I am reading an article which uses Scheme for describing an implementation. I know a bit of Common Lisp but no Scheme.
I am hoping you will be so kind as to explain two Scheme code samples and show me how they correspond to Common Lisp.
First, what does this Scheme mean:
(define (content cell)
(cell ’content))
I believe it means this: define a function named content which has one argument named cell. In Common Lisp it is written as:
(defun content (cell)
(...))
Am I right so far?
I am uncertain what the function's body is doing. Is the argument (cell) actually a function and the body is invoking the function, passing it a symbol, which happens to be the name of the current function? Is this the corresponding Common Lisp:
(defun content (cell)
(funcall cell ’content))
Here is the second Scheme code sample:
(define nothing #(*the-nothing*))
I believe it is creating a global variable and initializing it to #(*the-number*)). So the corresponding Common Lisp is:
(defvar nothing #(*the-nothing*))
Is that right? Does the pound symbol (#) have a special meaning? I'm guessing that *the-nothing* is referring to a global variable, yes?
Broadly speaking: yes to both, with one major caveat. More specifically, the first one is accepting an argument called cell and calling it with the symbol 'content. (BTW, your unicode quotation mark is freaking me out a bit. Is that just a copy-paste issue?)
In the second case, the hash is a shortcut for defining a vector. So, for instance:
(vector? #(abc)) ;; evaluates to #t
However, the hash also has quoting behavior. Just as the first element of
'(a b c)
is the symbol 'a (and not the value of the variable named a), the first value in the vector
#(*the-nothing*)
is the symbol '*the-nothing*, rather than the value of a global variable.
Related
Is there a way in Chicken Scheme to determine at run-time if a variable is currently defined?
(let ((var 1))
(print (is-defined? var)) ; #t
(print (is-defined? var)) ; #f
EDIT: XY problem.
I'm writing a macro that generates code. This generated code must call the macro in mutual recursion - having the macro simply call itself won't work. When the macro is recursively called, I need it to behave differently than when it is called initially. I would use a nested function, but uh....it's a macro.
Rough example:
(defmacro m (nested)
(if nested
BACKQUOTE(print "is nested")
BACKQUOTE(m #t)
(yes, I know scheme doesn't use defmacro, but I'm coming from Common Lisp. Also I can't seem to put backquotes in here without it all going to hell.)
I don't want the INITIAL call of the macro to take an extra argument that only has meaning when called recursively. I want it to know by some other means.
Can I get the generated code to call a macro that is nested within the first macro and doesn't exist at the call site, maybe? For example, generating code that calls (,other-macro) instead of (macro)?
But that shouldn't work, because a macro isn't a first-class object like a function is...
When you write recursive macros I get the impression that you have an macro expansion (m a b ...) that turns into a (m-helper a (b ...)) that might turn into (let (a ...) (m b ...)). That is not directly recursive since you are turning code into code that just happens to contain a macro.
With destructuring-bind you really only need to keep track of two variables. One for car and one for cdr and with an implicit renaming macro the stuff not coming from the form is renamed and thus hygenic:
(define-syntax destructuring-bind
(ir-macro-transformer
(lambda (form inject compare?)
(define (parse-structure structure expression optional? body)
;;actual magic happens here. Returns list structure with a mix of parts from structure as well as introduced variables and globals
)
(match form
[(structure expression) . body ]
`(let ((tmp ,expression))
,(parse-structure structure 'tmp #f body))))))
To check if something from input is the same symbol you use the supplied compare? procedure. eg. (compare? expression '&optional).
There's no way to do that in general, because Scheme is lexically scoped. It doesn't make much sense to ask if a variable is defined if an referencing an undefined variable is an error.
For toplevel/global variables, you can use the symbol-utils egg but it is probably not going to work as you expect, considering that global variables inside modules are also rewritten to be something else.
Perhaps if you can say what you're really trying to do, I can help you with an alternate solution.
(Though this is indeed simple question, I find sometimes it is common mistakes that I made when writing Scheme program as a beginner.)
I encountered some confusion about the define special form. A situation is like below:
(define num1
2)
(define (num2)
2)
I find it occurs quite often that I call num2 without the parentheses and program fails. I usually end up spending hours to find the cause.
By reading the r5rs, I realized that definition without parenthesis, e.g. num1, is a variable; while definition with parenthesis, e.g. num2, is a function without formal parameters.
However, I am still blurred about the difference between a "variable" and "function".
From a emacs lisp background, I can only relate above difference to similar idea as in emacs lisp:
In Emacs Lisp, a symbol can have a value attached to it just as it can
have a function definition attached to it.
[here]
Question: Is this a correct way of understanding the difference between enclosed and non-enclosed definitions in scheme?
There is no difference between a value and a function in Scheme. A function is just a value that can be used in a particular way - it can be called (as opposed to other kinds of value, such as numbers, which cannot be called, but can be e.g. added, which a function cannot).
The parentheses are just a syntactic shortcut - they're a faster, more readable (to experts) way of writing out the definition of the name as a variable containing a function:
(define (num)
2)
;is exactly the same as
(define num
(lambda () 2) )
The second of these should make it more visually obvious that the value being assigned to num is not a number.
