evaluate:
(let ((a 0))
(lambda () (incf a)))
You obtain a function object that has a local state, namely here the variable named a.
This is a lexical binding to a location that is exclusive to this function, if you evaluate a second time the same expression, you'll obtain a different anonymous function that has its own local state.
When you call the closure, the value stored in a in incremented and its value is returned.
Let's bind this closure to a variable named counter, call it multiple times and store the successive results in a list:
(let ((counter (let ((a 0))
(lambda () (incf a)))))
(list (funcall counter)
(funcall counter)
(funcall counter)
(funcall counter)))
The resulting list is:
(1 2 3 4)
Simple iterator
In your case, you want to have an iterator that starts counting from 5 when writing:
(inf 5)
This can implemented as follows:
(defun inf (n)
(lambda ()
(shiftf n (1+ n))))
Here is there is no need to add a let, the lexical binding of an argument to n is done when calling the function.
We assign n to a different value within the body over time.
More precisely, SHIFTF assigns n to (1+ n), but returns the previous value of n.
For example:
(let ((it (inf 5)))
(list (funcall it)
(funcall it)
(funcall it)
(funcall it)))
Which gives:
(5 6 7 8)
Generic iterator
The standard dolist expects a proper list as an input, there is no way you can put another kind of data and expect it to work (or maybe in an implementation-specific way).
We need a similar macro to iterate over all the values in an arbitrary iterator.
We also need to specify when iteration stops.
There are multiple possibilities here, let's define a basic iteration protocol as follows:
we can call make-iterator on any object, along with arbitrary arguments, to obtain an iterator
we can call next on an iterator to obtain the next value.
More precisely, if there is a value, next returns the value and T as a secondary value; otherwise, next returns NIL.
Let's define two generic functions:
(defgeneric make-iterator (object &key)
(:documentation "create an iterator for OBJECT and arguments ARGS"))
(defgeneric next (iterator)
(:documentation "returns the next value and T as a secondary value, or NIL"))
Using generic functions allows the user to define custom iterators, as long as they respect the specified behaviour above.
Instead of using dolist, which only works with eager sequences, we define our own macro: for.
It hides calls to make-iterator and next from the user.
In other words, for takes an object and iterates over it.
We can skip iteration with (return v) since for is implemented with loop.
(defmacro for ((value object &rest args) &body body)
(let ((it (gensym)) (exists (gensym)))
`(let ((,it (make-iterator ,object ,#args)))
(loop
(multiple-value-bind (,value ,exists) (next ,it)
(unless ,exists
(return))
,#body)))))
We assume any function object can act as an iterator, so we specialize next for values f of class function, so that the function f gets called:
(defmethod next ((f function))
"A closure is an interator"
(funcall f))
Also, we can also specialize make-iterator to make closures their own iterators (I see no other good default behaviour to provide for closures):
(defmethod make-iterator ((function function) &key)
function)
Vector iterator
For example, we can built an iterator for vectors as follows. We specialize make-iterator for values (here named vec) of class vector.
The returned iterator is a closure, so we will be able to call next on it.
The method accepts a :start argument defaulting to zero:
(defmethod make-iterator ((vec vector) &key (start 0))
"Vector iterator"
(let ((index start))
(lambda ()
(when (array-in-bounds-p vec index)
(values (aref vec (shiftf index (1+ index))) t)))))
You can now write:
(for (v "abcdefg" :start 2)
(print v))
And this prints the following characters:
#\c
#\d
#\e
#\f
#\g
List iterator
Likewise, we can build a list iterator.
Here to demonstrate other kind of iterators, let's have a custom cursor type.
(defstruct list-cursor head)
The cursor is an object which keeps a reference to the current cons-cell in the list being visited, or NIL.
(defmethod make-iterator ((list list) &key)
"List iterator"
(make-list-cursor :head list))
And we define next as follows, specializeing on list-cursor:
(defmethod next ((cursor list-cursor))
(when (list-cursor-head cursor)
(values (pop (list-cursor-head cursor)) t)))
Ranges
Common Lisp also allows methods to be specialized with EQL specializers, which means the object we give to for might be a specific keyword, for example :range.
(defmethod make-iterator ((_ (eql :range)) &key (from 0) (to :infinity) (by 1))
(check-type from number)
(check-type to (or number (eql :infinity)))
(check-type by number)
(let ((counter from))
(case to
(:infinity
(lambda () (values (incf counter by) t)))
(t
(lambda ()
(when (< counter to)
(values (incf counter by) T)))))))
A possible call for make-iterator would be:
(make-iterator :range :from 0 :to 10 :by 2)
This also returns a closure.
