In Section 12.4 of On Lisp, Paul Graham writes, "Unfortunately, we can't define a correct _f with define-modify-macro, because the operator to be applied to the generalized variable is given as an argument."
But what's wrong with something like this?
(define-modify-macro _f (op operand)
(lambda (x op operand)
(funcall op x operand)))
(let ((lst '(1 2 3)))
(_f (second lst) #'* 6)
lst)
=> (1 12 3)
Has there perhaps been a change made to define-modify-macro in ANSI Common Lisp that wasn't valid at the time On Lisp was written? Or are there reasons other than the one stated for not using define-modify-macro here?
It appears that Graham want's to be able to make a call such as
(_f * (second lst) 6)
rather than
(_f #'* (second lst) 6)
But surely that's not in keeping with a Lisp2 such as Common Lisp?
According to both Lispworks's Hyperspec and CLtL2 (look for define-modify-macro), the function is assumed to be a symbol (to a function or a macro). As far as I know, the following definition might not be conforming the specification:
(define-modify-macro _f (op operand)
(lambda (x op operand)
(funcall op x operand)))
But of course, it is possible that an implementation allows it.
To be sure you are conforming to the standard, you can define your own function, or even a macro:
(defmacro funcall-1 (val fun &rest args)
`(funcall ,fun ,val ,#args))
(define-modify-macro _ff (&rest args) funcall-1)
(let ((x (list 1 2 3 4)))
(_ff (third x) #'+ 10)
x)
If you wanted to have the function as a second argument, you could define another macro:
(defmacro ff (fun-form place &rest args)
`(_ff ,place ,fun-form ,#args))
Basically, your approach consists in wrapping funcall in define-modify-macro, and give the desired function as an argument of that function. At first sight, it looks like a hack, but as we can see below, this gives the same macroexanded code as the one in On Lisp, assuming we modify the latter a little.
The macroexpansion of the above is:
(LET ((X (LIST 1 2 3 4)))
(LET* ((#:G1164 X) (#:G1165 (FUNCALL #'+ (THIRD #:G1164) 10)))
(SB-KERNEL:%RPLACA (CDDR #:G1164) #:G1165))
X)
The version in On Lisp behaves as follows:
(defmacro _f (op place &rest args)
(multiple-value-bind (vars forms var set access)
(get-setf-expansion
place)
`(let* (,#(mapcar #'list vars forms)
(, (car var) (,op ,access ,#args)))
,set)))
(let ((x (list 1 2 3 4)))
(_f * (third x) 10)
x)
Macroexpansion:
(LET ((X (LIST 1 2 3 4)))
(LET* ((#:G1174 X) (#:G1175 (* (THIRD #:G1174) 10)))
(SB-KERNEL:%RPLACA (CDDR #:G1174) #:G1175))
X)
Here, the * is injected directly by the macroexpansion, which means that the resulting code has no possible runtime overhead (though compilers would probably handle your (funcall #'+ ...) equally well). If you pass #'+ to the macro, it fails to macroexpand. This is the major difference with your approach, but not a big limitation. In order to allow the On Lisp version to accept #'*, or even (create-closure) as an operator, it should be modified as follows:
(defmacro _f (op place &rest args)
(multiple-value-bind (vars forms var set access)
(get-setf-expansion
place)
`(let* (,#(mapcar #'list vars forms)
(, (car var) (funcall ,op ,access ,#args)))
,set)))
(see the call to funcall)
The previous example is then expanded as follows, for #'*:
(LET ((X (LIST 1 2 3 4)))
(LET* ((#:G1180 X) (#:G1181 (FUNCALL #'* (THIRD #:G1180) 10)))
(SB-KERNEL:%RPLACA (CDDR #:G1180) #:G1181))
X)
Now, it is exactly as your version. On Lisp uses _f to demonstrate how to use get-setf-expansion, and _f is a good example for that. But on the other hand, your implementation seems equally good.
