Below I have pasted the code from my drawmycoil.lsp file.
(defun drwcoil ()
(setq helices_center '(0 0)) ; this is the center of the helices
(setq inner_radius 1e3) ; 1000 microns
(setq outer_radius 2.4e3) ; 2400 microns
(setq coil_thickness 35) ; 35 microns
(setq turns (\ (- outer_radius inner_radius)
(* coil_thickness 2))) ; coil_thickness is multiplied by 2 to
; account for coil spacing
(while (< inner_radius outer_radius)
(command "HELIX"
helices_center
inner_radius
(+ inner_radius (* coil_thickness 2))
"t"
"1"
"0") ;draws one helix per iteration
(setq inner_radius
(+ inner_radius (* coil_thickness 2))))) ; increments the inner radius
; for the next iteration
(defun c:drawmycoil ()
(drwcoil)
(princ))
Here's the output of the lisp file after it is loaded and executed:
Command: _appload coil35um_intricate.lsp successfully loaded.
Command:
Command:
Command: drawmycoil ;
error: no function definition: \\\\
In spite of having clearly defined my functions, I am unable to make AutoCAD/AutoLISP recognize them. I am using Autodesk Education Master Suite 2011. I saw quite a few tutorial sites and most of them are focused on the LISP part of the functions. I am more focused on making AutoCAD plot my drawing for me through LISP.
Look at this code from your post:
(setq turns (\ (-
I suspect you meant to use / instead of \.
Related
I am new to Common Lisp and I am using SBCL, Slime and Emacs to learn.
While reading the book Common Lisp: A Gentle Introduction to Symbolic Computation, the author mentions the STEP tool which is helpful for debugging and able to do this:
It is not 100% clear if the italic text is coming from the author or the tool itself. Probably, just the author's comments.
However, even if I do not consider the italic, I am unable to generate descriptive info like this.
If I use just SBCL's REPL, I get:
* (step (if (oddp 5) 'yes 'no))
YES
If I use the REPL inside Emacs with Slime on, I get:
CL-USER> (step (if (oddp 5) 'yes 'no))
YES
The author says that:
Each implementation of Common Lisp provides its own version of this
tool; only the name has been standardized.
If I try the same thing in Emacs/Slime with a function, I get more info:
(defun my-abs (x)
(cond ((> x 0) x)
((< x 0) (- x))
(t 0)))
Using the definition above and the command bellow on the REPL:
CL-USER> (step (my-abs 10))
I get:
Evaluating call:
(MY-ABS 10)
With arguments:
10
[Condition of type STEP-FORM-CONDITION]
Restarts:
0: [STEP-CONTINUE] Resume normal execution
1: [STEP-OUT] Resume stepping after returning from this function
2: [STEP-NEXT] Step over call
3: [STEP-INTO] Step into call
4: [RETRY] Retry SLIME REPL evaluation request.
5: [*ABORT] Return to SLIME's top level.
--more--
Backtrace:
0: ((LAMBDA ()))
1: (SB-INT:SIMPLE-EVAL-IN-LEXENV (LET ((SB-IMPL::*STEP-OUT* :MAYBE)) (UNWIND-PROTECT (SB-IMPL::WITH-STEPPING-ENABLED #))) #S(SB-KERNEL:LEXENV :FUNS NIL :VARS NIL :BLOCKS NIL :TAGS NIL :TYPE-RESTRICTIONS ..
2: (SB-INT:SIMPLE-EVAL-IN-LEXENV (STEP (MY-ABS 10)) #<NULL-LEXENV>)
3: (EVAL (STEP (MY-ABS 10)))
--more--
Unfortunately, none of those options seem to give me what I want (which could be an error of interpretation on my side).
I would like to see something like:
SLIME seems to be a thorough tool. I might be missing something.
Is there a way to generate the same output described by the book using SLIME or SBCL?
As someone suggested above, SBCL does optimize away a lot by default, and compiles by default as well. Here's what I did to create an example:
I first made up "a bad value" for the "suggested optimization" by running
(declaim (optimize (debug 3) (space 0) (speed 0)))
Then, I defined a function with something more than an if condition, since that sort of thing is always inlined, unfortunately (though you might try (declaim (notinline ...)), I haven't. One way is to create a function that calls another one, like:
(defun foo () "hey!")
