I'm going through this Scheme tutorial and in section 3 (Making Lists) the guy says you should write '() to represent an empty list. But for every test I've wrote seems that it has the very same effect as using just ().
Also, as far as I understand, the quote means the interpreter won't evaluate the expression, but seems that the interpreter knows what's after the ' symbol, because doing this (cons 1 '()) yields (1), while doing this (cons 1 'abc) yields (1 . abc), so it knows '() is an empty list but 'abc is not.
Some Scheme implementations permit bare () as a synonym for '(), but only the quoted form is standard.
As for your second question: consider
(define abc '(1 2 3))
(define def '(1 2 3))
(cons 0 'abc)
(cons 0 'def)
(cons 0 abc)
(cons 0 def)
In the first two expressions, abc and def are not evaluated, so they stay symbols. In the latter two, they are evaluated to the objects they stand for, which are both equal to the list (1 2 3).
TL;DR: To make sure your applications work as as designed you should quote the empty list since it's unsure if it will work otherwise. see the long answer below.
As for how Scheme works for quoted values, quoting '(+ 3 4 5) makes an expression a constant that is not to be evaluated. It much like making a string with code in it, like "if( a == 0 ) return 4;" in Java or C. The difference is that a quoted expression are structured data rather than byte sequences.
(cons 1 'abc) and (cons 1 '()) does the same. A cons has two placeholders for values and those two expressions sets two values in the exact same manner. It's only display (and the repl) that knows that a list that ends with () should display differently and not (1 . ()) like it actually is stored.
The long answer about the need to quote the empty list
It all boils down to the standard you're using. Most implementations today are R5RS and it requires the empty list be quoted since the empty list is not an expression. Implementations might still allow it though since it won't interfere with a proper Scheme application. Heres a quote from the R5RS report:
Note: In many dialects of Lisp, the empty combination, (), is a
legitimate expression. In Scheme, combinations must have at least one
subexpression, so () is not a syntactically valid expression.
This actually happened in R3RS (under Procedure calls) so it's been around for a while. When looking for it in R6RS however it seems to have disappeared from the section making me think they have reverted it so that it would be self evaluating. However, I cannot find it in the language changes part.
When looking at the R7RS draft (NB: PDF), the part from R5RS is back so I guess this was an error in the R6RS report. This might be the reason racket (and probably other implementors) allow () as an expression in R6RS to be sure it will work even when the report is ambiguous about it.
Related
From my understanding, it is not legal to modify a list created using quote:
(let ((numbers '(3 2 1)))
(set-car! numbers 99) ; Illegal.
numbers)
What about lists created using quasiquote? Is it legal to modify lists created using quasiquote?
(let ((numbers `(3 2 1)))
(set-car! numbers 99) ; Legal?
numbers)
(let ((numbers `(,(+ 1 2) 2 1)))
(set-car! numbers 99) ; Legal?
numbers)
The short answer is no, this isn't "legal", and certainly this should never be done in a program that aims to be portable. R6RS and R7RS have almost identical language around this, so I'll just quote from R6RS, Section 11.17 Quasiquotation:
A quasiquote expression may return either fresh, mutable objects or literal structure for any structure that is constructed at run time during the evaluation of the expression. Portions that do not need to be rebuilt are always literal.
Section 4.2.8 of R7RS has the same language, except that it says "newly allocated" instead of "fresh".
Since it is an error to attempt to modify literals in Scheme, it is an error to modify the result of a quasiquote form. This is something that you may seem get away with sometimes, but it will bite you sooner or later. The real catch here is that "portions that do not need to be rebuilt are always literal". Other portions may or may not be literal.
More specifically for OP posted code, `(3 2 1) is guaranteed to evaluate to a list literal by the semantics of quasiquote described in Section 11.17 of R6RS:
Semantics: If no unquote or unquote-splicing forms appear within the <qq template>, the result of evaluating (quasiquote <qq template>) is equivalent to the result of evaluating (quote <qq template>).
R7RS contains similar language in Section 4.2.8. Since (quote (3 2 1)) creates a list literal, the expression `(3 2 1) must also evaluate to a list literal.
On the other hand, OP code `(,(+ 1 2) 2 1) must evaluate (+ 1 2) and insert that result into the resulting structure. In this case, unquote is used via the , operator, so the resulting list structure may or may not be a list literal.
