Is it true that this is an S-expression?
xyz
asks The Little Schemer. but how to test?
syntactically, i get how to test other statements like
> (atom? 'turkey)
and
> (list? '(atom))
not entirely sure how to test this...
> (list? '(atom turkey) or)
as it just returns...
or: bad syntax in: or
but anyway, knowing how to test for S-expressions is foxing me
so, as per usual, any illumination much appreciated
An "S-expression" is built of atoms via several (possibly zero) cons applications:
(define (sexp? expr)
(or
; several cases:
(atom? expr)
; or
(and (pair? expr) ; a pair is built by a cons
(sexp? (car expr)) ; from a "car"
(sexp? .........)) ; and a "cdr"
)))
This is practically in English. Nothing more to say about it (in code, I mean). Except, after defining the missing
(define (atom? x)
(not (pair? x)))
we see that (sexp? ...) can only return #t. This is the whole point to it: in Lisp, everything is an S-expression – either an atom, or a pair of S-expressions.
The previous answer is correct -- Scheme (and Lisp) are languages that are based on S-expressions. And the provided code is a great start.
But it's not quite correct that everything is an S-expression in those languages. In this case you have an expression that isn't syntactically correct, so the language chokes when it tries to read it in. In other words, it's not an S-expression.
I know it's frustrating, and honestly, not that great of an answer, but this is a great lesson in one of the Golden Rules of Computer Programming: Garbage In, Garbage Out. The fat is that you are going to get these kinds of errors simply because it isn't really feasible, when starting out programming, to test for every possible way that something isn't an S-expression without using the language itself.
Related
Can any case of using call/cc be rewritten equivalently without using it?
For example
In (g (call/cc f)), is the purpose of f to evaluate the value of
some expression, so that g can be applied to the value?
Is (g (call/cc f)) always able to be rewritten equivalently
without call/cc e.g. (g expression)?
In ((call/cc f) arg), is the purpose of f to evaluate the
definition of some function g, so that function g can be
applied to the value of arg?
Is ((call/cc f) arg) always able to be rewritten equivalently
without call/cc e.g. (g arg)?
If the answers are yes, why do we need to use call/cc?
I am trying to understand the purpose of using call/cc, by contrasting it to not using it.
The key to the direct answer here is the notion of "Turing equivalence". That is, essentially all of the commonly used programming languages (C, Java, Scheme, Haskell, Lambda Calculus etc. etc.) are equivalent in the sense that for any program in one of these languages, there is a corresponding program in each of the other languages which has the same meaning.
Beyond this, though, some of these equivalences may be "nice" and some may be really horrible. This suggests that we reframe the question: which features can be rewritten in a "nice" way into languages without that feature, and which cannot?
A formal treatment of this comes from Matthias Felleisen, in his 1991 paper "On the Expressive Power of Programming Languages" (https://www.sciencedirect.com/science/article/pii/016764239190036W), which introduces a notion of macro expressibility, pointing out that some features can be rewritten in a local way, and some require global rewrites.
The answer to your original question is obviously yes. Scheme is Turing-complete, with or without call/cc, so even without call/cc, you can still compute anything that is computable.
Why "it is more convenient than writing the equivalent expression using lambda"?
The classic paper On the Expressive Power of Programming Languages by Matthias Felleisen gives one answer to this question. Pretty much, to rewrite a program with call/cc to one without it, you might potentially need to transform your whole program (global transformation). This is to contrast some other constructs that only need a local transformation (i.e., can be written as macro) to remove them.
The key is: If your program is written in continuation passing style, you don't need call/cc. If not, good luck.
I whole-heartedly recommend:
Daniel P. Friedman. "Applications of Continuations: Invited Tutorial". 1988 Principles of Programming Languages (POPL88). January 1988
https://cs.indiana.edu/~dfried/appcont.pdf
If you enjoy reading that paper, then check out:
https://github.com/scheme-live/bibliography/blob/master/page6.md
Of course anything that is written with call/cc can be written without it, because everything in Scheme is ultimately written using lambda. You use call/cc because it is more convenient than writing the equivalent expression using lambda.
There are two senses to this question: an uninteresting one and an interesting one:
The uninteresting one. Is there some computation that you can do with call/cc that you can't do in a language which does not have it?
No, there isn't: call/cc doesn't make a language properly more powerful: it is famously the case that a language with only λ and function application is equivalent to a universal Turing machine, and thus there is no (known...) more powerful computational system.