Here, for example, you would iterate over a range as follows:
(for (v :range :from 0 :to 10 :by 2)
(print v))
The above expands as:
(let ((#:g1463 (make-iterator :range :from 0 :to 10 :by 2)))
(loop
(multiple-value-bind (v #:g1464)
(next #:g1463)
(unless #:g1464 (return))
(print v))))
Finally, if we add small modification to inf (adding secondary value):
(defun inf (n)
(lambda ()
(values (shiftf n (1+ n)) T)))
We can write:
(for (v (inf 5))
(print v)
(when (= v 7)
(return)))
Which prints:
5
6
7
I'll show it with a library:
How to create and consume an infinite list of integers with the GTWIWTG generators library
This library, called "Generators The Way I Want Them Generated", allows to do three things:
create generators (iterators)
combine them
consume them (once).
It is not unsimilar to the nearly-classic Series.
Install the lib with (ql:quickload "gtwiwtg"). I will work in its package: (in-package :gtwiwtg).
Create a generator for an infinite list of integers, start from 0:
GTWIWTG> (range)
#<RANGE-BACKED-GENERATOR! {10042B4D83}>
We can also specify its :from, :to, :by and :inclusive parameters.
Combine this generator with others: not needed here.
Iterate over it and stop:
GTWIWTG> (for x *
(print x)
(when (= x 7)
(return)))
0
1
2
3
4
5
6
7
T
This solution is very practical :)

How deterministic is Racket's evaluation order?

I would like to known how deterministic Racket's evaluation order is when set! is employed. More specifically,
Does #%app always evaluates its arguments from left to right?
If no, can the evaluation of different arguments be intertwined?
Take, for instance, this snippet:
#lang racket
(define a 0)
(define (++a) (set! a (add1 a)) a)
(list (++a) (++a)) ; => ?
Could the last expression evaluate to something different than '(1 2), such as '(1 1), '(2 2) or '(2 1)?
I failed to find a definite answer on http://docs.racket-lang.org/reference.
Unlike Scheme, Racket is guaranteed left to right. So for the example call:
(proc-expr arg-expr ...)
You can read the following in the Guide: (emphasis mine)
A function call is evaluated by first evaluating the proc-expr and all
arg-exprs in order (left to right).
That means that this program:
(define a 0)
(define (++a) (set! a (add1 a)) a)
(list (++a) (++a))
; ==> (1 2)
And it is consistent. For Scheme (2 1) is an alternative solution. You can force order by using bindings and can ensure the same result like this:
(let ((a1 (++ a)))
(list a1 (++ a)))
; ==> (1 2)

function (OccurencesOfPrimes < list >) which counts the number of primes in a (possibly nested) list

I am working on problem to get the occurence of Prime in a list in lisp.
Input:
Write a function (OccurencesOfPrimes < list >) which counts the number of primes in a (possibly nested) list.
Output: Example: (OccurencesOfPrimes (((1)(2))(5)(3)((8)3)) returns 4.
I am using the below code but getting the error like:
(
defun OccurencesOfPrimes (list)
(loop for i from 2 to 100
do ( setq isPrime t)
(loop for j from 2 to i
never (zerop (mod i j))
(setq isPrime f)
(break)
)
)
(if (setq isPrime t)
(append list i)
)
)
)
LOOP: illegal syntax near (SETQ ISPRIME F) in
(LOOP FOR J FROM 2 TO I NEVER (ZEROP (MOD I J)) (SETQ ISPRIME F) (BREAK)
)
Any help.
It is important to keep the format consistent with the expected conventions of the language. It helps when reading the code (in particular with other programmers), and can help you see errors.
Also, you should use an editor which, at the minimum, keep tracks of parentheses. In Emacs, when you put the cursor in the first opening parenthesis, the matching parenthesis is highlighted. You can spot that you have one additional parenthesis that serves no purpose.
(
defun OccurencesOfPrimes (list)
(loop for i from 2 to 100
do ( setq isPrime t)
(loop for j from 2 to i
never (zerop (mod i j))
(setq isPrime f)
(break)
)
)
(if (setq isPrime t)
(append list i)
)
) ;; <- end of defun
) ;; <- closes nothing
In Lisp, parentheses are for the computer, whereas indentation is for humans. Tools can automatically indent the code according to the structure (the parenthesis), and any discrepancy between what indentation you expect and the one being computed is a hint that your code is badly formed. If you look at the indentation of your expressions, you can see how deep you are in the form, and that alone helps you understand the code.
