问题描述
由于我是Common Lisp的新手，因此我尝试使用Common Lisp解决上的问题（ ）。 是读取数字直到找到数字42的简单任务。这是我的解决方案：
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)
该解决方案被接受。但是，当我查看结果的详细信息时，发现它使用了57M的MEM！血腥的不合理，但我不知道为什么。我该怎么做才能进行优化？
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?
推荐答案
您正在重复进行递归调用，而没有启用足够的优化以启用tail-消除呼叫（SBCL会执行此操作，但仅当您将针对速度优化设置为高，而将针对调试信息优化设置为低时）。
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).
Common Lisp标准将消除尾音作为实现质量问题，并提供了其他循环构造（例如LOOP或DO，两者都可能适用于此应用程序）。
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).
此外，由于需要引入运行时环境和基本映像，因此新启动的SBCL可能会比您预期的大。
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.
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