(defun bar () (foo))
Now, when I run (step (bar)), I see the debugger pane that you shared in your question above, and if I now select option #3, step into, I get the same pane but now focussed, as hopefully expected, on the call to foo.
Good luck!
Some references:
SBCL manual on "Single stepping": http://www.sbcl.org/manual/#Single-Stepping
SBCL manual on "Debugger policy control": http://www.sbcl.org/manual/#Debugger-Policy-Control
It looks like the author is using LispWorks' stepper.
With LispWorks
Here's my step session, using :s to step the current form and all its subforms.
CL-USER 5 > (step (my-abs -5))
(MY-ABS -5) -> :s
-5 -> :s
-5
(COND ((> X 0) X) ((< X 0) (- X)) (T 0)) <=> (IF (> X 0) (PROGN X) (IF (< X 0) (- X) (PROGN 0)))
(IF (> X 0) (PROGN X) (IF (< X 0) (- X) (PROGN 0))) -> :s
(> X 0) -> :s
X -> :s
-5
0 -> :s
0
NIL
(IF (< X 0) (- X) (PROGN 0)) -> :s
(< X 0) -> :s
X -> :s
-5
0 -> :s
0
T
(- X) -> :s
X -> :s
-5
5
5
5
5
5
Help is on :?:
:?
:s Step this form and all of its subforms (optional +ve integer arg)
:st Step this form without stepping its subforms
:si Step this form without stepping its arguments if it is a function call
:su Step up out of this form without stepping its subforms
:sr Return a value to use for this form
:sq Quit from the current stepper level
:bug-form <subject> &key <filename>
Print out a bug report form, optionally to a file.
:get <variable> <command identifier>
Get a previous command (found by its number or a symbol/subform within it) and put it in a variable.
:help Produce this list.
:his &optional <n1> <n2>
List the command history, optionally the last n1 or range n1 to n2.
:redo &optional <command identifier>
Redo a previous command, found by its number or a symbol/subform within it.
:use <new> <old> &optional <command identifier>
Do variant of a previous command, replacing old symbol/subform with new symbol/subform.
For compiled code it also has a visual stepper, where you can press a red button to set a breakpoints, see the intermediate variables changing etc. It looks like this:
LispWorks is a proprietary implementation and IDE that has a free but limited version. I just wrote a review that should be merged on the Cookbook.
trace and printv
Do you know trace? printv, an external library, is a trace on steroids. They resemble the output you appreciate.
(defun factorial (n)
(if (plusp n)
(* n (factorial (1- n)))
1))
(trace factorial)
(factorial 2)
0: (FACTORIAL 3)
1: (FACTORIAL 2)
2: (FACTORIAL 1)
3: (FACTORIAL 0)
3: FACTORIAL returned 1
2: FACTORIAL returned 1
1: FACTORIAL returned 2
0: FACTORIAL returned 6
6
(untrace factorial)
printv prints the code and the returned values.
(printv:printv
(+ 2 3)
*print-case*
*package*
'symbol
(let* ((x 0) (y (1+ x)) (z (1+ y)))
(values x y z)))
;;; (+ 2 3) => 5
;;; *PRINT-CASE* => :UPCASE
;;; *PACKAGE* => #<PACKAGE "ISSR-TEST">
;;; 'SYMBOL => SYMBOL
;;; (LET* ((X 0) (Y (1+ X)) (Z (1+ Y)))
(VALUES X Y Z)) =>
[ [X=0] [Y=1] [Z=2] ]
;;; => 0, 1, 2
This is not a homework assignment. In the following code:
(defparameter nums '())
(defun fib (number)
(if (< number 2)
number
(push (+ (fib (- number 1)) (fib (- number 2))) nums))
return nums)
(format t "~a " (fib 100))
Since I am quite inexperienced with Common Lisp, I am at a loss as to why the function does not return an value. I am a trying to print first 'n' values, e.g., 100, of the Fibonacci Sequence.
Thank you.
An obvious approach to computing fibonacci numbers is this:
(defun fib (n)
(if (< n 2)
n
(+ (fib (- n 1)) (fib (- n 2)))))
(defun fibs (n)
(loop for i from 1 below n
collect (fib i)))
A little thought should tell you why no approach like this is going to help you compute the first 100 Fibonacci numbers: the time taken to compute (fib n) is equal to or a little more than the time taken to compute (fib (- n 1)) plus the time taken to compute (fib (- n 2)): this is exponential (see this stack overflow answer).