Symbol names are dash-separated, not camlCased.
Your code, with remarks:
(defun occurences-of-primes (list)
;; You argument is likely to be a LIST, given its name and the way
;; you call APPEND below. But you never iterate over the list. This
;; is suspicious.
(loop
for i from 2 to 100
do
(setq is-prime t) ;; setting an undeclared variable
(loop
for j from 2 to i
never (zerop (mod i j))
;; the following two forms are not expected here according
;; to LOOP's grammar; setting IS-PRIME to F, but F is not
;; an existing variable. If you want to set to false, use
;; NIL instead.
(setq is-prime f)
;; BREAK enters the debugger, maybe you wanted to use
;; LOOP-FINISH instead, but the NEVER clause above should
;; already be enough to exit the loop as soon as its
;; sub-expression evaluates to NIL.
(break)))
;; The return value of (SETQ X V) is V, so here your test would
;; always succeed.
(if (setq is-prime t)
;; Append RETURNS a new list, without modifying its
;; arguments. In particular, LIST is not modified. Note that "I"
;; is unknown at this point, because the bindings effective
;; inside the LOOP are not visible in this scope. Besides, "I"
;; is a number, not a list.
(append list i)))
Original question
Write one function which counts all the occurrences of a prime number in a (possibly nested) list.
Even though the homework questions says "write one function", it does not say that you should write one big function that compute everything at once. You could write one such big function, but if you split your problem into sub-problems, you will end with different auxiliary functions, which:
are simpler to understand (they do one thing)
can be reused to build other functions
The sub-problems are, for example: how to determine if a number is a prime? how to iterate over a tree (a.k.a. a possibly nested list)? how to count
the occurrences?
The basic idea is to write an "is-prime" function, iterate over the tree and call "is-prime" on each element; if the element is prime and was never seen before, add 1 to a counter, local to your function.
You can also flatten the input tree, to obtain a list, then sort the resulting
list; you iterate over the list while keeping track of the last
value seen: if the value is the same as the previous one, you
already know if the number is prime; if the previous number differs, then
you have to test if the number is prime first.
You could also abstract things a little more, and define a higher-order tree-walker function, which calls a function on each leaf of the tree. And write another higher-order function which "memoizes" calls: it wraps around a
function F so that if you call F with the same arguments as before,
it returns the result that was stored instead of recomputing it.
Example
I'll combine the above ideas because if you give that answer to a teacher you are likely to have to carefully explain what each part does (and if you can, great for you); this is not necessarily the "best" answer, but it covers a lot of things.
(defun tree-walk-leaves (tree function)
(typecase tree
(null nil)
(cons
(tree-walk-leaves (car tree) function)
(tree-walk-leaves (cdr tree) function))
(t (funcall function tree))))
(defun flatten (tree &optional keep-order-p)
(let ((flat nil))
(tree-walk-leaves tree (lambda (leaf) (push leaf flat)))
(if keep-order-p
(nreverse flat)
flat)))
(defun prime-p (n)
(or (= n 2)
(and (> n 2)
(oddp n)
(loop
for d from 3 upto (isqrt n) by 2
never (zerop (mod n d))))))
(defun count-occurences-of-prime (tree)
(count-if #'prime-p (remove-duplicates (flatten tree))))
(count-occurences-of-prime '(((1)(2))(5)(3)((8)3)))
=> 4
If, instead, you don't want to remove duplicates but count the multiple times a prime number occurs, you can do:
(count-if (memoize #'prime-p) (flatten tree))
... where memoize is:
(defun memoize (function &key (test #'equalp) (key #'identity))
(let ((hash (make-hash-table :test test)))
(lambda (&rest args)
(let ((args (funcall key args)))
(multiple-value-bind (result exists-p) (gethash args hash)
(values-list
(if exists-p
result
(setf (gethash args hash)
(multiple-value-list (apply function args))))))))))
(memoize is useless if there are no duplicates)

Why does function apply complain about long lists?

As part of some Eulerian travails, I'm trying to code a Sieve of Eratosthenes with a factorization wheel. My code so far is:
(defun ring (&rest content)
"Returns a circular list containing the elements in content.
The returned list starts with the first element of content."
(setf (cdr (last content)) content))
(defun factorization-wheel (lst)
"Returns a circular list containing a factorization
wheel using the list of prime numbers in lst"
(let ((circumference (apply #'* lst)))
(loop for i from 1 to circumference
unless (some #'(lambda (x) (zerop (mod i x))) lst)
collect i into wheel
finally (return (apply #'ring
(maplist
#'(lambda (x) ; Takes exception to long lists (?!)
