I know Emacs 24 introduced lexical scoping; that is great but I work with lexical scoping all the time, and I'm trying to see from a different point of view with dynamic scoping.
Originally I just did what I would always do, rely on a closure:
(add-hook 'cider-mode-hook
(lambda ()
(dolist (p '(("M-l" . cider-load-current-buffer)
("M-e" . cider-eval-last-expression)))
(local-set-key
(kbd (car p))
(lambda () (interactive) (save-buffer) (cdr p))))))
After finally understanding why p is undefined when the lambda runs, I came up with this to force the evaluation of p in the context of the dolist rather than when the lambda runs.
(add-hook 'cider-mode-hook
(lambda ()
(dolist (p '(("M-l" . cider-load-current-buffer)
("M-e" . cider-eval-last-expression)))
(local-set-key
(kbd (car p))
(cons 'lambda `(() (interactive) (save-buffer) (funcall (quote ,(cdr p)))))))))
Is this the classical solution for solving the problem without closures and lexical scoping?
I would do this like this if I wanted to keep bindings in a list like you.
I actually prefer to spell out local-set-key for each command.
(defun save-before-call (f)
`(lambda()
(interactive)
(save-buffer)
(funcall #',f)))
(add-hook 'cider-mode-hook
(lambda ()
(mapc (lambda(x)(local-set-key
(kbd (car x))
(save-before-call (cdr x))))
'(("M-l" . cider-load-current-buffer)
("M-e" . cider-eval-last-expression)))))
Related
SICP indicates cdr is opened up:
In section 3.5.4 , i saw this block:
(define (integral delayed-integrand initial-value dt)
(define int
(cons-stream initial-value
(let ((integrand (force delayed-integrand)))
(add-streams (scale-stream integrand dt)
int))))
int)
Normally if this was something like:
(define (stream-map proc s)
(if (stream-null? s)
the-empty-stream
(cons-stream (proc (stream-car s))
(stream-map proc (stream-cdr s)))))
The stream-cdr s would be evaluated as (cons-stream (stream-car (cdr s)) delay<>) even when the actual call would be in a delay. ie even though the stream-map function itself is delayed, the arguments are pre-computed. [Is this correct? - By the applicative model, the arguments should be substituted for ,before the function is "called", but is the call evaluation when delay is forced or when it's just specified]
Then why is let not pre-computed?
What i think? I think let is a lambda function with the variable as the arguments, so it's execution is delayed
(let ((var1 e1) (var2 e2)) e3)
is same as
Lambda (var1 var2) e3 (with var1 bound to e1 and var2 bound to e2)
Can someone please help me confirm this? Thanks
In SICP type streams the car of a stream is not delayed, but the cdr is.
The whole expression,
(let ((integrand (force delayed-integrand)))
(add-streams (scale-stream integrand dt)
int))
, is delayed since it is the second argument to cons-stream. What kind of expression is delayed doesn't matter so you can have a call, evaluation of variable or even a let there.
"the arguments are pre-computed. is this correct?"
No. (cons-stream a (func (stream-cdr b))) is just like
(cons-stream a
(lambda ()
;; our code is placed here, verbatim, as a whole:
(func (stream-cdr b))
)
)
const-stream is a macro, it just moves pieces of code around.
The lambda might be enclosed in a memo-proc call for the memoization, to do call-by-need, but it'll still have the code we wrote, like (func (stream-cdr b)), placed inside the lambda, "textually", by cons-stream. Which is a macro, just moving pieces of code around.
Regarding the snippet which you've added to the question, the authors were just being imprecise. What is meant there is, when (stream-cdr stream) in (stream-filter pred (stream-cdr stream)) will be called, it will produce (cons 10008 (delay .... )), as shown. Not before.
Where it says "which in this case is" is should have said "which in this case is the same as".
In section 3.5.1 Streams Are Delayed Lists the book says:
To make the stream implementation automatically and transparently interleave the construction of a stream with its use, we will arrange for the cdr of a stream to be evaluated when it is accessed by the stream-cdr procedure rather than when the stream is constructed by cons-stream.
I'm currently attempting to create a stream using a macro, as shown below:
(define-syntax create-stream
(syntax-rules (using starting at with increment )
[(create-stream name using f starting at i0 with increment delta)
(letrec
([name (lambda (f current delta)
(cons current (lambda () (name (f (+ current delta) delta)))))])
(lambda () (name f i0 delta)))
]))
What happens though is I get a compilation error later on when I try to pass it the following lambda function, which says "lambda: not an identifier, identifier with default, or keyword".
(create-stream squares using (lambda (x) (* x x)) starting at 5 with increment 2)
What I suspect is happening is that by attempting to use lambda in a macro, it's shadowing the actual lambda function in Racket. If this is true, I'm wondering in what way one can create a stream without using lambda, since as far as I can tell, there needs to be a function somewhere to create said stream.
