Is it possible to define a procedure f such that it prints Exiting... if it is the last thing to do before exiting, and prints Not done yet... otherwise?
For examples,
(display "hello, world\n")
(f)
should give
hello, world
Exiting...
While
(f)
(display "bye, world\n")
should give
Not done yet...
bye, world
I have thought about using control operators such as shift / reset, but with no success. The key difficulty seems to be that there is no way to tell if the current continuation is terminating. Any idea?
A continuation is never empty. What happens after the end is implementation specific but there is always some sort of resource deallocation and shutdown.
So imagine you have the following code which I had high hopes for:
(call/cc (lambda (end)
(define (f)
(call/cc (lambda (c)
(if (eq? c end)
(display "bye, world\n")
(display "Not done yet...")))))
(f)
(display "hello, world\n")
(f)))
Now you are not guaranteed that the continuation c and end can be compared even if they ae the same continuation. This has to do with the language details that upto R6RS there were no way to compare two procedures and that we aren't really comparing procedures so the implementation might have open coded their halt continuation such that it gets wrapped in a lambda and thus you are really comparing (eq? (lambda (v) (halt)) (lambda (v) (halt))) and it is not guaranteed to be #t or #f.
Related
I wonder how do you, experienced lispers / functional programmers usually make decision what to use. Compare:
(define (my-map1 f lst)
(reverse
(let loop ([lst lst] [acc '()])
(if (empty? lst)
acc
(loop (cdr lst) (cons (f (car lst)) acc))))))
and
(define (my-map2 f lst)
(if (empty? lst)
'()
(cons (f (car lst)) (my-map2 f (cdr lst)))))
The problem can be described in the following way: whenever we have to traverse a list, should we collect results in accumulator, which preserves tail recursion, but requires list reversion in the end? Or should we use unoptimized recursion, but then we don't have to reverse anything?
It seems to me the first solution is always better. Indeed, there's additional complexity (O(n)) there. However, it uses much less memory, let alone calling a function isn't done instantly.
Yet I've seen different examples where the second approach was used. Either I'm missing something or these examples were only educational. Are there situations where unoptimized recursion is better?
When possible, I use higher-order functions like map which build a list under the hood. In Common Lisp I also tend to use loop a lot, which has a collect keyword for building list in a forward way (I also use the series library which also implements it transparently).
I sometimes use recursive functions that are not tail-recursive because they better express what I want and because the size of the list is going to be relatively small; in particular, when writing a macro, the code being manipulated is not usually very large.
For more complex problems I don't collect into lists, I generally accept a callback function that is being called for each solution. This ensures that the work is more clearly separated between how the data is produced and how it is used.
This approach is to me the most flexible of all, because no assumption is made about how the data should be processed or collected. But it also means that the callback function is likely to perform side-effects or non-local returns (see example below). I don't think it is particularly a problem as long the the scope of the side-effects is small (local to a function).
For example, if I want to have a function that generates all natural numbers between 0 and N-1, I write:
(defun range (n f)
(dotimes (i n)
(funcall f i)))
The implementation here iterates over all values from 0 below N and calls F with the value I.
If I wanted to collect them in a list, I'd write:
(defun range-list (N)
(let ((list nil))
(range N (lambda (v) (push v list)))
(nreverse list)))
But, I can also avoid the whole push/nreverse idiom by using a queue. A queue in Lisp can be implemented as a pair (first . last) that keeps track of the first and last cons cells of the underlying linked-list collection. This allows to append elements in constant time to the end, because there is no need to iterate over the list (see Implementing queues in Lisp by P. Norvig, 1991).
(defun queue ()
(let ((list (list nil)))
(cons list list)))
(defun qpush (queue element)
(setf (cdr queue)
(setf (cddr queue)
(list element))))
(defun qlist (queue)
(cdar queue))
And so, the alternative version of the function would be:
(defun range-list (n)
(let ((q (queue)))
(range N (lambda (v) (qpush q v)))
(qlist q)))
The generator/callback approach is also useful when you don't want to build all the elements; it is a bit like the lazy model of evaluation (e.g. Haskell) where you only use the items you need.
Imagine you want to use range to find the first empty slot in a vector, you could do this:
(defun empty-index (vector)
(block nil
(range (length vector)
(lambda (d)
(when (null (aref vector d))
(return d))))))
Here, the block of lexical name nil allows the anonymous function to call return to exit the block with a return value.