On the question of whether one might prefer to pass * or #'*, we can also note that the define-modify-macro version of _f and #coredump's adapted version (with funcall) both accept lambda forms in the op position with or without #' e.g. both (lambda (x y) (* x y)) and #'(lambda (x y) (* x y)), whereas Graham's original version accepts only the former.
Interestingly in his book Let over Lambda, Doug Hoyte draws attention to a remark by Graham in his book ANSI Common Lisp that being able to omit the #' before a lambda form provides "a specious form of elegance at best" before going on to prefer to omit it.
I'm not taking a stand either way, merely pointing out that given Graham's choice for _f, the absence of the #' is no longer specious but necessary.
Related
I have code like this:
(define-syntax macron
(syntax-rules ()
((_ name)
(lambda (x)
(eval (cons 'name x) (interaction-environment))))))
(define x (map (macron lambda)
'(((x) (display x)) ((a b) (+ a b)))))
(let ((square (car x))
(sum (cadr x)))
(display (square 10))
(newline)
(display (sum 1 2 3))
(newline))
the code is working it use macro as value by wrapping it with lambda. My question is how can I put inside syntax-rule macro literal symbol 'name instead of (cons 'lambda ...) so the output code is:
(lambda (x)
(eval (cons 'name x) (interaction-environment)))
so it work with code like this:
(define (name x)
(display x)
(newline))
(for-each (macron lambda) ;; lambda can be anything
'((1) (2) (3)))
and it print all the numbers.
I know that I can change the name in pattern into something else, but I want to know more about syntax-rules and it's edge cases. So is it possible to have name if I use it as input pattern?
I'm looking for answers with R7RS, that have more of this type of edge cases covered.
All macros happens in compile time so runtime stuff might not exist. That means that you should think of it as syntax sugar and use it as susch. eg.
(for-each (macron something) '((1) (2) (3)))
Should then have an expansion based on that. Your current expansion is that it turns into this:
(for-each (lambda (x)
(eval (cons 'someting x) (interaction-environment))
'((1) (2) (3)))
For something being a macro this will apply the macro in runtime. It is bad. It also removes the need for the macro in the first place. You could do this instead:
(define (macron-proc name)
(lambda (x)
(eval (cons name x) (interaction-environment))))
(for-each (macron-proc 'something) '((1) (2) (3)))
I made a programming language that had passable macros:
(define xor (flambda (a b) `(if ,a (not ,b) ,b)))
(define (fold comb init lst)
(if (null? lst)
init
(fold comb (comb (car lst) init) (cdr lst))))
(fold xor #f '(#t #t)) ; ==> #f
It's not a very good approach if you are targeting an efficient compiled end product. The first macros were indeed like this and they removed it in LISP 1.5 before Common Lisp. Scheme avoided macros for many years and opted for syntax-rules in R4RS as an optional feature. R6RS is the only version that has full power macros.
With a procedure instead of macros this is actually the same as the following code with the bad eval removed:
(for-each (lambda (x)
(apply something x))
'((1) (2) (3)))
Which means you can implement macron much easier:
(define-syntax macron
(syntax-rules ()
((_ name)
(lambda (x)
(apply name x)))))
But from looking at this now you don't need a macro at all. This is partial application.
(define (partial proc arg)
(lambda (lst)
(apply proc arh lst)))
(map (partial + 3) '((1 2) (3 4) (4 5)))
; ==> (6 10 12)
There is actually a SRFI-26 called cut/cute which allows us to do something similar where it wraps it in a lambda:
(map (cut apply + 3 <>) '((1 2) (3 4) (4 5)))
The syntax-rules are the macros with the least power. You cannot do anything unhygienic and you cannot make new identifiers based on other ones. Eg. it' impossible to implement a racket style struct where you can do (struct complex [real imag]) and have the macro create complex?, complex-real, and complex-imag as procedures. You need to do as SRFI-57 does and require th euser to specify all the names such that you don't need to concatenate to new identifiers.
