My goal is to be able to apply a macro to an argument list in the same way the apply primitive procedure applies a procedure to an argument list.
The list will already be evaluated at the time of application of the macro, there is no way around that and that’s fine; I am wondering if there is any way to programmatically “splice” the list into the macro application (in the same sense as with unquote-splicing). The difficulty resides in that one cannot pass the macro identifier as an argument.
One use case would be
(apply and list)
which would be equivalent to
(not (memq #f list))
to see if there is a #f in list.
Preferably this would be R7RS conformant.
One sort of hacky way would be (as suggested on reddit)
(eval (cons 'and list))
but this is not R7RS conformant, as eval must take an environment argument and it seems to me the standard doesn’t specify how to snatch the environment in effect at the call to eval.
Another half solution is the following, which only works if the list is given directly as a parenthesized sequence of values:
(syntax-rules ()
((_ identifier (val ...))
(identifier val ...)))
I'm posting this as a partial answer I found to my own question, and I'll accept it in a few days if nothing new pops up.
The following works, but only if the macro to apply is contained in a library.
(import (scheme base)
(scheme eval)
(scheme write))
(define (apply-macro mac args . libs)
(eval (cons mac args)
(apply environment libs)))
(define list '(#f #t #t #t))
(display (apply-macro 'and list '(scheme base))) ; => #f
(display (apply-macro 'and (cdr list) '(scheme base))) ; => #t
You can't do that; macros apply to syntax, transforming code fragments into other code fragments, not to values.
Even if you could do it, it would not be equivalent to applying and, since all the elements of list would be evaluated.
For instance, if you define the non-terminating procedure,
(define (forever) (forever))
then (and #f (forever)) is #f, but (apply and (list #f (forever))) would not terminate.
You cannot do this without eval. You would need to implement a procedure version of AND.
The reason it's impossible is because macro expansion is one phase and evaluation is a later phase. The list is piece of dynamic data existing only in the later phase, so a macro cannot use that.
Related
Here is a task I have to solve: Write an iterative function, that receives a list with numbers and creates a new list, which only consists of the even numbers from the received list.
I even found a few posts with similar questions and solutions, but I couldn't use them as help, because in these solution they were using the "car" and "cdr" commands, which we haven't had in our schedule yet. How can do this? I would start like this:
(define filter-odds
(lambda(x)
(if (odd? x)
...)
Terrible oversight: no one has mentioned How To Design Programs, here. This kind of program is right smack in the middle of HtDP's path. Specifically, this is a totally straightforward application of the List template from section 9.3, "Programming with Lists." Perhaps your class is already using HtDP?
Well, in the absence of car ⁄ cdr accessors, your only other choice is using reduce from SRFI1:
(define filter-odds
(lambda (lst)
(reduce-right
(lambda (e ; list's element
acc) ; accumulated results from the right
(if (odd? e)
.... ; combine e and the results-so-far
.... )) ; ignore e
'()
lst )))
But, right reduce most likely won't be iterative. Left reduce usually is. It will build the result in reverse order though, but you can reverse a list iteratively in another pass, using the same approach, without filtering at all (i.e. using the test procedure which always succeeds).
With the first ⁄ rest accessors available though, all you need to do is to write down the implementation for reduce yourself, then fuse it into the definitions above:
(define reduce
(lambda (f acc ls)
(if (null? ls)
acc
(reduce f (f acc (first ls)) .....))))
In other words, just follow the above code skeleton and add replace the call to f with some checks for odd? or something, to get the filter-odd?-reversed; then figure out the f to use so as to define the reverse-list function.
So here is my problem (generalized to an abstract situation). It is generally an interpreter.
I have got a program that needs to parse an input list and according to its elements sequentially call some functions that need to modify variables. Functions are defined separately and chosen by (cond . Those 3 variables that I need to update contain information about current situation (exactly - robot position, maze and robot orientation). The result of previous function is used in the next function (i.e. the updated variables are used).
(define func-a ... )
(define func-b ... )
(define (main-func <maze> <orientation> <coordinates> <list with commands>)
;; here I parse the list and choose which function to call
;; output must contain (list <new-maze> <new-orientation> <new-coordinates>))
)
What scheme tools can I use to update those variables? I have considered several options:
use (define and then call set! (this is bad style cause it is not pure functional programming);
call functions from back to beginning (this won't work: I also have to check if movement is valid);
don't make those variables constant at all, try passing them as arguments to each function;
Is there any other proper way to do it?
You have to keep some state (as well as read a file), so there will not be a pure functional programming, and you have to accept some deviations.
The general approach is to keep the shared object as a local in some meta-function, say parse, and update it by calling those your function, say parse-one, parse-two, and so on.
Now you need a way to update them.
You can make them visible for parse-one and parse-two by defining them inside the scope:
(let ((state (make-state)))
(let ((parse-one (lambda () ...))
(parse-two (lambda () ...)))
....))
Or you use the return value:
(let ((state (make-state)))
...
(set! state (parse-one state))
...)