A good solution to this is memoization: the calculation of (fib n) repeats subcalculations a huge number of times, and if we can just remember the answer we computed last time we can avoid doing so again.
(An earlier version of this answer has an overcomplex macro here: something like that may be useful in general but is not needed here.)
Here is how you can memoize fib:
(defun fib (n)
(check-type n (integer 0) "natural number")
(let ((so-far '((2 . 1) (1 . 1) (0 . 0))))
(labels ((fibber (m)
(when (> m (car (first so-far)))
(push (cons m (+ (fibber (- m 1))
(fibber (- m 2))))
so-far))
(cdr (assoc m so-far))))
(fibber n))))
This keeps a table – an alist – of the results it has computed so far, and uses this to avoid recomputation.
With this memoized version of the function:
> (time (fib 1000))
Timing the evaluation of (fib 1000)
User time = 0.000
System time = 0.000
Elapsed time = 0.000
Allocation = 101944 bytes
0 Page faults
43466557686937456435688527675040625802564660517371780402481729089536555417949051890403879840079255169295922593080322634775209689623239873322471161642996440906533187938298969649928516003704476137795166849228875
The above definition uses a fresh cache for each call to fib: this is fine, because the local function, fibber does reuse the cache. But you can do better than this by putting the cache outside the function altogether:
(defmacro define-function (name expression)
;; Install EXPRESSION as the function value of NAME, returning NAME
;; This is just to avoid having to say `(setf ...)`: it should
;; probably do something at compile-time too so the compiler knows
;; the function will be defined.
`(progn
(setf (fdefinition ',name) ,expression)
',name))
(define-function fib
(let ((so-far '((2 . 1) (1 . 1) (0 . 0))))
(lambda (n)
(block fib
(check-type n (integer 0) "natural number")
(labels ((fibber (m)
(when (> m (car (first so-far)))
(push (cons m (+ (fibber (- m 1))
(fibber (- m 2))))
so-far))
(cdr (assoc m so-far))))
(fibber n))))))
This version of fib will share its cache between calls, which means it is a little faster, allocates a little less memory but may be less thread-safe:
> (time (fib 1000))
[...]
Allocation = 96072 bytes
[...]
> (time (fib 1000))
[...]
Allocation = 0 bytes
[...]
Interestingly memoization was invented (or at least named) by Donald Michie, who worked on breaking Tunny (and hence with Colossus), and who I also knew slightly: the history of computing is still pretty short!
Note that memoization is one of the times where you can end up fighting a battle with the compiler. In particular for a function like this:
(defun f (...)
...
;; no function bindings or notinline declarations of F here
...
(f ...)
...)
Then the compiler is allowed (but not required) to assume that the apparently recursive call to f is a recursive call into the function it is compiling, and thus to avoid a lot of the overhead of a full function call. In particular it is not required to retrieve the current function value of the symbol f: it can just call directly into the function itself.
What this means is that an attempt to write a function, memoize which can be used to mamoize an existing recursive function, as (setf (fdefinition 'f) (memoize #'f)) may not work: the function f still call directly into the unmemoized version of itself and won't notice that the function value of f has been changed.
This is in fact true even if the recursion is indirect in many cases: the compiler is allowed to assume that calls to a function g for which there is a definition in the same file are calls to the version defined in the file, and again avoid the overhead of a full call.
The way to deal with this is to add suitable notinline declarations: if a call is covered by a notinline declaration (which must be known to the compiler) then it must be made as a full call. From the spec:
A compiler is not free to ignore this declaration; calls to the specified functions must be implemented as out-of-line subroutine calls.
What this means is that, in order to memoize functions you have to add suitable notinline declarations for recursive calls, and this means that memoizing either needs to be done by a macro, or must rely on the user adding suitable declarations to the functions to be memoized.
This is only a problem because the CL compiler is allowed to be smart: almost always that's a good thing!
Your function unconditionally returns nums (but only if a variable called return exists). To see why, we can format it like this:
(defun fib (number)
(if (< number 2)
number
(push (+ (fib (- number 1)) (fib (- number 2))) nums))
return
nums)
If the number is less than 2, then it evaluates the expression number, uselessly, and throws away the result. Otherwise, it pushes the result of the (+ ....) expression onto the nums list. Then it uselessly evaluates return, throwing away the result. If a variable called return doesn't exist, that's an error situation. Otherwise, it evaluates nums and that is the return value.