Right now R7RS-small only has syntax-rules. I think it was a mistake not to have a more powerful macro as an alternative since now the R7RS-large cannot be implemented with R7RS-small.
Is it possible to use/implement tacit programming (also known as point-free programming) in Lisp? And in case the answer is yes, has it been done?
This style of programming is possible in CL in principle, but, being a Lisp-2, one has to add several #'s and funcalls. Also, in contrast to Haskell for example, functions are not curried in CL, and there is no implicit partial application. In general, I think that such a style would not be very idiomatic CL.
For example, you could define partial application and composition like this:
(defun partial (function &rest args)
(lambda (&rest args2) (apply function (append args args2))))
(defun comp (&rest functions)
(flet ((step (f g) (lambda (x) (funcall f (funcall g x)))))
(reduce #'step functions :initial-value #'identity)))
(Those are just quick examples I whipped up – they are not really tested or well thought-through for different use-cases.)
With those, something like map ((*2) . (+1)) xs in Haskell becomes:
CL-USER> (mapcar (comp (partial #'* 2) #'1+) '(1 2 3))
(4 6 8)
The sum example:
CL-USER> (defparameter *sum* (partial #'reduce #'+))
*SUM*
CL-USER> (funcall *sum* '(1 2 3))
6
(In this example, you could also set the function cell of a symbol instead of storing the function in the value cell, in order to get around the funcall.)
In Emacs Lisp, by the way, partial application is built-in as apply-partially.
In Qi/Shen, functions are curried, and implicit partial application (when functions are called with one argument) is supported:
(41-) (define comp F G -> (/. X (F (G X))))
comp
(42-) ((comp (* 2) (+ 1)) 1)
4
(43-) (map (comp (* 2) (+ 1)) [1 2 3])
[4 6 8]
There is also syntactic threading sugar in Clojure that gives a similar feeling of "pipelining":
user=> (-> 0 inc (* 2))
2
You could use something like (this is does a little more than -> in
Clojure):
(defmacro -> (obj &rest forms)
"Similar to the -> macro from clojure, but with a tweak: if there is
a $ symbol somewhere in the form, the object is not added as the
first argument to the form, but instead replaces the $ symbol."
(if forms
(if (consp (car forms))
(let* ((first-form (first forms))
(other-forms (rest forms))
(pos (position '$ first-form)))
(if pos
`(-> ,(append (subseq first-form 0 pos)
(list obj)
(subseq first-form (1+ pos)))
,#other-forms)
`(-> ,(list* (first first-form) obj (rest first-form))
,#other-forms)))
`(-> ,(list (car forms) obj)
,#(cdr forms)))
obj))
(you must be careful to also export the symbol $ from the package in
which you place -> - let's call that package tacit - and put
tacit in the use clause of any package where you plan to use ->, so -> and $ are inherited)
Examples of usage:
(-> "TEST"
string-downcase
reverse)
(-> "TEST"
reverse
(elt $ 1))
This is more like F#'s |> (and the shell pipe) than Haskell's ., but they
are pretty much the same thing (I prefer |>, but this is a matter of personal taste).
To see what -> is doing, just macroexpand the last example three times (in SLIME, this is accomplished by putting the cursor on the first ( in the example and typing C-c RET three times).
YES, it's possible and #danlei already explained very well. I am going to add up some examples from the book ANSI Common Lisp by Paul Graham, chapter 6.6 on function builders:
you can define a function builder like this:
(defun compose (&rest fns)
(destructuring-bind (fn1 . rest) (reverse fns)
#'(lambda (&rest args)
(reduce #'(lambda (v f) (funcall f v))
rest
:initial-value (apply fn1 args)))))
(defun curry (fn &rest args)
#'(lambda (&rest args2)
(apply fn (append args args2))))
and use it like this
(mapcar (compose #'list #'round #'sqrt)
'(4 9 16 25))
returns
((2) (3) (4) (5))
The compose function call:
(compose #'a #'b #'c)
is equlvalent to
#'(lambda (&rest args) (a (b (apply #'c args))))
This means compose can take any number of arguments, yeah.
