问题
since I'm a newbie to Common Lisp I tried to solve problems on SPOJ by using Common Lisp (SBCL). The first problem is a simple task of reading numbers until number 42 is found. Here's my solution:
(defun study-num ()
(let ((num (parse-integer (read-line t))))
(when (not (= num 42))
(format t "~A~%" num)
(study-num))))
(study-num)
The solution is accepted. But when I looked into the details of the result I found it used 57M of MEM! It's bloody unreasonable but I can't figure out why. What can I do to make an optimization?
回答1:
You are making repeated recursive calls, without enough optimization switched on to enable tail-call elimination (SBCL does do this, but only when you have "optimize for speed" set high and "optimize for debug info" set low).
The Common Lisp standard leaves tail-call elimination as an implementation quality issue and provides other looping constructs (like LOOP or DO, both possibly suitable for this application).
In addition, a freshly started SBCL is probably going to be larger than you expect, due to needing to pull in its runtime environment and base image.
回答2:
I think yo are not realizing that Common Lisp is an online language environment with full library and compiler loaded into RAM just to give you the first prompt. After that, load in your program is probably a hardly even noticeable increase in size. Lisp does not compile and link an independent executable file made of only your code and whatever lib routines reachable from your code. That's what C and similar languages do. Instead, Lisp adds your code into its already sizeable online environment. As a new user it seams horrible. But if you have a modern general purpose computer with 100's MB of RAM, it quickly becomes something you can forget about as you enjoy the benefits of the online environment. Thins is also called a "dynamic language environment."
回答3:
Various Lisp implementations have different ways to create programs. One is to dump an image of the memory of a Lisp system and to write that to disk. On restart this image is loaded with a runtime and then started again. This is quite common.
This is also what SBCL does when it saves an executable. Thus this executable includes the full SBCL.
Some other implementations are creating smaller executables using images (CLISP), some can remove unused code from executables (Allegro CL, LispWorks) and others are creating very small programs via compilation to C (mocl).
SBCL has only one easy way to reduce the size of an executable: one can compress the image.
来源：https://stackoverflow.com/questions/9001821/why-would-common-lisp-sbcl-use-so-much-memory-for-a-simple-program