In Common Lisp, there is a return operator for terminating and returning out of anonymous named blocks (blocks whose name is the symbol nil). If you define a named function with defun, then an invisible block exists which is not anonymous: it has the same name as that function. In that case, return-from can be used:
(defun function ()
(return-from function 42) ;; function terminates, returns 42
(print 'notreached)) ;; this never executes
Certain standard control flow and looping constructs establish a hidden anonymous block, so return can be used:
(dolist (x '(1 2 3))
(return 42)) ;; loop terminates, yields 42 as its result
If we use (return ...) but there is no enclosing anonymous block, that is an error.
The expression (return ...) is different from just return, which evaluates a variable named by the symbol return, retrieving its contents.
It is not clear how to repair your fib function, because the requirements are unknown. The side effect of pushing values into a global list normally doesn't belong inside a mathematical function like this, which should be pure (side-effect-free).
So you might know that if you know the two previous numbers you can compute the next. What comes after 3, 5? If you guess 8 you have understood it. Now if you start with 0, 1 and roll 1, 1, 1, 2, etc you collect the first variable until you have the number of numbers you'd like:
(defun fibs (elements)
"makes a list of elements fibonacci numbers starting with the first"
(loop :for a := 0 :then b
:for b := 1 :then c
:for c := (+ a b)
:for n :below elements
:collect a))
(fibs 10)
; ==> (0 1 1 2 3 5 8 13 21 34)
Every form in Common Lisp "returns" a value. You can say it evaluates to. eg.
(if (< a b)
5
10)
This evaluates either to 5 or 10. Thus you can do this and expect that it evaluates to either 15 or 20:
(+ 10
(if (< a b)
5
10))
You basically want your functions to have one expression that calculates the result. eg.
(defun fib (n)
(if (zerop n)
n
(+ (fib (1- n)) (fib (- n 2)))))
This evaluates to the result og the if expression... loop with :collect returns the list. You also have (return expression) and (return-from name expression) but they are usually unnecessary.
Your global variable num is actually not that a bad idea.
It is about to have a central memory about which fibonacci numbers were already calculated. And not to calculate those already calculated numbers again.
This is the very idea of memoization.
But first, I do it in bad manner with a global variable.
Bad version with global variable *fibonacci*
(defparameter *fibonacci* '(1 1))
(defun fib (number)
(let ((len (length *fibonacci*)))
(if (> len number)
(elt *fibonacci* (- len number 1)) ;; already in *fibonacci*
(labels ((add-fibs (n-times)
(push (+ (car *fibonacci*)
(cadr *fibonacci*))
*fibonacci*)
(cond ((zerop n-times) (car *fibonacci*))
(t (add-fibs (1- n-times))))))
(add-fibs (- number len))))))
;;> (fib 10)
;; 89
;;> *fibonacci*
;; (89 55 34 21 13 8 5 3 2 1 1)
Good functional version (memoization)
In memoization, you hide the global *fibonacci* variable
into the environment of a lexical function (the memoized version of a function).
(defun memoize (fn)
(let ((cache (make-hash-table :test #'equal)))
#'(lambda (&rest args)
(multiple-value-bind (val win) (gethash args cache)
(if win
val
(setf (gethash args cache)
(apply fn args)))))))
(defun fib (num)
(cond ((zerop num) 1)
((= 1 num) 1)
(t (+ (fib (- num 1))
(fib (- num 2))))))
The previously global variable *fibonacci* is here actually the local variable cache of the memoize function - encapsulated/hidden from the global environment,
accessible/look-up-able only through the function fibm.
Applying memoization on fib (bad version!)
(defparameter fibm (memoize #'fib))
Since common lisp is a Lisp 2 (separated namespace between function and variable names) but we have here to assign the memoized function to a variable,
we have to use (funcall <variable-name-bearing-function> <args for memoized function>).
(funcall fibm 10) ;; 89
Or we define an additional
(defun fibm (num)
(funcall fibm num))
and can do
(fibm 10)
However, this saves/memoizes only the out calls e.g. here only the
Fibonacci value for 10. Although for that, Fibonacci numbers
for 9, 8, ..., 1 are calculated, too.
To make them saved, look the next section!