Make a function which add 3 to argument:
(curry #'+ 3)
See more in the book.
Yes, this is possible in general with the right functions. For example, here is an example in Racket implementing sum from the Wikipedia page:
#lang racket
(define sum (curry foldr + 0))
Since procedures are not curried by default, it helps to use curry or write your functions in an explicitly curried style. You could abstract over this with a new define macro that uses currying.
So I thought one of the advantages of lisp (among other languages) is its ability to implement function factories (accept functions as arguments; return new functions). I want to use this capability to make small changes to a function and save it as a new function so that if changes are made to the original function, they are also reflected in the new function on which it is based. Note: I am not the one writing the original function so I can't necessarily encapsulate the common parts in a separate function to be called by both, which would be the obvious answer otherwise.
Toy example in emacs lisp (may not be the most ideal as it is a lisp-2):
I have a function, foo that is provided to me:
(defun foo (x y)
(+ x y)))
I want my new function to include a statement that allows me to change the value of a variable if a certain condition is met. For instance:
(defun newfoo (x y)
(if (condition-met-p x)
(setq x (transform x)))
(+ x y))
Please disregard that I could use defadvice in this particular example as I am more interested in the general task of modifying functions where defadvice may not apply. I believe I can modify the body with this form:
(setq conditional-transformation
'(if (condition-met x) (setq x (transform x))))
(setq newbody (append conditional-transformation
(nth 2 (symbol-function 'foo)))))
My questions are specifically how to
create a copy of foo to newfoo
and replace the body with the value
of newbody defined above. (I've
looked into fset, setf, and
function but perhaps not using
them properly.)
possibly wrap this in a function
called makenewfoo() or something
like this so I can invoke
makenewfoo(foo) and allow this to
create newfoo().
And, more generally,
is something like this is commonly
done or there is a more idiomatic
way to modify functions?
this is a very simple case, but is
there a more general way than
specifying the list element number
to nth for the modification. For
instance, the actual function is
more complex so is there a way to
recursively search down this
s-expression tree and test for a
particular syntax and insert this
conditional-transformation
expression before or after it
(possibly using equal), so it is
less sensitive to changes made in
the original function?
It does work in Emacs Lisp:
elisp> (defun foo (x y)
(+ x y))
foo
elisp> (fset 'newfoo
(append (lambda (x y)
(when (< x 2)
(setq x (* x 2))))
(cddr (symbol-function 'foo))))
(lambda
(x y)
(when
(< x 2)
(setq x
(* x 2)))
(+ x y))
elisp> (newfoo 1 3)
5
elisp> (newfoo 3 3)
6
But I really don't think that it is commonly done or idiomatic. You should use defadvice if you want to modify the behavior of functions.
As far as CL is concerned: Some implementations provide similar functions/macros (for example in CCL: ccl:advise), and you can specify :before, :after, and :around methods for generic functions.
Example code for insertion of expressions:
(defun find-node (elt tree)
(cond ((null tree) nil)
((equal (car tree) elt) tree)
((consp (car tree)) (let ((node (find-node elt (car tree))))
(if node node (find-node elt (cdr tree)))))
(t (find-node elt (cdr tree)))))
(defun insert-before (node elt)
(setcdr node (cons (car node) (cdr node)))
(setcar node elt))
(let* ((function (copy-tree (symbol-function 'foo)))
(node (find-node '(+ x y) function)))
(when node
(insert-before node '(if (< x 2) (setq x (* x 2))))
(fset 'newfoo function)))
I'm currently working on a LISP exercise for a small project and need severe help. This may be more or less of a beginner's question but I'm absolutely lost on writing a certain function that takes in two unevaluated functions and spits out the result dependent on if the variables were given an assignment or not.
