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无法理解混淆的 C 代码

转载 作者:太空狗 更新时间:2023-10-29 15:49:13 24 4
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我在 http://www.cise.ufl.edu/~manuel/obfuscate/obfuscate.html 中看到了这段代码(http://www.cise.ufl.edu/~manuel/obfuscate/savastio) 网站。但是这段代码非常疯狂。我无法理解正在发生的事情。请帮我。让我知道每行执行时发生了什么。例如printf 函数在那里,如果我们看到没有名为“Enter the number”的语句。但仍然在执行时询问“Enter the number”。它来自哪里。还有为什么代码要这样写。它看起来像“n!”(尽管它表示阶乘)。请帮助我。

#include <stdio.h>

#define l11l 0xFFFF
#define ll1 for
#define ll111 if
#define l1l1 unsigned
#define l111 struct
#define lll11 short
#define ll11l long
#define ll1ll putchar
#define l1l1l(l) l=malloc(sizeof(l111 llll1));l->lll1l=1-1;l->ll1l1=1-1;
#define l1ll1 *lllll++=l1ll%10000;l1ll/=10000;
#define l1lll ll111(!l1->lll1l){l1l1l(l1->lll1l);l1->lll1l->ll1l1=l1;}\
lllll=(l1=l1->lll1l)->lll;ll=1-1;
#define llll 1000




l111 llll1 {
l111 llll1 *
lll1l,*ll1l1 ;l1l1 lll11 lll [
llll];};main (){l111 llll1 *ll11,*l1l,*
l1, *ll1l, * malloc ( ) ; l1l1 ll11l l1ll ;
ll11l l11,ll ,l;l1l1 lll11 *lll1,* lllll; ll1(l
=1-1 ;l< 14; ll1ll("\t\"8)>l\"9!.)>vl" [l]^'L'),++l
);scanf("%d",&l);l1l1l(l1l) l1l1l(ll11 ) (l1=l1l)->
lll[l1l->lll[1-1] =1]=l11l;ll1(l11 =1+1;l11<=l;
++l11){l1=ll11; lll1 = (ll1l=( ll11=l1l))->
lll; lllll =( l1l=l1)->lll; ll=(l1ll=1-1
);ll1(;ll1l-> lll1l||l11l!= *lll1;){l1ll
+=l11**lll1++ ;l1ll1 ll111 (++ll>llll){
l1lll lll1=( ll1l =ll1l-> lll1l)->lll;
}}ll1(;l1ll; ){l1ll1 ll111 (++ll>=llll)
{ l1lll} } * lllll=l11l;}
ll1(l=(ll=1- 1);(l<llll)&&
(l1->lll[ l] !=l11l);++l); ll1 (;l1;l1=
l1->ll1l1,l= llll){ll1(--l ;l>=1-1;--l,
++ll)printf( (ll)?((ll%19) ?"%04d":(ll=
19,"\n%04d") ):"%4d",l1-> lll[l] ) ; }
ll1ll(10); }

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