Applying memoization on fib (better version by #Sylwester - thank you!)
(setf (symbol-function 'fib) (memoize #'fib))
Now the original fib function is the memoized function,
so all fib-calls will be memoized.
In addition, you don't need funcall to call the memoized version,
but just do
(fib 10)
Ok, I'm been learning COMMON LISP programming and I'm working on a very simple program to calculate a factorial of a given integer. Simple, right?
Here's the code so far:
(write-line "Please enter a number...")
(setq x (read))
(defun factorial(n)
(if (= n 1)
(setq a 1)
)
(if (> n 1)
(setq a (* n (factorial (- n 1))))
)
(format t "~D! is ~D" n a)
)
(factorial x)
Problem is, when I run this on either CodeChef or Rexter.com, I get a similar error: "NIL is NOT a number."
I've tried using cond instead of an if to no avail.
As a side note, and most bewildering of all, I've seen a lot of places write the code like this:
(defun fact(n)
(if (= n 1)
1
(* n (fact (- n 1)))))
Which doesn't even make sense to me, what with the 1 just floating out there with no parentheses around it. However, with a little tinkering (writing additional lines outside the function) I can get it to execute (equally bewildering!).
But that's not what I want! I'd like the factorial function to print/return the values without having to execute additional code outside it.
What am I doing wrong?
One actually needs to flush the I/O buffers in portable code with FINISH-OUTPUT - otherwise the Lisp may want to read something and the prompt hasn't yet been printed. You better replace SETQ with LET, as SETQ does not introduce a variable, it just sets it.
(defun factorial (n)
(if (= n 1)
1
(* n (factorial (- n 1)))))
(write-line "Please enter a number...")
(finish-output) ; this makes sure the text is printed now
(let ((x (read)))
(format t "~D! is ~D" x (factorial x)))
Before answering your question, I would like to tell you some basic things about Lisp. (Neat fix to your solution at the end)
In Lisp, the output of every function is the "last line executed in the function". Unless you use some syntax manipulation like "return" or "return-from", which is not the Lisp-way.
The (format t "your string") will always return 'NIL as its output. But before returning the output, this function "prints" the string as well.
But the output of format function is 'NIL.
Now, the issue with your code is the output of your function. Again, the output would be the last line which in your case is:
(format t "~D! is ~D" n a)
This will return 'NIL.
To convince yourself, run the following as per your defined function:
(equal (factorial 1) 'nil)
This returns:
1! is 1
T
So it "prints" your string and then outputs T. Hence the output of your function is indeed 'NIL.
So when you input any number greater than 1, the recursive call runs and reaches the end as input 1 and returns 'NIL.
and then tries to execute this:
(setq a (* n (factorial (- n 1))))
Where the second argument to * is 'NIL and hence the error.
A quick fix to your solution is to add the last line as the output:
(write-line "Please enter a number...")
(setq x (read))
(defun factorial(n)
(if (= n 1)
(setq a 1)
)
(if (> n 1)
(setq a (* n (factorial (- n 1))))
)
(format t "~D! is ~D" n a)
a ;; Now this is the last line, so this will work
)
(factorial x)
Neater code (with Lisp-like indentation)
(defun factorial (n)
(if (= n 1)
1
(* n (factorial (- n 1)))))
(write-line "Please enter a number...")
(setq x (read))
(format t "~D! is ~D" x (factorial x))
Common Lisp is designed to be compiled. Therefore if you want global or local variables you need to define them before you set them.
On line 2 you give x a value but have not declared the existence of a variable by that name. You can do so as (defvar x), although the name x is considered unidiomatic. Many implementations will give a warning and automatically create a global variable when you try to set something which hasn’t been defined.
In your factorial function you try to set a. This is a treated either as an error or a global variable. Note that in your recursive call you are changing the value of a, although this wouldn’t actually have too much of an effect of the rest of your function were right. Your function is also not reentrant and there is no reason for this. You can introduce a local variable using let. Alternatively you could add it to your lambda list as (n &aux a). Secondarily your factorial function does not return a useful value as format does not return a useful value. In Common Lisp in an (implicit) progn, the value of the final expression is returned. You could fix this by adding a in the line below your format.
For tracing execution you could do (trace factorial) to have proper tracing information automatically printed. Then you could get rid of your format statement.