(if (cdr x)
(- (cadr x) (car x))
(- circumference (car x) -1)))
wheel))))))
(defun eratosthenes (n &optional (wheel (ring 4 2)))
"Returns primes up to n calculated using
a Sieve of Eratosthenes and a factorization wheel"
(let* ((candidates (loop with s = 1
for i in wheel
collect (setf s (+ i s))
until (> s n))))
(maplist #'(lambda (x)
(if (> (expt (car x) 2) n)
(return-from eratosthenes candidates))
(delete-if
#'(lambda (y) (zerop (mod y (car x))))
(cdr x)))
candidates)))
I got the following result for wheels longer than 6 elements. I didn't really understand why:
21 > (factorization-wheel '(2 3 5 7 11 13))
(16 2 4 6 2 6 4 2 4 6 6 2 6 4 2 6 4 6 8 4 ...)
21 > (factorization-wheel '(2 3 5 7 11 13 17))
> Error: Too many arguments.
> While executing: FACTORIZATION-WHEEL, in process listener(1).
The algorithm seems to be working OK otherwise and churns out primes with wheels having 6 or fewer elements.
Apparently apply or ring turn up their noses when long lists are passed to them.
But shouldn't the list count as a single argument? I admit I'm thoroughly flummoxed. Any input is appreciated.
ANSI Common Lisp allows implementations to constrain the maximum number of arguments which can be passed to a function. This limit is given by call-arguments-limit can be as small as 50.
For functions which behave like algebraic group operators obeying the
associative property (+, list, and others), we can get around the limit by using reduce to decimate the input list while treating the function as binary.
For instance to add a large list of numbers: (reduce #'+ list) rather than (apply #'+ list).
Notes on reduce
In Common Lisp, reduce will appear to work even if the list is empty. Few other languages give you this, and it doesn't actually come from reduce: it won't work for all functions. But with + we can write (reduce #'+ nil) and it calculates zero, just like (apply #'+ nil).
Why is that? Because the + function can be called with zero arguments, and when called with zero arguments, it yields the identity element for the additive group: 0. This dovetails with the reduce function.
In some other languages the fold or reduce function must be given an initial seed value (like 0), or else a nonempty list. If it is given neither, it is an error.
The Common Lisp reduce, if it is given an empty list and no :initial-value, will call the kernel function with no arguments, and use the return value as the initial value. Since that value is then the only value (the list is empty), that value is returned.
Watch out for functions with special rules for the leftmost argument. For instance:
(apply #'- '(1)) -> -1 ;; same as (- 1), unary minus semantics.
(reduce #'- '(1)) -> 1 ;; what?
What's going on is that when reduce is given a one-element list, it just returns the element without calling the function.
Basically it is founded on the mathematical assumption mentioned above that if no :initial-value is supplied then f is expected to support (f) -> i, where i is some identity element in relation to f, so that (f i x) -> x. This is used as the initial value when reducing the singleton list, (reduce #'f (list x)) -> (f (f) x) -> (f i x) -> x.
The - function doesn't obey these rules. (- a 0) means "subtract zero from a" and so yields a, whereas (- a) is the additive inverse of a, probably for purely pragmatic, notational reasons (namely, not making Lisp programmers write (- 0 a) just to flip a sign, just for the sake of having - behave more consistently under reduce and apply). The - function also may not be called with zero arguments.
If we want to take a list of numbers and subtract them all from some value x, the pattern for that is:
(reduce #'- list-of-numbers :initial-value x)

Is there a straightforward lisp equivalent of Python's generators?

In Python you can write this:
def firstn(n):
num = 0
while num < n:
yield num
num += 1
What is the lisp equivalent of this?
Existing package
Download, install and load the GENERATORS system with Quicklisp. Then, use package :generators (or preferably, define your own package first).
(ql:quickload :generators)
(use-package :generators)
Define an infinite generator for random values:
(defun dice (n)
(make-generator ()
;; repeatedly return a random value between 1 and N
(loop (yield (1+ (random n))))))
Use the generator:
(loop
with dice = (dice 6)
repeat 20
collect (next dice))
=> (1 2 6 1 1 4 4 2 4 3 6 2 1 5 6 5 1 5 1 2)
Note however what the author of the library says:
This library is more of an interesting toy, though as far as I know it
does work. I dont think I have ever used this in application code,
though I think that with care, it could be.
See also
The ITERATE package provides a way to define generators for use inside its iteration facility.