If you wanted the function to take arguments, they would either go within the parentheses (after num, e.g. (num x y) in the first form, or within lambda's parentheses (e.g. (lambda (x y)... in the second.
Most tutorials for the total beginner actually don't introduce the first form for several exercises, in order to drive home the point that it isn't separate and doesn't really provide any true functionality on its own. It's just a shorthand to reduce the amount of repetition in your program's text.
In Scheme, all functions are values; variables hold any one value.
In Scheme, unlike Common Lisp and Emacs Lisp, there are no different namespaces for functions and other values. So the statement you quoted is not true for Scheme. In Scheme a symbol is associated with at most one value and that value may or may not be a function.
As to the difference between a non-function value and a nullary function returning that value: In your example the only difference is that, as you know, num2 must be applied to get the numeric value and num1 does not have to be and in fact can't be applied.
In general the difference between (define foo bar) and (define (foo) bar) is that the former evaluated bar right now and foo then refers to the value that bar has been evaluated to, whereas in the latter case bar is evaluated each time that (foo) is used. So if the expression foo is costly to calculate, that cost is paid when (and each time) you call the function, not at the definition. And, perhaps more importantly, if the expression has side effects (like, for example, printing something) those effects happen each time the function is called. Side effects are the primary reason you'd define a function without parameters.
Even though #sepp2k has answered the question, I will make it more clearer with example:
1 ]=> (define foo1 (display 23))
23
;Value: foo1
1 ]=> foo1
;Unspecified return value
Notice in the first one, foo1 is evaluated on the spot (hence it prints) and evaluated value is assigned to name foo1. It doesn't evaluate again and again
1 ]=> (define (foo2) (display 23))
;Value: foo2
1 ]=> foo2
;Value 11: #[compound-procedure 11 foo2]
1 ]=> (foo2)
23
;Unspecified return value
Just foo2 will return another procedure (which is (display 23)). Doing (foo2) actually evaluates it. And each time on being called, it re-evaluates again
1 ]=> (foo1)
;The object #!unspecific is not applicable.
foo1 is a name that refers a value. So Applying foo1 doesn't make sense as in this case that value is not a procedure.
So I hope things are clearer. In short, in your former case it is a name that refers to value returned by evaluating expression. In latter case, each time it is evaluated.
I've started trying to learn the innards of Scheme evaluation, and one aspect of quasiquotation, unquoting, evaluation and cons-cells is confusing me. If you can recommend any good references on the subject I'd be very grateful.
The R7RS draft has this example in section 4.2.8 on quasiquotation.
`(( foo ,(- 10 3)) ,#(cdr '(c)) . ,(car '(cons)))
(It's in the R4RS spec too, so this isn't a new thing.)
According to the spec this should evaluate to:
((foo 7) . cons)
I'm having some trouble understanding why. To my mind, the . removes the unquote from the start of the inner list, meaning it won't be evaluated as a procedure.
Here's a simpler expression that demonstrates the same problem:
`(foo . ,(car '(bar)))
Using the same logic as above, this should evaluate to:
(foo . bar)
And indeed it does evaluate to that on the Scheme systems I've tried.
However, to my understanding it shouldn't evaluate to that, so I want to find out where I'm going wrong.
My understanding of Scheme evaluation is (OK, simplified) if it's the first keyword after an open-bracket, call that procedure with the remainder of the list as the parameters.
My understanding of the spec is that ',' is exactly equivalent to wrapping the next expression in an '(unquote' procedure.
My understanding of the dot notation is that, for general display purposes, you remove the dot and opening parenthesis (and matching closing parenthesis), as described here:
In general, the rule for printing a pair is as follows: use the dot
notation always, but if the dot is immediately followed by an open
parenthesis, then remove the dot, the open parenthesis, and the
matching close parenthesis.
So:
`(foo . ,(car '(bar)))
Could equally be rendered as:
(quasiquote (foo unquote (car (quote (bar)))))
(In fact, this is how jsScheme will render the input in its log window.)
However, when it comes to evaluating this:
(quasiquote (foo unquote (car (quote (bar)))))
Why is the 'unquote' evaluated (as a procedure?), unquoting and evaluating the (car...) list? Surely it should just be treated as a quoted symbol, since it's not after an opening bracket?
I can think of a number of possible answers - 'unquote' isn't a regular procedure, the 'unquote' is evaluated outside of the regular evaluation process, there's a different way to indicate a procedure to be called other than a '(' followed by the procedure's symbol - but I'm not sure which is right, or how to dig for more information.
Most of the scheme implementations I've seen handle this using a macro rather than in the same language as the evaluator, and I'm having difficulty figuring out what's supposed to be going on. Can someone explain, or show me any good references on the subject?
You are correct in that there are macros involved: in particular, quasiquote is a macro, and unquote and unquote-splicing are literals. None of those are procedures, so normal evaluation rules do not apply.