In other languages, the same behaviour is often reversed inside-out: we use iterator objects with a cursor and next operations. I tend to think it is simpler to write the iteration plainly and call a callback function, but this would be another interesting approach too.
Tail recursion with accumulator
Traverses the list twice
Constructs two lists
Constant stack space
Can crash with malloc errors
Naive recursion
Traverses list twice (once building up the stack, once tearing down the stack).
Constructs one list
Linear stack space
Can crash with stack overflow (unlikely in racket), or malloc errors
It seems to me the first solution is always better
Allocations are generally more time-expensive than extra stack frames, so I think the latter one will be faster (you'll have to benchmark it to know for sure though).
Are there situations where unoptimized recursion is better?
Yes, if you are creating a lazily evaluated structure, in haskell, you need the cons-cell as the evaluation boundary, and you can't lazily evaluate a tail recursive call.
Benchmarking is the only way to know for sure, racket has deep stack frames, so you should be able to get away with both versions.
The stdlib version is quite horrific, which shows that you can usually squeeze out some performance if you're willing to sacrifice readability.
Given two implementations of the same function, with the same O notation, I will choose the simpler version 95% of the time.
There are many ways to make recursion preserving iterative process.
I usually do continuation passing style directly. This is my "natural" way to do it.
One takes into account the type of the function. Sometimes you need to connect your function with the functions around it and depending on their type you can choose another way to do recursion.
You should start by solving "the little schemer" to gain a strong foundation about it. In the "little typer" you can discover another type of doing recursion, founded on other computational philosophy, used in languages like agda, coq.
In scheme you can write code that is actually haskell sometimes (you can write monadic code that would be generated by a haskell compiler as intermediate language). In that case the way to do recursion is also different that "usual" way, etc.
false dichotomy
You have other options available to you. Here we can preserve tail-recursion and map over the list with a single traversal. The technique used here is called continuation-passing style -
(define (map f lst (return identity))
(if (null? lst)
(return null)
(map f
(cdr lst)
(lambda (r) (return (cons (f (car lst)) r))))))
(define (square x)
(* x x))
(map square '(1 2 3 4))
'(1 4 9 16)
This question is tagged with racket, which has built-in support for delimited continuations. We can accomplish map using a single traversal, but this time without using recursion. Enjoy -
(require racket/control)
(define (yield x)
(shift return (cons x (return (void)))))
(define (map f lst)
(reset (begin
(for ((x lst))
(yield (f x)))
null)))
(define (square x)
(* x x))
(map square '(1 2 3 4))
'(1 4 9 16)
It's my intention that this post will show you the detriment of pigeonholing your mind into a particular construct. The beauty of Scheme/Racket, I have come to learn, is that any implementation you can dream of is available to you.
I would highly recommend Beautiful Racket by Matthew Butterick. This easy-to-approach and freely-available ebook shatters the glass ceiling in your mind and shows you how to think about your solutions in a language-oriented way.
I am attempting to write a loop in Scheme using named let. I would like to be able to break out of the iteration early based on various criteria, rather than always looping right at the end. Effectively, I would like to have while, break and continue. I am constrained to use guile 1.8 for strong reasons, and guile 1.8 does not implement the R6RS while construct. My question is, does recursing using named let have to be tail recursive, and why can't you restart the loop before the end? [Does this need a code example?] When I do attempt to recurse using an early exit at several point with IO operations I invariably read past EOF and get unpredictable corruption of data.
(let name ((iter iter-expr) (arg1 expr1) (arg2 expr2))
(cond
(continue-predicate (name (next iter) arg1 arg2)))
(break-predicate break-expression-value)
(else (name (next iter) next-arg1-expression next-ar2-expression))))
A continue is just calling again using most of the same arguments unchanged except the iterated parts which will change to the next thing.
A break is a base case.
So what is a while? It is a named let with a break predicate and a default case.
Scheme doesn't really have a while construct. If you read the report you'll see that it is just a syntax sugar (macro) that turns into something like a named let.
You need it to be tail recursive if you want to be able to exit it without all the previous calculations to be done. You can also use call/cc to supply an exit continuation which is basically having Scheme doing it for you under the hood. Usually call/cc is quite far out for beginners and it takes some time to master so making your procedures tail recursive is easier to understand than doing something like this:
(define (cars lists-of-pair)
(call/cc (lambda (exit)
(fold (lambda (e a)
(if (pair? e)
(cons (car e) a)
(exit '()))) ; throw away continuations to make current result make it ()
'()
lists-of-pair)))
(cars '((1 2) (a b))) ; ==> (1 a)
(cars '((1 2) ())) ; ==> ()
Here is my code. As I call continuation twice, I expect "world" be displayed twice.