Also this is a homework problem, so I do have to use a macro. My equivalent function for creating the stream would be:
(define create-stream
(letrec ([name (lambda (f current delta) (cons current (lambda () (name (f (+ current delta) delta)))))])
(lambda () (name f i0 delta))))
What I suspect is happening is that by attempting to use lambda in a macro, it's shadowing the actual lambda function in Racket
No, that's not correct.
The problem here is that you use f and delta in binding positions, in (lambda (f current delta) ...). That means after (create-stream squares using (lambda (x) (* x x)) starting at 5 with increment 2) is expanded, you will get code like:
(lambda ((lambda (x) (* x x)) current 2) ...)
which is obviously a syntax error.
You can see this yourself by using the macro stepper in DrRacket.
The fix is to rename the identifiers to something else. E.g.,
(define-syntax create-stream
(syntax-rules (using starting at with increment )
[(create-stream name using f starting at i0 with increment delta)
(letrec
([name (lambda (f* current delta*)
(cons current (lambda () (name (f* (+ current delta*) delta*)))))])
(lambda () (name f i0 delta)))
]))
This will make your code start running, but there are still other several bugs in your code. Since this is your homework, I'll leave them to you.
Your current problem essentially boils down to this (broken) code
(let ((x 0))
(define y x))
This is not equivalent to (define y 0) - y's definition is local to the let-binding.
The solution is to flip the definitions around and make the letrec local to the define:
(define name
(letrec (... fn definition ...)
(lambda () (fn i0))))
This is a problem related to ex3.51 in SICP, here is the code
(define (cons-stream x y)
(cons x (delay y)))
(define (stream-car stream) (car stream))
(define (stream-cdr stream) (force (cdr stream)))
(define (stream-map proc s)
(if (stream-null? s)
the-empty-stream
(cons-stream
(proc (stream-car s))
(stream-map proc (stream-cdr s)))))
(define (stream-enumerate-interval low high)
(if (> low high)
the-empty-stream
(cons-stream
low
(stream-enumerate-interval (+ low 1) high))))
(define (stream-ref s n)
(if (= n 0)
(stream-car s)
(stream-ref (stream-cdr s) (- n 1))))
(define (show x)
(display x)
x)
;test
(stream-map show (stream-enumerate-interval 0 10))
the output is 012345678910(0 . #<promise>).
but I thought the delay expression in cons-stream delayed the evaluation, if i use a different processing function in stream-map like lambda (x) (+ x 1) the output (1 . #<promise>) is more reasonable, so why does display print all the numbers?
The problem is with this definition:
(define (cons-stream x y)
(cons x (delay y)))
It defines cons-stream as a function, since it uses define.
Scheme's evaluation is eager: the arguments are evaluated before the function body is entered. Thus y is already fully calculated when it is passed to delay.
Instead, cons-stream should be defined as a macro, like
(define-syntax cons-stream
(syntax-rules ()
((_ a b) (cons a (delay b)))))
or we can call delay explicitly, manually, like e.g.
(define (stream-map proc s)
(if (stream-null? s)
the-empty-stream
(cons
(proc (stream-car s))
(delay
(stream-map proc (stream-cdr s))))))
Then there'd be no calls to cons-stream in our code, only the (cons A (delay B)) calls. And delay is a macro (or special form, whatever), it does not evaluate its arguments before working but rather goes straight to manipulating the argument expressions instead.
And we could even drop the calls to delay, and replace (cons A (delay B)) with (cons A (lambda () B)). This would entail also reimplementing force (which is built-in, and goes together with the built-in delay) as simply (define (force x) (x)) or just calling the (x) manually where appropriate to force a stream's tail.
You can see such lambda-based streams code towards the end of this answer, or an ideone entry (for this RosettaCode entry) without any macros using the explicit lambdas instead. This approach can change the performance of the code though, as delay is memoizing but lambda-based streams are not. The difference will be seen if we ever try to access a stream's element more than once.
See also this answer for yet another take on streams implementation, surgically modifying list's last cons cell as a memoizing force.
Is it possible in (Common) Lisp to jump to another function instead of call another?
I mean, that the current function is broken and another is called, without jumping back through thousands of functions, as if I'd decide myself if tail call optimization is done, even if it is not the tail.
I'm not sure if "(return-from fn x)" does, what I want.
Example:
(defun fn (x)
(when x
(princ x)
(jump 'fn (cdr x)))
(rest))
'jump' should be like calling the following function without saving the position of this function, instead returning to, where the original funcall was, so that there will be no stack overflow.
'rest' should only be executed if x is nil.