An example would be
(setq p1 '(+ x (* x (- y (/ z 2)))))
Where
(evalexp p1 '( (x 2) (z 8) ))
returns (+ 2 (* 2 (- y 4)))
My goal is to write the evalexp function but I can't even think of where to start.
So far I have
(defun evalexp (e b) )
.. not very much. If anyone could please help or lead me in a good direction I'd be more than appreciative.
Here's a full solution. It's pretty straightforward, so I'll leave out a full explanation. Ask me in the comments if there's anything you can't figure out yourself.
(Using eval to do the actual evaluation might not be what you want in your exercise/project. Look up "meta-circular interpreter" for another way.)
(defun apply-env (exp env)
(reduce (lambda (exp bdg) (subst (cadr bdg) (car bdg) exp))
env :initial-value exp))
(defun try-eval (exp)
(if (atom exp)
exp
(let ((exp (mapcar #'try-eval exp)))
(if (every #'numberp (cdr exp))
(eval exp)
exp))))
(defun evalexp (exp env)
(try-eval (apply-env exp env)))
Here's a hint, this is how you might do it (in pseudocode):
function replace(vars, list):
for each element of list:
if it's an atom:
if there's an association in vars:
replace atom with value in vars
else:
leave atom alone
else:
recursively apply replace to the sublist
There will certainly be some details to work out as you convert this to Lisp code.
I'm a Lisp beginner. I'm trying to memoize a recursive function for calculating the number of terms in a Collatz sequence (for problem 14 in Project Euler). My code as of yet is:
(defun collatz-steps (n)
(if (= 1 n) 0
(if (evenp n)
(1+ (collatz-steps (/ n 2)))
(1+ (collatz-steps (1+ (* 3 n)))))))
(defun p14 ()
(defvar m-collatz-steps (memoize #'collatz-steps))
(let
((maxsteps (funcall m-collatz-steps 2))
(n 2)
(steps))
(loop for i from 1 to 1000000
do
(setq steps (funcall m-collatz-steps i))
(cond
((> steps maxsteps)
(setq maxsteps steps)
(setq n i))
(t ())))
n))
(defun memoize (fn)
(let ((cache (make-hash-table :test #'equal)))
#'(lambda (&rest args)
(multiple-value-bind
(result exists)
(gethash args cache)
(if exists
result
(setf (gethash args cache)
(apply fn args)))))))
The memoize function is the same as the one given in the On Lisp book.
This code doesn't actually give any speedup compared to the non-memoized version. I believe it's due to the recursive calls calling the non-memoized version of the function, which sort of defeats the purpose. In that case, what is the correct way to do the memoization here? Is there any way to have all calls to the original function call the memoized version itself, removing the need for the special m-collatz-steps symbol?
EDIT: Corrected the code to have
(defvar m-collatz-steps (memoize #'collatz-steps))
which is what I had in my code.
Before the edit I had erroneously put:
(defvar collatz-steps (memoize #'collatz-steps))
Seeing that error gave me another idea, and I tried using this last defvar itself and changing the recursive calls to
(1+ (funcall collatz-steps (/ n 2)))
(1+ (funcall collatz-steps (1+ (* 3 n))))
This does seem to perform the memoization (speedup from about 60 seconds to 1.5 seconds), but requires changing the original function. Is there a cleaner solution which doesn't involve changing the original function?
