3

我在http://www.cise.ufl.edu/~manuel/obfuscate/obfuscate.html (http://www.cise.ufl.edu/~manuel/obfuscate/savastio) 网站上看到了这段代码。但是这段代码非常非常疯狂。当它发生时，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，如果我们将看到没有称为“ Enter the number”的语句。但仍然在执行时询问“ Enter the number”。它从哪里来。以及为什么代码是这样写的。它看起来像“ n!”（虽然它意味着阶乘）。请帮我。

#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); }
4

2 回答 2

10

首先，您可以使代码正确缩进并消除由#defines 引起的间接性。缩进可以由GNUindent完成，gcc -E并将进行预处理。假设代码在factorial.c（这需要使用命令行）：

> gcc -E factorial.c | indent > clean_factorial.c

需要注意的是预处理步骤会将所有内容转储stdio.h到clean_factorial.c; #include<stdio.h>但那是无关紧要的信息，所以我们应该在运行之前注释掉/删除gcc -E。这基本上给出了：

struct llll1
{
  struct llll1 *lll1l, *ll1l1;
  unsigned short lll[1000];
};
main ()
{
  struct llll1 *ll11, *l1l, *l1, *ll1l, *malloc ();
  unsigned long l1ll;
  long l11, ll, l;
  unsigned short *lll1, *lllll;
  for (l = 1 - 1; l < 14; putchar ("\t\"8)>l\"9!.)>vl"[l] ^ 'L'), ++l);
  scanf ("%d", &l);
  l1l = malloc (sizeof (struct llll1));
  l1l->lll1l = 1 - 1;
  l1l->ll1l1 = 1 - 1;
  ll11 = malloc (sizeof (struct llll1));
  ll11->lll1l = 1 - 1;
  ll11->ll1l1 = 1 - 1;
  (l1 = l1l)->lll[l1l->lll[1 - 1] = 1] = 0xFFFF;
  for (l11 = 1 + 1; l11 <= l; ++l11)
    {
      l1 = ll11;
      lll1 = (ll1l = (ll11 = l1l))->lll;
      lllll = (l1l = l1)->lll;
      ll = (l1ll = 1 - 1);
      for (; ll1l->lll1l || 0xFFFF != *lll1;)
        {
          l1ll += l11 ** lll1++;
          *lllll++ = l1ll % 10000;
          l1ll /= 10000;
          if (++ll > 1000)
            {
              if (!l1->lll1l)
                {
                  l1->lll1l = malloc (sizeof (struct llll1));
                  l1->lll1l->lll1l = 1 - 1;
                  l1->lll1l->ll1l1 = 1 - 1;;
                  l1->lll1l->ll1l1 = l1;
                }
              lllll = (l1 = l1->lll1l)->lll;
              ll = 1 - 1;
              lll1 = (ll1l = ll1l->lll1l)->lll;
            }
        }
      for (; l1ll;)
        {
          *lllll++ = l1ll % 10000;
          l1ll /= 10000;
          if (++ll >= 1000)
            {
              if (!l1->lll1l)
                {
                  l1->lll1l = malloc (sizeof (struct llll1));
                  l1->lll1l->lll1l = 1 - 1;
                  l1->lll1l->ll1l1 = 1 - 1;;
                  l1->lll1l->ll1l1 = l1;
                }
              lllll = (l1 = l1->lll1l)->lll;
              ll = 1 - 1;
            }
        }
      *lllll = 0xFFFF;
    }
  for (l = (ll = 1 - 1); (l < 1000) && (l1->lll[l] != 0xFFFF); ++l);
  for (; l1; l1 = l1->ll1l1, l = 1000)
    {
      for (--l; l >= 1 - 1; --l, ++ll)
        printf ((ll) ? ((ll % 19) ? "%04d" : (ll =
                                              19, "\n%04d")) : "%4d",
                l1->lll[l]);
    }
  putchar (10);
}

这有点可读性，我们可以做一些事情，比如重命名变量，这样我们就可以很容易地把它们区分开来，即使我们还不知道它们做了什么。例如，这会将结构重命名为structure，并调用其中的指针left和right（您也可以使用编辑器中的查找替换工具来执行此操作）：

