C/C++ Source Code Obfuscation Show

SourceFormatX is not only the multi-language code formatter but also a code obfuscator that can obfuscate your C and C++ source code for the purposes of source code security and intellectual property protection.

  C/C++ Source Code Obfuscation Examples:     Example 1   |   Example 2   |   Example 3

This is C code obfuscating sample, the purpose of it is to show the power of SourceFormatX C/C++ Code Formatter's syntax parse engine.

  /* Before C/C++ Code Obfuscating */

  #include <stdio.h>
  #include <stdlib.h>

  int main(int a, char** A)
  {
    FILE* B;
    typedef unsigned long C;
    C b[8];
    if (!(a == 7 && (B = fopen(1[A], "rb"))))
      return 1;
    for (7[b] = 0; 7[b] < 5; 7[b]++)
      b[7[b]] = strtoul(A[2 + 7[b]], 0, 16 - !7[b]* 6);
    5[b] = 3[b];
    while ((6[b] = getc(B)) != (C) - 1)
    {
      if (2[b])
        for (7[b] = 0; 7[b] < 4; 7[b]++)
          if (((6[b] >> 7[b]) ^ (6[b] >> (7 - 7[b]))) & 1)
            6[b] ^= (1 << 7[b]) ^ (1 << (7 - 7[b]));
      5[b] ^= 6[b] << (0[b] - 8);
      for (7[b] = 0; 7[b] < 8; 7[b]++)
        if ((5[b] >> (0[b] - 1)) & 1)
          5[b] = (5[b] << 1) ^ 1[b];
        else
          5[b] <<= 1;
    }
    5[b] &= ((((C)1 << (0[b] - 1)) - 1) << 1) | 1;
    if (2[b])
      for (7[b] = 0; 7[b] < (0[b] >> 1); 7[b]++)
        if (((5[b] >> 7[b]) ^ (5[b] >> (0[b] - 1 - 7[b]))) & 1)
          5[b] ^= ((C)1 << 7[b]) ^ ((C)1 << (0[b] - 1 - 7[b]));
    5[b] ^= 4[b];
    fclose(B);
    printf("%0*lX\n", (int)(0[b] + 3) >> 2, 5[b]);
    return 0;
  }

  /* After C/C++ Code Obfuscating */

  #include <stdio.h>
  #include <stdlib.h>
  int main(int a,char**A){FILE*B;typedef unsigned long C;C b[8];if(!(a==7&&(B=
  fopen(1[A],"rb"))))return 1;for(7[b]=0;7[b]<5;7[b]++)b[7[b]]=strtoul(A[2+7[b]],
  0,16-!7[b]*6);5[b]=3[b];while((6[b]=getc(B))!=(C)-1){if(2[b])for(7[b]=0;7[b]<4;
  7[b]++)if(((6[b]>>7[b])^(6[b]>>(7-7[b])))&1)6[b]^=(1<<7[b])^(1<<(7-7[b]));5[b]
  ^=6[b]<<(0[b]-8);for(7[b]=0;7[b]<8;7[b]++)if((5[b]>>(0[b]-1))&1)5[b]=(5[b]<<1)^
  1[b];else 5[b]<<=1;}5[b]&=((((C)1<<(0[b]-1))-1)<<1)|1;if(2[b])for(7[b]=0;7[b]<(
  0[b]>>1);7[b]++)if(((5[b]>>7[b])^(5[b]>>(0[b]-1-7[b])))&1)5[b]^=((C)1<<7[b])^((
  C)1<<(0[b]-1-7[b]));5[b]^=4[b];fclose(B);printf("%0*lX\n",(int)(0[b]+3)>>2,5[b]
  );return 0;}

Example 1   |   Example 2   |   Example 3

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