There is a third approach, called OOP. Define all of them in a single closure, so they can share some data:
(define (make-parser)
(let ((state (make-state))
(let (((parse-one (lambda () ...))
((parse-two (lambda () ...))
((get-state (lambda () state)))
(list parse-one parse-two get-state))))
(destructuring-bind (parse-one parse-two get-state) (make-parser)
...
(parse-one)
...
(get-state))
(destructuring-bind is just an easy way to destruct a list, see it's scheme implementation) But it seems to be a complicated version of the first.
Just because Scheme is considered a 'functional language' doesn't forbid you from using 'set!' - after all it exists in the language to be used. Thus there is nothing wrong with:
(define-record-type position (fields x y z))
(define robot-position (make-position 0.0 0.0 0.0))
(define update-robot-position (new-x new-y new-z)
(set! robot-position (make-position new-x new-y new-x)))
[I've chosen to define positions as invariant.]
You can chose another approach if you want but fundamentally the position of the robot changed and that change will be in your code is some fashion. Why not use the simplest, most straight-forward approach?
I have a list that is used within a loop, and on each iteration I apply a function that will alter the list permanently (popping and adding elements). The problem is, the original list is never changed whenever it is nil. How may I solve this problem?. My code is shown below
(defun looping-func ()
(let ((queue '(2)))
(loop while (not (null queue)) do
(let ( (num (pop queue)))
(if (oddp num)
(format t "~%~A success" num)
(progn (format t "~%fail")
(add-to-list (1+ num) queue)))))))
(defun add-to-list (elem l)
(nconc l (list elem)))
The code works as intended if the list contains more than 1 element. if it contains exactly 1 element, once that element is popped and the list becomes nil, the applied changes aren't permanent to the list anymore. I suppose this is because of how nconc is defined, if the first argument is nil, just return the second one without any alteration. Any ideas on how to go about this?
PS: I know the code above is useless, but I am using the same concept for a school project that I unfortunately can't post code for.
Change
(add-to-list (1+ num) queue)
to
(setq queue (add-to-list (1+ num) queue))
You can't "extend" nil with nconc
(nconc nil . lists)
is equivalent to
(nconc . lists)
so, you need to put the result of add-to-list in queue
Don't add elements to the end of a list.
Never.
Lists in Lisp are designed in a way that adding an element to the head is cheap. Adding to the end is potentially costly.
To implement a LIFO queue you need a different implementation.
Don't alter constant literal data in source code during runtime.
Indent your code properly.
Because I assumed this to be an exercise, here's an example, which you shouldn't use in your daily practice, you should use push macro, which probably does something similar to it:
(defmacro push-example (item list)
(let ((the-list list)) ; we do this to prevent
; multiple evaluations
; of the `list' argument
`(setq ,the-list (cons ,item ,the-list))))
(defparameter *test* nil)
(push-example 'foo *test*) ;; (foo)
*test* ;; (foo)
While you didn't ask for a macro (you asked for the function), Doug's answer is technically more correct, this illustrates how you could have done it using code generation through macro. Notice how this is basically doing the same thing as your function does, except that it can encapsulate the call to setq you would have to make otherwise.
At a conceptual level a macro in LISP (and dialects) take a piece of code (as list) and returns another piece of code (again as list).
Based on above principle a simple macro could be:
(defmacro zz [a] (list print a))
;macroexpand says : (#<core$print clojure.core$print#749436> "Hello")
But in clojure this can also be written as:
(defmacro zz [a] `(print ~a))
;macroexpand says : (clojure.core/print "Hello")
I am not exactly sure about the difference here and which should be the preferred way. The first one look simple as we are supposed to return list and avoid using weird characters like back tick.
No one has pointed this out yet...the difference between your 2 macros is this: your second form (using backtick)
(defmacro zz [a] `(print ~a))
is equivalent to:
(defmacro zz [a] (list 'print a))
Which is different from your first example:
(defmacro zz [a] (list print a))
Note the missing single quote -- that is why your macroexpand is different. I agree with the other people posting: using backquote is more conventional if your macro has a fairly simple 'shape'. If you have to do code walking or dynamic construction (i.e. a complex macro), then using lists and building it up is often what's done.
I hope this explanation makes sense.
Constructing lists explicitly is "simplest", in a way, because there are few core concepts you need to know: just accept a list and change it around till you have a new list. Backtick is a convenient shortcut for "templating" chunks of code; it is possible to write any macro without it, but for any large macro it quickly becomes very unpleasant. For example, consider two ways of writing let as a macro over fn:
(defmacro let [bindings & body]
(let [names (take-nth 2 bindings)
vals (take-nth 2 (rest bindings))]
`((fn [~#names]
(do ~#body))
~#vals)))
(defmacro let [bindings & body]
(let [names (take-nth 2 bindings)
vals (take-nth 2 (rest bindings))]
(cons (list `fn (vec names) (cons `do body))
vals)))
In the first case, using backtick makes it fairly clear that you're writing a function of the names containing the body, and then calling it with the values - the macro code is "shaped" the same as the expansion code, so you can imagine what it will look like.