Thus, it's possible to give (quasiquote (foo bar baz unquote x)) the special treatment it needs, despite unquote not being the first syntax element.
I want to write a procedure (function) that checks if a string contains another string. I read the documentation of string library from http://sicp.ai.mit.edu/Fall-2004/manuals/scheme-7.5.5/doc/scheme_7.html
According from to them,
Pattern must be a string. Searches
string for the rightmost occurrence of
the substring pattern. If successful,
the index to the right of the last
character of the matched substring is
returned; otherwise, #f is returned.
This seemed odd to me, cause the return value is either integer or boolean, so what should I compare my return value with?
I tried
(define (case-one str)
(if (= #f (string-search-forward "me" str))
#t
#f))
DrScheme don't like it,
expand: unbound identifier in module in: string-search-forward
Thanks,
string-search-forward is not a standardized Scheme procedure; it is an extension peculiar to the MIT-Scheme implementation (that's why your link goes to the "MIT Scheme Reference Manual.") To see only those procedures that are guaranteed, look at the R5RS document.
In Scheme, #f is the only value that means "false," anything else when used in a conditional expression will mean "true." There is therefore no point in "comparing" it to anything. In cases like string-search-forward that returns mixed types, you usually capture the return value in a variable to test it, then use it if it's non-false:
(let ((result (string-search-forward "me" str)))
(if result
(munge result) ; Execute when S-S-F is successful (result is the index.)
(error "hurf") ; Execute when S-S-F fails (result has the value #f.)
))
A more advanced tactic is to use cond with a => clause which is in a sense a shorthand for the above:
(cond ((string-search-forward "me" str) => munge)
(else (error "hurf")))
Such a form (<test> => <expression>) means that if <test> is a true value, then <expression> is evaluated, which has to be a one-argument procedure; this procedure is called with the value of <test> as an argument.
Scheme has a very small standard library, which is both a blessing (you can make small scheme implementations to embed in an application or device, you can learn the language quickly) and a curse (it's missing a lot of useful functions). string-search-forward is a non-standard function of MIT Scheme, it's not present in DrScheme.
Many library additions are available in the form of SRFIs. An SRFI is a community-adopted extension to the base language — think of it as an optional part of a Scheme implementation. DrScheme (or at least its successor Racket) implements many SRFIs.
DrScheme has a number of string functions as part of SRFI 13. Amongst the string searching functions, there is string-contains, which is similar except that it takes its arguments in the opposite order.
(require srfi/13)
(define (case-one str)
(integer? (string-contains str "me")))
You'll notice that the two implementations used a different argument order (indicating that they were developed independently), yet use the same return value. This illustrates that it's quite natural in Scheme to have a function return different types depending on what it's conveying. In particular, it's fairly common to have a function return a useful piece of information if it can do its job, or #f if it can't do its job. That way, the function naturally combines doing its job (here, returning the index of the substring) with checking whether the job is doable (here, testing whether the substring occurs).
Error message seems a little odd (I don't have drscheme installed unfortunately so can't investigate too much).
Are you sure str is a string?
Additionally = is for integer comparisons only, you can use false? instead.
As for the return value of string-search-forward having mixed types, scheme has the mindset that if any useful value can be returned it should be returned, so this means different return types are common for functions.
Try using srfi-13's string-index: http://docs.racket-lang.org/srfi-std/srfi-13.html#Searching The documentation you are looking at isn't specifically for PLT. and probably corresponds to some other version of Scheme.
I am reading DrRacket document http://docs.racket-lang.org/guide/binding.html
There is a function
(define f
(lambda (append)
(define cons (append "ugly" "confusing"))
(let ([append 'this-was])
(list append cons))))
> (f list)
'(this-was ("ugly" "confusing"))
I see that we define function f, inside we define lambda that takes (append), why ?
Procedure (body) for lambda is another function called cons, that appends two strings.
I don't understand this function at all.
Thanks !
The section that you're referring to demonstrates lexical scope in Racket. As in other Scheme implementations, the main point is that you can "shadow" every binding in the language. Unlike most "mainstream" languages, there are no real keywords that are "sacred" in the sense that they can never be shadowed by a local binding.
Note that a really good tool to visualize what is bound where is DrRacket's "check syntax" button: click it, and you'll see your code with highlights that shows which parts are bindings, which are special forms -- and if you hover the mouse over a specific name, you'll see an arrow that tells you where it came from.
Scheme takes some getting used to :)
f is assigned the function returned by the lambda.
lambda defines the function that takes a parameter (called append).
(define cons (append "ugly" "confusing")) is not a function per se, but calls append with the two strings as parameter and assigns the result to cons.
inside the let block, append is re-assigned a different value, the symbol this-was.
the let block creates a list of append (which now contains 'this-was) and cons (which contains '("ugly" "confusing") from 3 above
since 5 is the last statement that value is returned by the whole function which is called f
f is called with the parameter list (the list function). which gets passed as the parameter append. And this is why 3 above creates a list '("ugly" "confusing") which gets assigned to cons.
Hope that cleared up things a bit.
Cheers!