But it output only once. What am I missing?
(define call-with-current-continuation call/cc)
(define (repeat arg)
(lambda (cc)
(cc arg)
(cc arg)))
(display (call/cc (repeat "world")))
The short version: the continuation you're capturing is bigger than you think it is. When scheme is evaluating the "call/cc", the continuation is "display this result and then exit the program". So, the first call to the continuation inside the body of the (lambda (cc) (cc arg) (cc arg)) will print the value ... and then exit the program.
To get the behavior you want, you probably want a delimited continuation, that captures only the portion of the continuation out to the top level. Different Scheme implementations differ in the availability of delimited continuations (though you can implement delimited continuations in terms of full continuations, at the expense of efficiency).
Racket, for instance, has a "call with composable continuation" that slices off only the portion of the continuation that reaches out to the nearest prompt... and there's an implicit prompt around each top-level expression. So this program:
#lang racket
(define (repeat arg)
(lambda (cc)
(cc arg)
(cc arg)))
(display (call-with-composable-continuation (repeat "world")))
prints "world" twice, as you'd hoped.
Does anyone know how I can figure out the free variables in a lambda expression? Free variables are the variables that aren't part of the lambda parameters.
My current method (which is getting me nowhere) is to simply use car and cdr to go through the expression. My main problem is figuring out if a value is a variable or if it's one of the scheme primitives. Is there a way to test if something evaluates to one of scheme's built-in functions? For example:
(is-scheme-primitive? 'and)
;Value: #t
I'm using MIT scheme.
For arbitrary MIT Scheme programs, there isn't any way to do this. One problem is that the function you describe just can't work. For example, this doesn't use the 'scheme primitive' and:
(let ((and 7)) (+ and 1))
but it certainly uses the symbol 'and.
Another problem is that lots of things, like and, are special forms that are implemented with macros. You need to know what all of the macros in your program expand into to figure out even what variables are used in your program.
To make this work, you need to restrict the set of programs that you accept as input. The best choice is to restrict it to "fully expanded" programs. In other words, you want to make sure that there aren't any uses of macros left in the input to your free-variables function.
To do this, you can use the expand function provided by many Scheme systems. Unfortunately, from the online documentation, it doesn't look like MIT Scheme provides this function. If you're able to use a different system, Racket provides the expand function as well as local-expand which works correctly inside macros.
Racket actually also provides an implementation of the free-variables function that you ask for, which, as I described, requires fully expanded programs as input (such as the output of expand or local-expand). You can see the source code as well.
For a detailed discussion of the issues involved with full expansion of source code, see this upcoming paper by Flatt, Culpepper, Darais and Findler.
[EDIT 4] Disclaimer; or, looking back a year later:
This is actually a really bad way to go about solving this problem. It works as a very quick and dirty method that accomplishes the basic goal of the OP, but does not stand up to any 'real life' use cases. Please see the discussion in the comments on this answer as well as the other answer to see why.
[/EDIT]
This solution is probably less than ideal, but it will work for any lambda form you want to give it in the REPL environment of mit-scheme (see edits). Documentation for the procedures I used is found at the mit.edu doc site. get-vars takes a quoted lambda and returns a list of pairs. The first element of each pair is the symbol and the second is the value returned by environment-reference-type.
(define (flatten lst)
(cond ((null? lst) ())
((pair? (car lst)) (append (flatten (car lst)) (flatten (cdr lst))))
(else
(cons (car lst) (flatten (cdr lst))))))
(define (get-free-vars proc-form)
(let ((env (ge (eval proc-form user-initial-environment))))
(let loop ((pf (flatten proc-form))
(out ()))
(cond ((null? pf) out)
((symbol? (car pf))
(loop (cdr pf) (cons (cons (car pf) (environment-reference-type env (car pf))) out)))
(else
(loop (cdr pf) out))))))
EDIT: Example usage:
(define a 100)
(get-vars '(lambda (x) (* x a g)))
=> ((g . unbound) (a . normal) (x . unbound) (* . normal) (x . unbound) (lambda . macro))
EDIT 2: Changed code to guard agains calling environment-reference-type being called with something other than a symbol.
