main 的三个参数以及其他令人困惑的技巧
以下混淆的 C 代码打印“12 days of Xmas”的单词。
我试图弄清楚它是如何工作的。 我基本上完全迷失了。 初始调用中 main 的三个无类型参数、第一次返回后的一系列字符、调用 main 的负数字参数有何意义? 哎呀!
我主要是在想也许我会学习 C 语言的一些有趣的角落,因此最受欢迎的回复是这样的。
#include <stdio.h>
main(t,_,a)char *a;{return!0<t?t<3?main(-79,-13,a+main(-87,1-_,
main(-86,0,a+1)+a)):1,t<_?main(t+1,_,a):3,main(-94,-27+t,a)&&t==2?_<13?
main(2,_+1,"%s %d %d\n"):9:16:t<0?t<-72?main(_,t,
"@n'+,#'/*{}w+/w#cdnr/+,{}r/*de}+,/*{*+,/w{%+,/w#q#n+,/#{l,+,/n{n+,/+#n+,/#\
;#q#n+,/+k#;*+,/'r :'d*'3,}{w+K w'K:'+}e#';dq#'l \
q#'+d'K#!/+k#;q#'r}eKK#}w'r}eKK{nl]'/#;#q#n'){)#}w'){){nl]'/+#n';d}rw' i;# \
){nl]!/n{n#'; r{#w'r nc{nl]'/#{l,+'K {rw' iK{;[{nl]'/w#q#n'wk nw' \
iwk{KK{nl]!/w{%'l##w#' i; :{nl]'/*{q#'ld;r'}{nlwb!/*de}'c \
;;{nl'-{}rw]'/+,}##'*}#nc,',#nw]'/+kd'+e}+;#'rdq#w! nr'/ ') }+}{rl#'{n' ')# \
}'+}##(!!/")
:t<-50?_==*a?putchar(31[a]):main(-65,_,a+1):main((*a=='/')+t,_,a+1)
:0<t?main(2,2,"%s"):*a=='/'||main(0,main(-61,*a,
"!ek;dc i@bK'(q)-[w]*%n+r3#l,{}:\nuwloca-O;m .vpbks,fxntdCeghiry"),a+1);}
The following obfuscated C code prints the words to the "12 days of Xmas".
I was trying to puzzle out how it works. I'm basically completely lost. What is the significance of the three untyped arguments to main in the initial call, the series of characters after the first return, the negative numeric arguments to the calls to main? Eek!
I'm mostly doing this thinking maybe I'll learn some interesting corners of the C language, so replies in that vein are the most welcome.
#include <stdio.h>
main(t,_,a)char *a;{return!0<t?t<3?main(-79,-13,a+main(-87,1-_,
main(-86,0,a+1)+a)):1,t<_?main(t+1,_,a):3,main(-94,-27+t,a)&&t==2?_<13?
main(2,_+1,"%s %d %d\n"):9:16:t<0?t<-72?main(_,t,
"@n'+,#'/*{}w+/w#cdnr/+,{}r/*de}+,/*{*+,/w{%+,/w#q#n+,/#{l,+,/n{n+,/+#n+,/#\
;#q#n+,/+k#;*+,/'r :'d*'3,}{w+K w'K:'+}e#';dq#'l \
q#'+d'K#!/+k#;q#'r}eKK#}w'r}eKK{nl]'/#;#q#n'){)#}w'){){nl]'/+#n';d}rw' i;# \
){nl]!/n{n#'; r{#w'r nc{nl]'/#{l,+'K {rw' iK{;[{nl]'/w#q#n'wk nw' \
iwk{KK{nl]!/w{%'l##w#' i; :{nl]'/*{q#'ld;r'}{nlwb!/*de}'c \
;;{nl'-{}rw]'/+,}##'*}#nc,',#nw]'/+kd'+e}+;#'rdq#w! nr'/ ') }+}{rl#'{n' ')# \
}'+}##(!!/")
:t<-50?_==*a?putchar(31[a]):main(-65,_,a+1):main((*a=='/')+t,_,a+1)
:0<t?main(2,2,"%s"):*a=='/'||main(0,main(-61,*a,
"!ek;dc i@bK'(q)-[w]*%n+r3#l,{}:\nuwloca-O;m .vpbks,fxntdCeghiry"),a+1);}
如果你对这篇内容有疑问,欢迎到本站社区发帖提问 参与讨论,获取更多帮助,或者扫码二维码加入 Web 技术交流群。
绑定邮箱获取回复消息
由于您还没有绑定你的真实邮箱,如果其他用户或者作者回复了您的评论,将不能在第一时间通知您!
发布评论
评论(2)
有人已经走了并扭转了这一点:http://research.microsoft .com/en-us/um/people/tball/papers/xmasgift/。 请通读一下。 它将解释这一切是如何运作的。
Someone's already gone and reversed this: http://research.microsoft.com/en-us/um/people/tball/papers/xmasgift/. Just read through that. It'll explain how it all works.
一些 Unix 系统不仅将参数计数和参数传递给 main,而且还将向量传递给环境(请参阅 http://en.wikipedia.org/wiki/Main_function_(programming))。 我很确定这就是这个令人困惑的例子所期望的。
我不认为你能从这样混乱的代码中学到很多东西。 它可能是混淆c 竞赛的参赛者。
Some Unix systems do not only pass the argument count and the arguments to main, but also a vector to the environment (see http://en.wikipedia.org/wiki/Main_function_(programming)). I am pretty sure that is what this obfuscated example did expect.
I don't think that you can learn to much from such obfuscated code. It was probably a contestant to the obfuscated c contest.