But that's kind of uninteresting from the point of view of programming-language design: subject to the normal constraints on memory &c, pretty much all programming languages are equivalent to UTMs, but people still prefer to use languages which don't involve punching holes in paper tape if they can.
The interesting one. Is it the case that call/cc makes some desirable features of a programming language easier to express?
The answer to this is yes, it does. I'll just give a couple of examples. Let's say you want to have some kind of non-local exit feature in your language, so some deeply-nested bit of program can just say 'to hell with this I want out', without having to climb back out through some great layer of functions. This is trivial with call/cc: the continuation procedure is the escape procedure. You can wrap it in some syntax if you want it to be nicer:
(define-syntax with-escape
(syntax-rules ()
[(_ (e) form ...)
(call/cc (λ (e) form ...))]))
(with-escape (e)
... code in here, and can call e to escape, and return some values ...)
Can you implement this without call/cc? Well, yes, but not without either relying on some other special construct (say block and return-from in CL), or without turning the language inside out in some way.
And you can build on things like this to implement all sorts of non-local escapes.
Or, well, let's say you want GO TO (the following example is Racket):
(define (test n)
(define m 0)
(define start (call/cc (λ (c) c)))
(printf "here ~A~%" m)
(set! m (+ m 1))
(when (< m n)
(start start)))
Or, with some syntax around this:
(define-syntax-rule (label place)
(define place (call/cc identity)))
(define (go place)
(place place))
(define (horrid n)
(define m 0)
(label start)
(printf "here ~A~%" m)
(set! m (+ m 1))
(when (< m n)
(go start)))
So, OK, this perhaps is not a desirable feature of a programming language. But, well, Scheme doesn't have GO TO right, and yet, here, it does.
So, yes, call/cc (especially when combined with macros) makes a lot of desirable features of a programming language possible to express. Other languages have all these special-purpose, limited hacks, Scheme has this universal thing from which all these special-purpose hacks can be built.
The problem is that call/cc doesn't stop with the good special-purpose hacks: you can also build all the awful horrors that used to blight programming languages out of it. call/cc is like having access to an elder god: it's really convenient if you want dread power, but you'd better be careful what comes with it when you call, because it may well be an unspeakable horror from beyond spacetime.
An easy use of call/cc is as a bail out. eg.
;; (1 2) => (2 4)
;; #f if one element is not a number
(define (double-numbers lst)
(call/cc
(lambda (exit)
(let helper ((lst lst))
(cond ((null? lst) '())
((not (number? (car lst))) (exit #f))
(else (cons (* 2 (car lst)) (helper (cdr lst)))))))))
So to understand this. If we are doing (double-numbers '(1 2 r)) the result is #f, but the helper has done (cons 1 (cons 2 (exit #f)))
Without call/cc we see the continuation would be whatever called double-numbers since it actually return normally from it. Here is an example without call/cc:
;; (1 2) => (2 4)
;; #f if one element is not a number
(define (double-numbers lst)
(define (helper& lst cont)
(cond ((null? lst) (cont '()))
((not (number? (car lst))) #f) ; bail out, not using cont
(else (helper& (cdr lst)
(lambda (result)
(cont (cons (* 2 (car lst)) result)))))))
(helper& lst values)) ; values works as an identity procedure
I imagine it gets harder pretty quick. Eg. my generator implementation. The generator relies on having access to continuations to mix the generator code with where it's used, but without call/cc you'll need to do CPS in both the generator, the generated generator and the code that uses it.
I've been learning Scheme through the Little Schemer book and it strikes me as odd that a file in Scheme / Lisp isn't a list of statements. I mean, everything is supposed to be a list in Lisp, but a file full of statements doesn't look like a list to me. Is it represented as a list underneath? Seems odd that it isn't a list in the file.
For instance...
#lang scheme
(define atom?
(lambda (x)
(and (not (pair? x)) (not (null? x)))))
(define sub1
(lambda (x y)
(- x y)))
(define add1
(lambda (x y)
(+ x y)))
(define zero?
(lambda (x)
(= x 0)))
Each define statement is a list, but there is no list of define statements.
It is not, because there is no practical reasons for it. In fact, series of define statements change internal state of the language. Information about the state can be accessible via functions. For example , you can ask Lisp if some symbol is bound to a function.
There is no practical benefit in traversing all entered forms (for example, define forms). I suppose that this approach (all statements are elements of a list) would lead to code that would be hard to read.
Also, I think it not quite correct to think that "everything is supposed to be a list in Lisp", since there are also some atomic types, which are quite self-sufficient.