I assume you're using Common-Lisp, which has separate namespaces for variable and function names. In order to memoize the function named by a symbol, you need to change its function binding, through the accessor `fdefinition':
(setf (fdefinition 'collatz-steps) (memoize #'collatz-steps))
(defun p14 ()
(let ((mx 0) (my 0))
(loop for x from 1 to 1000000
for y = (collatz-steps x)
when (< my y) do (setf my y mx x))
mx))
Here is a memoize function that rebinds the symbol function:
(defun memoize-function (function-name)
(setf (symbol-function function-name)
(let ((cache (make-hash-table :test #'equal)))
#'(lambda (&rest args)
(multiple-value-bind
(result exists)
(gethash args cache)
(if exists
result
(setf (gethash args cache)
(apply fn args)))))))
You would then do something like this:
(defun collatz-steps (n)
(if (= 1 n) 0
(if (evenp n)
(1+ (collatz-steps (/ n 2)))
(1+ (collatz-steps (1+ (* 3 n)))))))
(memoize-function 'collatz-steps)
I'll leave it up to you to make an unmemoize-function.
something like this:
(setf collatz-steps (memoize lambda (n)
(if (= 1 n) 0
(if (evenp n)
(1+ (collatz-steps (/ n 2)))
(1+ (collatz-steps (1+ (* 3 n))))))))
IOW: your original (non-memoized) function is anonymous, and you only give a name to the result of memoizing it.
Note a few things:
(defun foo (bar)
... (foo 3) ...)
Above is a function that has a call to itself.
In Common Lisp the file compiler can assume that FOO does not change. It will NOT call an updated FOO later. If you change the function binding of FOO, then the call of the original function will still go to the old function.
So memoizing a self recursive function will NOT work in the general case. Especially not if you are using a good compiler.
You can work around it to go always through the symbol for example: (funcall 'foo 3)
(DEFVAR ...) is a top-level form. Don't use it inside functions. If you have declared a variable, set it with SETQ or SETF later.
For your problem, I'd just use a hash table to store the intermediate results.
Changing the "original" function is necessary, because, as you say, there's no other way for the recursive call(s) to be updated to call the memoized version.
Fortunately, the way lisp works is to find the function by name each time it needs to be called. This means that it is sufficient to replace the function binding with the memoized version of the function, so that recursive calls will automatically look up and reenter through the memoization.
huaiyuan's code shows the key step:
(setf (fdefinition 'collatz-steps) (memoize #'collatz-steps))
This trick also works in Perl. In a language like C, however, a memoized version of a function must be coded separately.
Some lisp implementations provide a system called "advice", which provides a standardized structure for replacing functions with enhanced versions of themselves. In addition to functional upgrades like memoization, this can be extremely useful in debugging by inserting debug prints (or completely stopping and giving a continuable prompt) without modifying the original code.
This function is exactly the one Peter Norvig gives as an example of a function that seems like a good candidate for memoization, but which is not.
See figure 3 (the function 'Hailstone') of his original paper on memoization ("Using Automatic Memoization as a Software Engineering Tool in Real-World AI Systems").
So I'm guessing, even if you get the mechanics of memoization working, it won't really speed it up in this case.
A while ago I wrote a little memoization routine for Scheme that used a chain of closures to keep track of the memoized state:
(define (memoize op)
(letrec ((get (lambda (key) (list #f)))
(set (lambda (key item)
(let ((old-get get))
(set! get (lambda (new-key)
(if (equal? key new-key) (cons #t item)
(old-get new-key))))))))
(lambda args
(let ((ans (get args)))
(if (car ans) (cdr ans)
(let ((new-ans (apply op args)))
(set args new-ans)
new-ans))))))
This needs to be used like so:
(define fib (memoize (lambda (x)
(if (< x 2) x
(+ (fib (- x 1)) (fib (- x 2)))))))
I'm sure that this can be ported to your favorite lexically scoped Lisp flavor with ease.
I'd probably do something like:
(let ((memo (make-hash-table :test #'equal)))
(defun collatz-steps (n)
(or (gethash n memo)
(setf (gethash n memo)
(cond ((= n 1) 0)
((oddp n) (1+ (collatz-steps (+ 1 n n n))))
(t (1+ (collatz-steps (/ n 2)))))))))
It's not Nice and Functional, but, then, it's not much hassle and it does work. Downside is that you don't get a handy unmemoized version to test with and clearing the cache is bordering on "very difficult".