To take one more example, consider the quasiquoted expression `(,(+ 1 2) (2 1)). Here the main result is a list structure which may or may not be a list literal, but the element (2 1) of the resulting structure is guaranteed to be a list literal since it does not need to be rebuilt in the construction of the final result.
Is it possible to quote result of function call?
For example
(quoteresult (+ 1 1)) => '2
At first blush, your question doesn’t really make any sense. “Quoting” is a thing that can be performed on datums in a piece of source code. “Quoting” a runtime value is at best a no-op and at worst nonsensical.
The example in your question illustrates why it doesn’t make any sense. Your so-called quoteresult form would evaluate (+ 1 1) to produce '2, but '2 evaluates to 2, the same thing (+ 1 1) evaluates to. How would the result of quoteresult ever be different from ordinary evaluation?
If, however, you want to actually produce a quote expression to be handed off to some use of dynamic evaluation (with the usual disclaimer that that is probably a bad idea), then you need only generate a list of two elements: the symbol quote and your function’s result. If that’s the case, you can implement quoteresult quite simply:
(define (quoteresult x)
(list 'quote x))
This is, however, of limited usefulness for most programs.
For more information on what quoting is and how it works, see What is the difference between quote and list?.
I'm trying to add a pattern matching function to an existing scheme meta circular evaluator (this is homework) and I'm a bit lost on the wording of the instructions. I was hoping someone more skilled in this regard could help me interpret this.
The syntax for match should look like the following: (match a ((p1 v1) (p2 v2) (p3 v3)))
And it could be used to find length like so:
(define length (lambda (x)
(match x (('() 0)
(('cons a b) (+ 1 (length b))))))
The pattern language in the function should contain numeric constants, quoted constants, variables, and cons. If patterns are exhausted without finding a match, an error should be thrown.
I thought I understood the concept of pattern matching but implementing it in a function this way has me a bit thrown off. Would anyone be willing to explain what the above syntax is doing (specifically, how match is used in length above) so I can get a better understanding of what this function should do?
(match x (('() 0)
(('cons a b) (+ 1 (length b)))))
It may be most helpful to consider what this code would need to expand into. For each pattern, you'd need a test to determine whether the object you're trying to match matches, and you'd need code to figure out how to bind variables to its subparts. In this case, you'd want an expansion roughly like:
(if (equal? '() x)
0
(if (pair? x)
(let ((a (car x))
(b (cdr x)))
(+ 1 (length b)))
;; indicate failure to match
'NO-MATCH))
You could write that with a cond, too, of course, but if you have to procedurally expand this, it might be easier to use nested if forms.
If you're actually implementing this as a function (and not as a macro (i.e., source transformation), then you'll need to specify exactly how you can work with environments, etc.
I suggest you read the chapter four, Structured Types and the Semantics of Pattern Matching, from The Implementation of Functional Languages. The chapter is written by Simon L. Peyton Jones and Philip Wadler.
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!
(define a 42)
(set! 'a 10)
(define a 42)
(define (symbol) 'a)
(set! (symbol) 10)
(define a (cons 1 2))
(set! (car a) 10)
I tried running them in DrScheme and they don't work. Why?
Think of set! is a special form like define which does not evaluate its first operand. You are telling the scheme interpreter to set that variable exactly how you write it. In your example, it will not evaluate the expression 'a to the word a. Instead, it will look for a variable binding named "'a" (or depending on your interpreter might just break before then since I think 'a is not a valid binding).
For the last set of expressions, if you want to set the car of a pair, use the function (set-car! pair val) which works just like any scheme function in that it evaluates all of its operands. It takes in two values, a pair and some scheme value, and mutates the pair so that the car is now pointing to the scheme value.
So for example.
>(define pair (cons 1 2))
>pair
(1 . 2)
>(set-car! pair 3)
(3 . 2)
Because the first argument of set! is a variable name, not an "lvalue", so to speak.
For the last case, use (set-car! a 10).
The issue is with (set! 'a 10), as you shouldn't be quoting the symbol a.
It sounds like you're trying to learn Scheme, and you don't know Lisp, yes? If so, I strongly recommend trying Clojure as an easier to learn Lisp. I failed to grasp the interaction between the reader, evaluation, symbols, special forms, macros, and so forth in both Common Lisp and Scheme because those things all seemed to interact in tangled ways, but I finally really understand them in Clojure. Even though it's new, I found Clojure documentation is actually clearer than anything I found for Scheme or CL. Start with the videos at http://clojure.blip.tv/ and then read the docs at clojure.org.