用于将复杂声明转换为lol，纯文本的实用程序 [英] Utility to convert complex declarations into, well, lol, plain-text
问题描述
是否有一个实用程序采用任意复杂的C语言
声明，检查它的有效性，并将其分解为某些内容
更易理解。
是否有wintel平台的可执行版本？
Is there a utility that takes an arbitrarily complex C language
declaration, checks it validity, and breaks it down into something
more understandable.
Is there a executable version for wintel platforms?
推荐答案
文章< n6 ******** ************************@4ax.com>，
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>是否有一个实用程序采用任意复杂的C语言
声明，检查它的有效性，并将其分解成一些东西
更容易理解。
>Is there a utility that takes an arbitrarily complex C language
declaration, checks it validity, and breaks it down into something
more understandable.
你的意思是旧的''cdecl''练习吗？
You mean something like the old ''cdecl'' exercise?
>有没有wintel平台的可执行版本？
>Is there a executable version for wintel platforms?
你可以自己编译。
http://linux.maruhn.com/sec/cdecl.html
-
" ;重要的是要记住，在法律方面，计算机不会制作副本，只有人类才能制作副本。计算机给出了
命令，而非许可。只有人才能获得许可。
- Brad Templeton
You could compile it yourself.
http://linux.maruhn.com/sec/cdecl.html
--
"It is important to remember that when it comes to law, computers
never make copies, only human beings make copies. Computers are given
commands, not permission. Only people can be given permission."
-- Brad Templeton
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是否有一个实用程序采用任意复杂的C语言
声明，检查它的有效性，并将其分解成某种东西
更容易理解。
是否有wintel平台的可执行版本？
Is there a utility that takes an arbitrarily complex C language
declaration, checks it validity, and breaks it down into something
more understandable.
Is there a executable version for wintel platforms?
这是在FAQ中。
第18节工具和资源
18.1：我需要：A：寻找程序（另见
问题18.16）命名：
a C交叉引用cflow，cxref，调用， cscope，
生成器xscope，或ixfw
a C美化/漂亮，缩进，GNU缩进或
打印机vgrind
a版本控制或CVS，RCS或SCCS
配置管理
工具
a C来源obfusator obfus，寿衣或opqcp
（shrouder）
a" make"依赖makedepend，或尝试cc -M或
generator cpp -M
工具来计算代码ccount，Meter，lcount或csize，
指标或查看网址 http://www.qucis.queensu.ca/
Software-Engineering / Cmetrics.html;
还有一个包销售
由McCabe and Associates
a C源代码这可以做得很好
粗略地反对标准
Unix实用程序wc，和
稍好一点
grep -c" ;;"
a C声明援助检查第14卷
（cdecl ）comp.sources.unix（参见
问题18.16）和K& R2
a原型发电机见问题11.31
a工具追踪
malloc问题请参阅问题18.2
a" selective" C $ / $
预处理器见问题10.18
语言翻译见问题11.31和
工具20.26
C验证器（lint）见问题18.7
a C编译器！请参阅问题18.3
（此工具列表绝不完整;如果您知道
工具未提及，欢迎您联系此列表''
维护者。）
-
我希望有一天能学会阅读。
似乎比写作更难。
- 理查德希思菲尔德
This is in the FAQ.
Section 18. Tools and Resources
18.1: I need: A: Look for programs (see also
question 18.16) named:
a C cross-reference cflow, cxref, calls, cscope,
generator xscope, or ixfw
a C beautifier/pretty- cb, indent, GNU indent, or
printer vgrind
a revision control or CVS, RCS, or SCCS
configuration management
tool
a C source obfuscator obfus, shroud, or opqcp
(shrouder)
a "make" dependency makedepend, or try cc -M or
generator cpp -M
tools to compute code ccount, Metre, lcount, or csize,
metrics or see URL http://www.qucis.queensu.ca/
Software-Engineering/Cmetrics.html ;
there is also a package sold
by McCabe and Associates
a C lines-of-source this can be done very
counter crudely with the standard
Unix utility wc, and
somewhat better with
grep -c ";"
a C declaration aid check volume 14 of
(cdecl) comp.sources.unix (see
question 18.16) and K&R2
a prototype generator see question 11.31
a tool to track down
malloc problems see question 18.2
a "selective" C
preprocessor see question 10.18
language translation see questions 11.31 and
tools 20.26
C verifiers (lint) see question 18.7
a C compiler! see question 18.3
(This list of tools is by no means complete; if you know of
tools not mentioned, you''re welcome to contact this list''s
maintainer.)
--
"I hope, some day, to learn to read.
It seems to be even harder than writing."
--Richard Heathfield
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>是否有一个实用程序需要任意复杂C语言
声明，检查它的有效性，并将其分解为更易理解的内容。
是否有wintel平台的可执行版本？
>Is there a utility that takes an arbitrarily complex C language
declaration, checks it validity, and breaks it down into something
more understandable.
Is there a executable version for wintel platforms?
这是在FAQ中。
第18节工具和资源
This is in the FAQ.
Section 18. Tools and Resources
他想要的是
输入：
y = 1 /（（x + 1）*（x + 2）+ 3）;
输出：
y =（x + 1）*（x + 2）;
$ b $ + = 3;
y = 1 / y;
现在障碍更加明显。我们需要在第3行之前插入一个测试。
-
免费游戏和编程好东西。
http://www.personal.leeds.ac.uk/~bgy1mm
What he wants is something like
input:
y = 1/((x + 1) * ( x + 2) + 3);
output:
y = (x+1) * (x+2);
y += 3;
y = 1/y;
Now the snag is more obvious. We need to insert a test before line 3.
--
Free games and programming goodies.
http://www.personal.leeds.ac.uk/~bgy1mm
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