When you evaluate a form, if the form defines something, that definition is added to the environment, and that environment is (or can be) a single list. You can build a program without using files, by just typing definitions into the REPL. In Lisp as in any language, the program “lives” in the run-time environment, not the source files.
Given an expression in the form of : (* 3 (+ x y)), how can I evaluate the expression so as to put it in the form (+ (* 3 x) (* 3 y))? (note: in the general case, 3 is any constant, and "x" or "y" could be terms of single variables or other s-expressions (e.g. (+ 2 x)).
How do I write a lambda expression that will recursively evaluate the items (atoms?) in the original expression and determine whether they are a constant or a term? In the case of a term, it would then need to be evaluated again recursively to determine the type of each item in that term's list.
Again, the crux of the issue for me is the recursive "kernel" of the definition.
I would obviously need a base case that would determine once I have reached the last, single atom in the deepest part of the expression. Then recursively work "back up", building the expression in the proper form according to rules.
Coming from a Java / C++ background I am having great difficulty in understanding how to do this syntactically in Scheme.
Let's take a quick detour from the original problem to something slightly related. Say that you're given the following: you want to write an evaluator that takes "string-building" expressions like (* 3 "hello") and "evaluates" it to "hellohellohello". Other examples that we'd like to make work include things like
(+ "rock" (+ (* 5 "p") "aper")) ==> "rockpppppaper"
(* 3 (+ "scis" "sors")) ==> "scissorsscissorsscissors"
To tackle a problem like this, we need to specify exactly what the shape of the inputs are. Essentially, we want to describe a data-type. We'll say that our inputs are going to be "string-expressions". Let's call them str-exprs for short. Then let's define what it means to be a str-expr.
A str-expr is either:
<string>
(+ <str-expr> <str-expr>)
(* <number> <str-expr>)
By this notation, we're trying to say that str-exprs will all fit one of those three shapes.
Once we have a good idea of what the shape of the data is, we have a better guide to write functions that process str-exprs: they must case-analyze those three alternatives!
;; A str-expr is either:
;; a plain string, or
;; (+ str-expr str-expr), or
;; (* number str-expr)
;; We want to write a definition to "evaluate" such string-expressions.
;; evaluate: str-expr -> string
(define (evaluate expr)
(cond
[(string? expr)
...]
[(eq? (first expr) '+)
...]
[(eq? (first expr) '*)
...]))
where the '...'s are things that we'll be filling in.
Actually, we know how to fill in a little more about the '...': we know that in the second and third cases, the inner parts are themselves str-exprs. Those are prime spots where recurrence will probably happen: since our data is described in terms of itself, the programs that process them will also probably need to refer to themselves. In short, programs that process str-exprs will almost certainly follow this shape:
(define (evaluate expr)
(cond
[(string? expr)
... expr
...]
[(eq? (first expr) '+)
... (evaluate (second expr))
... (evaluate (third expr))
...]
[(eq? (first expr) '*)
... (second expr)
... (evaluate (third expr))
...]))
That's all without even doing any real work: we can figure this part out just purely because that's what the data's shape tells us. Filling in the remainder of the '...'s to make this all work out is actually not too bad, especially when we also consider the test cases we've cooked up. (Code)
It's this kind of standard data-analysis/case-analysis that's at the heart of your question, and it's one that's covered extensively by curricula such as HTDP. This is not Scheme or Racket specific: you'd do the same kind of data analysis in Java, and you see the same kind of approach in many other places. In Java, the low-mechanism used for the case analysis might be done differently, perhaps with dynamic dispatch, but the core ideas are all the same. You need to describe the data. Once you have a data definition, use it to help you sketch out what the code needs to look like to process that data. Use test cases to triangulate how to fill in the sketch.
You need to break down your problem. I would start by following the HtDP (www.htdp.org) approach. What are your inputs? Can you specify them precisely? In this case, those inputs are going to be self-referential.
Then, specify the output form. In fact, your text above is a little fuzzy on this: I think I know what your output form looks like, but I'm not entirely sure.
Next, write a set of tests. These should be based on the structure of your input terms; start with the simplest ones, and work upward from there.
Once you have a good set of tests, implementing the function should be pretty straightforward. I'd be glad to help if you get stuck!
Can someone please explain or link me to any helpful resources ( I couldn't find any threads on google) that could help me understand what atoms are.
Nowadays we consider an atom an element that's not a cons-pair and that is not null. That includes:
Numbers
Strings
Symbols
Booleans
Characters
This is best expressed with the following procedure, taken from the book The Little Schemer:
(define atom?