Finally it is worth noting that the whole function is quite unidiomatic. Your syntax is not normal. Common Lisp implementations come with a pretty printer. Emacs does too (bound to M-q). One does not normally do lots of reading and setting of global variables (except occasionally at the repl). Lisp isn’t really used for scripts in this style and has much better mechanisms for controlling scope. Secondarily one wouldn’t normally use so much mutating of state in a function like this. Here is a different way of doing factorial:
(defun factorial (n)
(if (< n 2)
1
(* n (factorial (1- n)))))
And tail recursively:
(defun factorial (n &optional (a 1))
(if (< n 2) a (factorial (1- n) (* a n))))
And iteratively (with printing):
(defun factorial (n)
(loop for i from 1 to n
with a = 1
do (setf a (* a i))
(format t “~a! = ~a~%” i a)
finally (return a)))
You can split it up into parts, something like this:
(defun prompt (prompt-str)
(write-line prompt-str *query-io*)
(finish-output)
(read *query-io*))
(defun factorial (n)
(cond ((= n 1) 1)
(t (* n
(factorial (decf n)))))
(defun factorial-driver ()
(let* ((n (prompt "Enter a number: "))
(result (factorial n)))
(format *query-io* "The factorial of ~A is ~A~%" n result)))
And then run the whole thing as (factorial-driver).
Sample interaction:
CL-USER 54 > (factorial-driver)
Enter a number:
4
The factorial of 4 is 24
I have a problem in understanding the performance of a Common Lisp function (I am still a novice). I have two versions of this function, which simply computes the sum of all integers up to a given n.
Non-tail-recursive version:
(defun addup3 (n)
(if (= n 0)
0
(+ n (addup (- n 1)))))
Tail-recursive version:
(defun addup2 (n)
(labels ((f (acc k)
(if (= k 0)
acc
(f (+ acc k) (- k 1)))))
(f 0 n)))
I am trying to run these functions in CLISP with input n = 1000000. Here is the result
[2]> (addup3 1000000)
500000500000
[3]> (addup2 1000000)
*** - Program stack overflow. RESET
I can run both successfully in SBCL, but the non-tail-recursive one is faster (only by a little, but that seems strange to me). I've scoured Stackoverflow questions for answers but couldn't find something similar. Why do I get a stack overflow although the tail-recursive function is designed NOT to put all recursive function calls on the stack? Do I have to tell the interpreter/compiler to optimise tail calls? (I read something like (proclaim '(optimize (debug 1)) to set the debug level and optimize at the cost of tracing abilities, but I don't know what this does).
Maybe the answer is obvious and the code is bullshit, but I just can't figure it out.
Help is appreciated.
Edit: danlei pointed out the typo, it should be a call to addup3 in the first function, so it is recursive. If corrected, both versions overflow, but not his one
(defun addup (n)
"Adds up the first N integers"
(do ((i 0 (+ i 1))
(sum 0 (+ sum i)))
((> i n) sum)))
While it may be a more typical way to do it, I find it strange that tail recursion is not always optimised, considering my instructors like to tell me it's so much more efficient and stuff.
There is no requirement for a Common Lisp implementation to have tail call optimization. Most do, however (I think that ABCL does not, due to limitations of the Java virtual machine).
The documentation of the implementation should tell you what optimization settings should be chosen to have TCO (if available).
It is more idiomatic for Common Lisp code to use one of the looping constructs:
(loop :for i :upto n
:sum i)
(let ((sum 0))
(dotimes (i n)
(incf sum (1+ i))))
(do ((i 0 (1+ i))
(sum 0 (+ sum i)))
((> i n) sum))
In this case, of course, it is better to use the "little Gauß":
(/ (* n (1+ n)) 2)
Well, your addup3 just isn't recursive at all.
(defun addup3 (n)
(if (= n 0)
0
(+ n (addup (- n 1))))) ; <--
It calls whatever addup is. Trying a corrected version in SBCL:
CL-USER> (defun addup3 (n)
(if (= n 0)
0
(+ n (addup3 (- n 1)))))
ADDUP3
CL-USER> (addup3 100000)
Control stack guard page temporarily disabled: proceed with caution
; ..
; Evaluation aborted on #<SB-SYS:MEMORY-FAULT-ERROR {C2F19B1}>.
As you'd expect.
Using GNU CommonLisp, GCL 2.6.12, compilation of addup2 will optimize tail calls, here is what I got:
>(compile 'addup2)
Compiling /tmp/gazonk_3012_0.lsp.