When you need a tail call optimization like structure in a language that doesn't (necessarily) provide it, but does provide closures, you can use a trampoline to achieve constant stack space (with a trade off for heap space for closure objects, of course). This isn't quite the same as what you asking for, but you might find it useful. It's pretty easy to implement in Common Lisp:
(defstruct thunk closure)
(defmacro thunk (&body body)
`(make-thunk :closure (lambda () ,#body)))
(defun trampoline (thunk)
(do ((thunk thunk (funcall (thunk-closure thunk))))
((not (thunk-p thunk)) thunk)))
To use the trampoline, you just call it with a thunk that performs the first part of your computation. That closure can either return another thunk, or a result. If it returns a thunk, then since it returned the initial stack frame is reclaimed, and then the closure of returned thunk is invoked. For instance, here's what an implementation of non-variadic mapcar might look like:
(defun t-mapcar1 (function list)
(labels ((m (list acc)
(if (endp list)
(nreverse acc)
(thunk
(m (rest list)
(list* (funcall function (first list)) acc))))))
(m list '())))
When the list is empty, we get an empty list immediately:
CL-USER> (t-mapcar1 '1+ '())
NIL
When it's not, we get back a thunk:
CL-USER> (t-mapcar1 '1+ '(1 2))
#S(THUNK :CLOSURE #<CLOSURE (LAMBDA #) {10033C7B39}>)
This means that we should wrap a call with trampoline (and this works fine for the base case too, since trampoline passes non-thunk values through):
CL-USER> (trampoline (t-mapcar1 '1+ '()))
NIL
CL-USER> (trampoline (t-mapcar1 '1+ '(1 2)))
(2 3)
CL-USER> (trampoline (t-mapcar1 '1+ '(1 2 3 4)))
(2 3 4 5)
Your example code isn't quite enough to be an illustrative example, but
(defun fn (x)
(when x
(princ x)
(jump 'fn (cdr x)))
(rest))
would become the following. The return provides the early termination from fn, and the thunk value that's returned provides the “next” computation that the trampoline would invoke for you.
(defun fn (x)
(when x
(princ x)
(return (thunk (fn (cdr x)))))
(rest))
How about you use a tail call?
(defun fn (x)
(if x
(progn
(princ x)
(fn (cdr x)))
(progn
(rest))))
It calls fn in a tail position. If an implementation provides tail call optimization, you won't get a stack overflow. If you don't want to rely on that, you would need to handle the problem in a non recursive way. There are no explicit 'remove this functions stack frame and then call function X' operators in Common Lisp.
Well, not really. I once did experiment with
(defmacro recurr (name bindings &body body)
(let* ((names (mapcar #'car bindings))
(restart (gensym "RESTART-"))
(temp1 (gensym))
(temp2 (gensym))
(shadows (mapcar (lambda (name) (declare (ignore name)) (gensym)) names)))
`(block ,name
(let ,bindings
(macrolet ((,name ,shadows
(list 'progn
(cons 'psetq
(loop
:for ,temp1 :in ',names
:for ,temp2 :in (list ,#shadows)
:nconcing (list ,temp1 ,temp2)))
(list 'go ',restart))))
(tagbody
,restart
(progn ,#body)))))))
and to be used like scheme's named-let, e.g.:
(recurr traverse ((list '(1 2 3 4)))
(if (null list) 'end
(traverse (cdr list))))
but:
The object defined (traverse in the example) is not a function, i.e., you cannot funcall or apply it
This kind of construct doesn't really cope with recursive structures (i.e., since no stack is kept, you cannot use it to traverse over arbitrary trees instead of sequences)
Another approach might be
(defmacro label (name (&rest bindings) &body body)
`(labels ((,name ,(mapcar #'first bindings) ,#body))
(,name ,#(mapcar #'second bindings))))
which actually addresses the points mentioned, but loses the "look ma, no stack space consing" property.
I am brushing up on my low Racket knowledge and came across this streams in racket excellent post. My question is about this snip:
(define powers-of-two
(letrec ([f (lambda (x) (cons x (lambda () (f (* x 2)))))])
(lambda () (f 2))))
I understand the reason for the 'inner lambdas' but why did the OP use a lambda for the whole function? Couldn't it be done like this just as effectively?
(define (powers-of-two)
(letrec ([f (lambda (x) (cons x (lambda () (f (* x 2)))))])
(f 2)))
I experimented and see no difference. My question is whether this is just a matter of style, or if there is some reason that the former is preferable.
There is no difference. Since f in the second example doesn't close over anything, it can be lifted outside the powers-of-two function, which is equivalent to the first example.
One reason why the first might be preferable is that there is only ever one f function that needs to be created. With the second one, a new f function would be created every time someone called (powers-of-two).
I tried timing them both though, and neither was significantly faster than the other.
(define (name arg)
arg)
This is the short form of writing:
(define name
(lambda (arg)
arg))
So what happens with the first example is that the letrect happens right away and the function returned would be (lambda () (f 2)) with f in it's closure.
The second makes a procedure called powers-of-two that , when applied (powers-of-two) will return the same as the first powers-of-two is.. Think of it as a powers-of-two-generator.
Thus:
(define powers-of-two
(letrec ([f (lambda (x) (cons x (lambda () (f (* x 2)))))])
(lambda () (f 2))))
(define (powers-of-two-generator)
(letrec ([f (lambda (x) (cons x (lambda () (f (* x 2)))))])
(f 2)))
(powers-of-two) ; ==> (2 . procedure)
(define powers-of-two2 (powers-of-two-generator)) ; ==> procedure
(powers-of-two2) ; ==> (2 . procedure)
Do you see the difference?