> sed -i 's/llll1/structure/g; s/lll1l/left/g; s/ll1l1/right/g;' clean_factorial.c

（例如，您必须注意订单，否则替换lll可能与 冲突lllll）。

还有一些其他简单的事情可以做：

  • 1 - 1经常发生：将其替换为0（以及1 + 1，除了将其替换为2而不是0）。
  • with 的"\t\"8)>l\"9!.)>vl"[l] ^ 'L'行在 xor 之后沿着字符串打印每个字符'L'（找出原因！这可能有助于将 放在putcharfor 循环的主体中，而不是放在逗号语句中）。
  • putchar(10)只是打印一个新行。
  • 了解逗号运算符。

除此之外，这只是很多艰苦的工作。您可以并且应该使用调试器之类的工具来跟踪执行流程并找出发生的情况。

于 2012-04-11T05:36:00.373 回答
2

这是我打开它的尝试，有点卡在提示逻辑上，所以我只是初始化longC为零，并手动打印出对数字的请求：

#include <stdio.h>

struct StructName
{
    struct StructName *structA, *structB;

    unsigned short unsignedShortArrayA[1000];
};

main ()
{
    struct StructName *structC, *structD, *structE, *structF, *malloc();

    unsigned long unsignedLongA;

    long longA, longB, longC = 0;

    unsigned short *unsignedShortA, *unsignedShortB;

    //for(longC=0; longC< 14; putchar("\t\"8)>longC\"9!.)>vl" [longC]^'longC'),++longC )
    //;

    printf("%s", "Enter a number: " );

    scanf("%d", &longC);

    structD = malloc(sizeof(struct StructName));
    structD->structA=0;
    structD->structB=0;

    structC=malloc(sizeof(struct StructName));
    structC->structA=0;
    structC->structB=0;

    (structE=structD)->unsignedShortArrayA[structD->unsignedShortArrayA[0] =1] = 0xFFFF;

    for( longA=1+1;longA<=longC;++longA)
    {
        structE=structC;
        unsignedShortA = (structF=( structC=structD))->unsignedShortArrayA;
        unsignedShortB =( structD=structE)->unsignedShortArrayA;
        longB=(unsignedLongA=0);

        for( ; structF->structA || 0xFFFF != *unsignedShortA; )
        {
            unsignedLongA+=longA**unsignedShortA++;
            *unsignedShortB++=unsignedLongA%10000;
            unsignedLongA/=10000;

            if( ++longB>1000 )
            {
                if( !structE->structA )
                {
                    structE->structA=malloc(sizeof(struct StructName));
                    structE->structA->structA=0;
                    structE->structA->structB=0;

                    structE->structA->structB=structE;
                }

                unsignedShortB=(structE=structE->structA)->unsignedShortArrayA;
                longB=0;
                unsignedShortA=( structF =structF-> structA)->unsignedShortArrayA;
            }
        }

        for( ; unsignedLongA; )
        {
            *unsignedShortB++=unsignedLongA%10000;
            unsignedLongA/=10000;

            if( ++longB>=1000 )
            {
                if( !structE->structA )
                {
                    structE->structA=malloc(sizeof(struct StructName));
                    structE->structA->structA=0;
                    structE->structA->structB=0;

                    structE->structA->structB=structE;
                }

                unsignedShortB=(structE=structE->structA)->unsignedShortArrayA;
                longB=0;
            }
        }

        * unsignedShortB=0xFFFF;
    }

    for( longC=(longB=1- 1); (longC<1000) && (structE->unsignedShortArrayA[ longC] !=0xFFFF); ++longC )
    {
        ;
    }

    for( ; structE; structE=structE->structB, longC=1000 )
    {
        for( --longC; longC>=0; --longC, ++longB)
        {
            printf( (longB)?((longB%19) ? "%04d" : (longB=19,"\n%04d") ):"%4d",structE-> unsignedShortArrayA[longC] );
        }
    }

    putchar(10);
}

编辑：清理了一下。

于 2012-04-11T05:42:21.533 回答