End of Pass 1.
;; Note: Tail-recursive call of F was replaced by iteration.
End of Pass 2.
OPTIMIZE levels: Safety=0 (No runtime error checking), Space=0, Speed=3
Finished compiling /tmp/gazonk_3012_0.lsp.
Loading /tmp/gazonk_3012_0.o
start address -T 0x9556e8 Finished loading /tmp/gazonk_3012_0.o
#<compiled-function ADDUP2>
NIL
NIL
>>(addup2 1000000)
500000500000
>(addup3 1000000)
Error: ERROR "Invocation history stack overflow."
Fast links are on: do (si::use-fast-links nil) for debugging
Signalled by IF.
ERROR "Invocation history stack overflow."
Broken at +. Type :H for Help.
1 Return to top level.
>>(compile 'addup3)
Compiling /tmp/gazonk_3012_0.lsp.
End of Pass 1.
End of Pass 2.
OPTIMIZE levels: Safety=0 (No runtime error checking), Space=0, Speed=3
Finished compiling /tmp/gazonk_3012_0.lsp.
Loading /tmp/gazonk_3012_0.o
start address -T 0x955a00 Finished loading /tmp/gazonk_3012_0.o
#<compiled-function ADDUP3>
NIL
NIL
>>(addup3 1000000)
Error: ERROR "Value stack overflow."
Hope it helps.
In SBCL User Manual:
The compiler is “properly tail recursive.” [...] The elimination of tail-recursive frames can be prevented by disabling tail-recursion optimization, which happens when the debug optimization quality is greater than 2.
And works as is in the REPL of a fresh image:
(defun sum-no-tail (n)
(if (zerop n)
0
(+ n (sum-no-tail (- n 1)))))
(defun sum-tail (n &key (acc 0))
(if (zerop n)
acc
(sum-tail (- n 1) :acc (+ n acc))))
CL-USER> (sum-no-tail 10000)
50005000 (26 bits, #x2FB0408)
CL-USER> (sum-no-tail 100000)
Control stack guard page temporarily disabled: proceed with caution
; Debugger entered on #<SB-KERNEL::CONTROL-STACK-EXHAUSTED {10026620A3}>
[1] CL-USER>
; Evaluation aborted on #<SB-KERNEL::CONTROL-STACK-EXHAUSTED {10026620A3}>
CL-USER> (sum-tail 100000)
5000050000 (33 bits, #x12A06B550)
CL-USER> (sum-tail 1000000)
500000500000 (39 bits, #x746A5A2920)
CL-USER> (sum-tail 10000000)
50000005000000 (46 bits, #x2D7988896B40)
Hope it helps in SBCL.
I'm a CommonLisp noob with a question. I have these two functions below.
A helper function:
(defun make-rests (positions rhythm)
"now make those positions negative numbers for rests"
(let ((resultant-rhythm rhythm))
(dolist (i positions resultant-rhythm)
(setf (nth i resultant-rhythm) (* (nth i resultant-rhythm) -1)))))
And a main function:
(defun test-return-rhythms (rhythms)
(let ((positions '((0 1) (0)))
(result nil))
(dolist (x positions (reverse result))
(push (make-rests x rhythms) result))))
When I run (test-return-rhythms '(1/4 1/8)), it evaluates to: ((1/4 -1/8) (1/4 -1/8))
However, I expected: (test-return-rhythms '(1/4 1/8)) to evaluate to: ((-1/4 -1/8) (-1/4 1/8)).
What am I doing wrong?
Your implementation of make-rests is destructive.
CL-USER> (defparameter *rhythm* '(1/4 1/4 1/4 1/4))
*RHYTHM*
CL-USER> (make-rests '(0 2) *rhythm*)
(-1/4 1/4 -1/4 1/4)
CL-USER> *rhythm*
(-1/4 1/4 -1/4 1/4)
So, if you run your test, the second iteration will see (-1/4 -1/8), and (make-rests '(0) '(-1/4 -1/8)) returns (1/4 -1/8). Your use of let in make-rests does not copy the list, it just creates a new binding that references it. Use copy-list in your let, or write a non-destructive version in the first place:
(defun make-rests (positions rhythm)
(loop for note in rhythm
for i from 0
collect (if (member i positions) (* note -1) note)))