The SERIES package provide stream-like data structures and operations on them.
The Snakes library (same approach as GENERATORS as far as I know).
Iterators in generic-cl
Closures
In practice, CL does not rely that much on generators as popularized by Python. What happens instead is that when people need lazy sequences, they use closures:
(defun dice (n)
(lambda ()
(1+ (random n))))
Then, the equivalent of next is simply a call to the thunk generated by dice:
(loop
with dice = (dice 6)
repeat 20
collect (funcall dice))
This is the approach that is preferred, in particular because there is no need to rely on delimited continuations like with generators. Your example involves a state, which the dice example does not require (there is a hidden state that influences random, but that's another story) . Here is how your counter is typically implemented:
(defun first-n (n)
(let ((counter -1))
(lambda ()
(when (< counter n)
(incf counter)))))
Higher-order functions
Alternatively, you design a generator that accepts a callback function which is called by your generator for each value. Any funcallable can be used, which allows the caller to retain control over code execution:
(defun repeatedly-throw-dice (n callback)
(loop (funcall callback (1+ (random n)))))
Then, you can use it as follows:
(prog ((counter 0) stack)
(repeatedly-throw-dice 6
(lambda (value)
(if (<= (incf counter) 20)
(push value stack)
(return (nreverse stack))))))
See documentation for PROG.
do-traversal idiom
Instead of building a function, data sources that provides a custom way of generating values (like matches of a regular expressions in a string) also regularly provide a macro that abstracts their control-flow. You would use it as follows:
(let ((counter 0) stack)
(do-repeatedly-throw-dice (value 6)
(if (<= (incf counter) 20)
(push value stack)
(return (nreverse stack))))))
DO-X macros are expected to define a NIL block around their body, which is why the return above is valid.
A possible implementation for the macro is to wrap the body in a lambda form and use the callback-based version defined above:
(defmacro do-repeatedly-throw-dice ((var n) &body body)
`(block nil (repeatedly-throw-dice ,n (lambda (,var) ,#body))))
Directly expanding into a loop would be possible too:
(defmacro do-repeatedly-throw-dice ((var n) &body body)
(let ((max (gensym)) (label (make-symbol "NEXT")))
`(prog ((,max ,n) ,var)
,label
(setf ,var (1+ (random ,max)))
(progn ,#body)
(go ,label))))
One step of macroexpansion for above form:
(prog ((#:g1078 6) value)
#:next
(setf value (1+ (random #:g1078)))
(progn
(if (<= (incf counter) 20)
(push value stack)
(return (nreverse stack))))
(go #:next))
Bindings
Broadly speaking, building a generator with higher-order functions or directly with a do- macro gives the same result. You can implement one with the other (personally, I prefer to define first the macro and then the function using the macro, but doing the opposite is also interesting, since you can redefine the function without recompiling all usages of the macro).
However, there is still a difference: the macro reuses the same variable across iterations, whereas the closure introduces a fresh binding each time. For example:
(let ((list))
(dotimes (i 10) (push (lambda () i) list))
(mapcar #'funcall list))
.... returns:
(10 10 10 10 10 10 10 10 10 10)
Most (if not all) iterators in Common Lisp tend to work like this1, and it should not come as a surprise for experienced users (the opposite would be surprising, in fact). If dotimes was implemented by repeatedly calling a closure, the result would be different:
(defmacro my-dotimes ((var count-form &optional result-form) &body body)
`(block nil
(alexandria:map-iota (lambda (,var) ,#body) ,count-form)
,result-form))
With the above definition, we can see that:
(let ((list))
(my-dotimes (i 10) (push (lambda () i) list))
(mapcar #'funcall list))
... returns:
(9 8 7 6 5 4 3 2 1 0)
In order to have the same result with the standard dotimes, you only need to create a fresh binding before building the closure:
(let ((list))
(dotimes (i 10)
(let ((j i))
(push (lambda () j) list))))
Here j is a fresh binding whose value is the current value of i at closure creation time; j is never mutated so the closure will constantly return the same value.
If you wanted to, you could always introduce that inner let from the macro, but this is rarely done.
1: Note that the specification for DOTIMES does not require that bindings are fresh at each iteration, or only mutates the same binding at each step: "It is implementation-dependent whether dotimes establishes a new binding of var on each iteration or whether it establishes a binding for var once at the beginning and then assigns it on any subsequent iterations." In order to write portably, it is necessary to assume the worst-case scenario (i.e. mutation, which happens to be what most (all?) implementations do) and manually rebind iteration variables if they are to be captured and reused at a later point.