传递命令行参数以Clozure Common Lisp

Question

我熟悉Python之前，现在我正在尝试在Windows系统下使用CCL（Clozure Common Lisp）来学习常见的Lisp。

我发现没有一个方便的方式运行LISP文件作为Python。所以我编写一个棒文件来编译并运行一个LISP文件。

@echo off  

set lisp_filename=%~1  
set ccl_path=D:\_play_\lispbox-0.7\ccl-1.6-windowsx86\wx86cl.exe  

IF "%PROCESSOR_ARCHITECTURE%" == "x86" (  
    set fsl_filename=%lisp_filename:.lisp=.wx32fsl%  
) ELSE (  
    set ccl_path=%ccl_path:wx86cl=wx86cl64%  
    set fsl_filename=%lisp_filename:.lisp=.wx64fsl%  
)  

IF NOT EXIST %fsl_filename% goto compile  

for %%a in ("%lisp_filename%") do (  
    set lisp_timestamp=%%~ta  
)  

for %%a in ("%fsl_filename%") do (  
    set fsl_timestamp=%%~ta  
)  

IF "%fsl_timestamp%" LSS "%lisp_timestamp%" (  
    goto compile  
) ELSE (  
    goto run  
)  


:compile  
REM echo "compile"  
%ccl_path% --eval "(progn (compile-file \"%lisp_filename:\=\\%\") (ccl:quit))"  

:run  
REM echo "run"  
%ccl_path% --eval "(progn (load \"%fsl_filename:\=\\%\") (ccl:quit))"  

:end  

.

一切顺利，但我无法找到将命令行参数传递到lisp脚本中。

我尝试了一个像这样的脚本（test.lisp） (defun main() (format t "~{~a~%~}" *command-line-argument-list*) 0 ) (main)
但结果是
D:\_play_\lispbox-0.7\ccl-1.6-windowsx86\wx86cl64.exe
--eval
(progn (load "D:\\_play_\\test.wx64fsl") (ccl:quit))
我知道这个输出是我蝙蝠文件的结果，但我找不到一个优雅的方式来将命令行参数传递到一个lisp脚本，也不是。

谁能告诉我更多关于如何通过参数的事情？ 我希望最好的答案可以实施：
test.lisp a b c
和产出
test.lisp
a
b
c

任何建议都非常欣赏。： - ）

Answer 1

I have get some suggestion from others, which I think it's really useful. I give the result here, hope to be useful for other ones.

CCL stops processing command-line arguments when it encounters a pseudoargument named "--"; any following arguments are then available as the value of CCL:UNPROCESSED-COMMAND-LINE-ARGUMENTS. That value is a list of strings.

console> ccl64 -- these arguments aren\'t processed
Welcome to Clozure Common Lisp Version 1.7-dev-r14704M-trunk (FreebsdX8664)!
? *unprocessed-command-line-arguments*
("these" "arguments" "aren't" "processed")