(lambda (x)
(and (not (pair? x)) (not (null? x)))))
The term "atom" is used by several authors (McCarthy and Friedman/Felleisen, among others) to refer to a datum that is not a "cons" pair. I claim that these days, you'd be more likely to invert that, and test for "cons"-hood rather than "atom"-hood. Where are you seeing the term used?
I recently started reading through Paul Graham's On Lisp with a friend, and we realized that we have very different opinions of reduce: I think it expresses a certain kind of recursive form very clearly and concisely; he prefers to write out the recursion very explicitly.
I suspect we're each right in some context and wrong in another, but we don't know where the line is. When do you choose one form over the other, and what do you think about when making that choice?
To be clear about what I mean by reduce vs. explicit recursion, here's the same function implemented twice:
(defun my-remove-if (pred lst)
(fold (lambda (left right)
(if (funcall pred left)
right
(cons left right)))
lst :from-end t))
(defun my-remove-if (pred lst)
(if lst
(if (funcall pred (car lst))
(my-remove-if pred (cdr lst))
(cons (car lst) (my-remove-if pred (cdr lst))))
'()))
I'm afraid I started out a Schemer (now we're Racketeers?) so please let me know if I've botched the Common Lisp syntax. Hopefully the point will be clear even if the code is incorrect.
If you have a choice, you should always express your computational intent in the most abstract terms possible. This makes it easier for a reader to figure out your intentions, and it makes it easier for the compiler to optimize your code. In your example, when the compiler trivially knows you are doing a fold operation by virtue of you naming it, it also trivially knows that it could possibly parallelize the leaf operations. It would be much harder for a compiler to figure that out when you write extremely low level operations.
I'm going to take a slightly-subjective question and give a highly-subjective answer, since Ira already gave a perfectly pragmatic and logical one. :-)
I know writing things out explicitly is highly valued in some circles (the Python guys make it part of their "zen"), but even when I was writing Python I never understood it. I want to write at the highest level possible, all the time. When I want to write things out explicitly, I use assembly language. The point of using a computer (and a HLL) is to get it to do these things for me!
For your my-remove-if example, the reduce one looks fine to me (apart from the Scheme-isms like fold and lst :-)). I'm familiar with the concept of reduce, so all I need to understand it is figure out your f(x,y) -> z. For the explicit variant, I had to think it for a second: I have to figure out the loop myself. Recursion isn't the hardest concept out there, but I think it is harder than "a function of two arguments".
I also don't care for a whole line being repeated -- (my-remove-if pred (cdr lst)). I think I like Lisp in part because I'm absolutely ruthless at DRY, and Lisp allows me to be DRY on axes that other languages don't. (You could put in another LET at the top to avoid this, but then it's longer and more complex, which I think is another reason to prefer the reduction, though at this point I might just be rationalizing.)
I think maybe the contexts in which the Python guys, at least, dislike implicit functionality would be:
when no-one could be expected to guess the behavior (like frobnicate("hello, world", True) -- what does True mean?), or:
cases when it's reasonable for implicit behavior to change (like when the True argument gets moved, or removed, or replaced with something else, since there's no compile-time error in most dynamic languages)
But reduce in Lisp fails both of these criteria: it's a well-understood abstraction that everybody knows, and that isn't going to change, at least not on any timescale I care about.
Now, I absolutely believe there are some cases where it'd be easier for me to read an explicit function call, but I think you'd have to be pretty creative to come up with them. I can't think of any offhand, because reduce and mapcar and friends are really good abstractions.
In Common Lisp one prefers the higher-order functions for data structure traversal, filtering, and other related operations over recursion. That's also to see from many provided functions like REDUCE, REMOVE-IF, MAP and others.
Tail recursion is a) not supported by the standard, b) maybe invoked differently with different CL compilers and c) using tail recursion may have side effects on the generated machine code for surrounding code.
Often, for certain data structures, many of these above operations are implemented with LOOP or ITERATE and provided as higher-order function. There is a tendency to prefer new language extensions (like LOOP and ITERATE) for iterative code over using recursion for iteration.
(defun my-remove-if (pred list)
(loop for item in list
unless (funcall pred item)
collect item))
Here is also a version that uses the Common Lisp function REDUCE:
(defun my-remove-if (pred list)
(reduce (lambda (left right)
(if (funcall pred left)
right
(cons left right)))
list
:from-end t
:initial-value nil))