In the second case, with just cons and list all over the place, it is a real headache to work out what the expansion will look like. This isn't always the case, of course: sometimes it can be clearer to write something without a backtick.
Another very important point was made by Kyle Burton: print is not the same as 'print! Your macro expansion should contain the symbol print, not its value (which is a function). Embedding objects (such as functions) in code is very fragile and only works by accident. So make sure your macros expand to code you could actually have written yourself, and let the evaluation system do the hard work - you could type in the symbol print, but you couldn't type in a pointer to the current value of the function print.
There's a style difference between them. Your example is very simple but in more complex macros the difference will be bigger.
For example the unless macro as defined in "The Joy of Clojure" book:
(defmacro unless [condition & body]
`(if (not ~condition)
(do ~#body)))
From the book:
Syntax-quote allows the following if-form to act as a sort of template for the expression
that any use of the macro become when it is expanded.
When creating a macro always choose the most readable and idiomatic style.
To contrast, the above code can equivalently be written like so:
(defmacro unless [condition & body]
(list 'if (list 'not condition)
(list* 'do body)))
In my experience they are equivalent. Though there may be some edge cases I'm not aware of.
#islon 's example can equivalently be written as:
To contrast, the above code can equivalently be written like so:
(defmacro unless [condition & body]
(list 'if (list 'not condition)
(list* 'do body)))
I have defined a list (in Racket/Scheme):
(define myList (cons 'data1 (cons 'data2 (cons 'data3 (cons 'data4 empty)))))
or
(list 'data1 'data2 'data3 'data4)
And I want to write a function that cycles through the list and outputs all values of the list.
(define (outputListData list)
(cond
[(null? list) list]
[else (getListData)]))
With what function can I cycle through the content of the list? I know one can use first & rest to get list data, but I guess that's not the right way here.
BTW: Is there a good, compact racket reference like php.net? I find the official Racket docs very confusing ...
You can use a for loop. Example:
(for ([x (list 1 2 3)])
(printf "~s -> ~s\n" x (* x x)))
There are more functional ways to do this, of course, but this way works too. You'll probably want to look at a textbook like How To Design Programs to do the recursive approach. See: http://www.ccs.neu.edu/home/matthias/HtDP2e/
dyoo's solution is nice and succinct in a Scheme like Racket that has useful iteration routines built in. Just FYI, though, your 'outputListData' is not far from being the standard recursive way to do this. You just need to change a couple of lines:
(define (outputListData list)
(cond
[(null? list) #f] ; actually doesn't really matter what we return
[else (printf "~s\n" (first list)) ; display the first item ...
(outputListData (rest list))])) ; and start over with the rest
Since this is an "imperative" kind of procedure that isn't designed to return a value, it doesn't really matter what we do with an empty list so long as we stop recurring (to avoid an infinite loop). If the list isn't empty, we output the first element and start over recursively with the rest of the list.
BTW, here's another way you could write something almost identical if you just needed a "for" loop in the middle of some other function:
(let loop ((l (list 'foo 'bar 'baz 'quux))) ; or put whatever input you need
(cond ((null? l) #f)
(else
(printf "~s\n" (first l))
(loop (rest l)))))
One way to think about this "named let" is that it defines a temporary function called loop, which works just like outputListData above. Scheme has the nice property that it won't grow the stack for "tail calls" like these, so you can always write what would be an "iterative" for or while loop in this recursive style.
I highly recommend The Little Schemer by Friedman and Felleisen for a quick intro to this style of function writing! I found it through Douglas Crockford's page here.
Edit as per comments: Use for-each
(for-each display myList)
Try this:
(void (map display myList))
Breaking it down:
(void x) causes x to be ignored instead of returned to the REPL/parent expression as a value.
(map function lst): For a list '(a1 a2 ... an) returns the list '((function a1) (function a2) ... (function an)).
So we use map to display all the items, but since we only care about the side-effect and not the return value, we call void on the returned list.
Official docs:
void
map
I think the solution that is the easiest to understand it to come up with a so called "list-eater" function. This is the way my university introduced recursion and lists in Racket. Also most books on Racket (i.e. "How To Design Programs" or "Realm Of Racket") explain it this way. This is the code:
(define my-list (list 'data1 'data2 'data3 'data4))
(define (print-list a-list-of-data)
(when (not (empty? a-list-of-data))
(print (first a-list-of-data))
(print-list (rest a-list-of-data))))
If you call the function with the example list my-list, you will get the following output:
(print-list my-list)
'data1'data2'data3'data4
The function does the following: As long as the given list is not empty, it grabs the first element of that list and passes it to the function print. Then, it tells itself to do the exact same thing with the rest of the list. (It calls itself on the rest of the list.) This second part is what they call recursion.
However, you can shorten that by using a function called map:
(define (print-list a-list-of-data)
(map print a-list-of-data))
This basically says that you want the function print to be called on each element of the given list. The output is exactly the same.
Hope it helped!