EDIT 3: As Sam has pointed out in the comments, this will not see the symbols bound in a let under the lambda as having any value.. not sure there is an easy fix for this. So, my statement about this taking any lambda is wrong, and should have read more like "Any simple lambda that doesn't contain new binding forms"... oh well.
Structure and Interpretation of Computer Programs (SICP) 3.5.2 introduces infinite streams:
(define ones
(cons-stream 1 ones))
This code doesn't work in DrRacket, with the error:
ones: undefined; cannot reference an identifier before its definition
Other code like this:
(define (integers-starting-from n)
(cons-stream n
(integers-starting-from (+ n 1))))
(define integers (integers-starting-from 1))
produce error:
Interactions disabled
(fall in infinite loop?)
So far as I read(SICP), the key point of implementing an infinite stream is delayed evaluation:
(define (delay exp)
(lambda () exp))
(define (cons-stream a b)
(cons a
(delay b)))
With this used in cons-stream, infinite stream still illegal.
Such data structure reminds me of recursive function in whose definition self-call is legal (in compiling) within or without an actual exit.
Why is it illegal for a value to reference itself? Even the reference has been delayed?
Could infinite stream supported by other programming language?
If not, is it about the way processor deal with assembly language? The data stack stuff?
When you make a procedure the arguments are already evaluated when you start the body of the procedure. Thus delay would not do anything since it's already computed at that time. cons-stream need to be a macro.
DrRacket is not one language implementation. It's an IDE that supports lots of languages. One of them is a SICP compatibility language. I mananged to run your code without errors using this code in DrRacket:
#!planet neil/sicp
(define ones
(cons-stream 1 ones))
(define (integers-starting-from n)
(cons-stream n
(integers-starting-from (+ n 1))))
(define integers (integers-starting-from 1))
It works like a charm. The default language in DrRacket, #!racket, has streams too, but the names are different:
#!racket
(define ones
(stream-cons 1 ones))
(define (integers-starting-from n)
(stream-cons n
(integers-starting-from (+ n 1))))
(define integers (integers-starting-from 1))
However for Scheme you should use SRFI-41 that you can use from both #!racket (using (require srfi/41)) and #!r6rs (and R7RS-large when it's finished)
(import (rnrs)
(srfi :41))
(define ones
(stream-cons 1 ones))
(define (integers-starting-from n)
(stream-cons n
(integers-starting-from (+ n 1))))
(define integers (integers-starting-from 1))
To roll your own SICP streams in both #!racket and #!r6rs you can use define-syntax
(define-syntax stream-cons
(syntax-rules ()
((_ a d) (cons a (delay d)))))
Note that RSFI-41 and rackets own stream library for #!racket delays both values in a stream-cons and not just the tail as in SICP version here.
It's because the expression in a define is evaluated before the name is bound to a value. It tries to evaluate (cons-stream 1 ones) before ones is defined, causing an error.
The reason this works fine for functions is that the body of a function is not evaluated when the function is. That is, to evaluate (lambda (x) (f x)), the language returns a function without looking at its body. Since
(define (f x) (f x))
is syntax sugar for defining a lambda, the same logic applies.
Did you define cons-stream yourself as above? If you make cons-stream a normal function it won't work properly. Since Scheme is strict by default, arguments are evaluated before the function gets called. If cons-stream is a normal function, b would get evaluated completely before being passed to delay, making you hit an infinite loop.
cons-stream in SICP is a "special form" rather than a function, which means that it can control how its arguments are evaluated.
If you use the stream-cons and other stream- operations built into Racket, you'd get the behavior you want.
Finally, some other languages do allow values to reference themselves. A great example is Haskell, where this works because everything is lazy by default. Here's a Haskell snippet defining ones to be an infinite list of ones. (Since everything is lazy, Haskell lists behave just like Scheme streams):
ones = 1 : ones
This definition works in Racket:
(define ones
(cons-stream 1 ones))
…As long as you provide a delayed implementation of cons-stream as a special form, which is the whole point of section 3.5 in SICP:
(define-syntax cons-stream
(syntax-rules ()
((_ head tail)
(cons head (delay tail)))))
Adding to the answers above, to make sure the code run, you should also define delay like this:
(define-syntax delay
(syntax-rules ()
((_ exp)
(lambda () exp))))
delay as well as cons-stream has to be defined as a macro.
In another option, you could just call the delay predefined in Racket rather than building a new one.
But don't do this:
(define (delay exp)
(lambda () exp))
The compiler would take your definition, thus the program crashes in evaluation:
Interactions disabled