• LostXOR@fedia.ioOP
    1·
    12 days ago

    If anyone wants to actually run this, here ya go:

              #include              <stdio.h>
      short i=0;long          b[]={1712,6400
    ,3668,14961,00116,      13172,10368,41600,
  12764,9443,112,12544,15092,11219,116,8576,8832
,12764,9461,99,10823,17,15092,11219,99,6103,14915,
69,1721,10190,12771,10065,16462,13172,10368,11776,
14545,10460,10063,99,12544,14434,16401,16000,8654,
12764,13680,10848,9204,113,10441,14306,9344,12404,
  32869,42996,12288,141129,12672,11234,87,10086,
    12655,99,22487,14434,79,10083,12750,10368,
      10086,14929,79,10868,14464,12357};long
        n=9147811012615426336;long main(){
          if(i<0230)printf("%c",(char)((
            0100&b[i++>>1]>>(i--&0x1)*
              007)+((n>>(b[i>>001]>>
                7*(0b1&01-i++)))&1
                  *main(111))));
                    return 69-
                      0b0110
                        ;}

    Bonus points if you can deobfuscate it!

    • Redkey@programming.dev
      1·
      12 days ago

      I don’t know if this will work or even compile, but I feel like I’m pretty close.

      long main () {
    char output;
    unsigned char shift;
    long temp;
    
    if (i < 152) {
        shift = (i & 1) * 7;
        temp = b[i >> 1] >> shift;
        i++;
        output = (char)(64 & temp);
        output += (char)((n >> (temp & 63)) & main());
        printf("%c", output);
    }

    return 63;
}
      • ulterno@lemmy.kde.socialEnglish
        1·
        12 days ago

        Here’s it with some amount of de-obfuscation:

        #include <stdio.h>
short i = 0;
const long b[]
	= { 0xd60,  0x3200,  0x1ca8, 0x74e2, 0x9c,   0x66e8, 0x5100,  0x14500,
		0x63b8, 0x49c6,  0xe0,   0x6200, 0x75e8, 0x57a6, 0xe8,    0x4300,
		0x4500, 0x63b8,  0x49ea, 0xc6,   0x548e, 0x22,   0x75e8,  0x57a6,
		0xc6,   0x2fae,  0x7486, 0x8a,   0xd72,  0x4f9c, 0x63c6,  0x4ea2,
		0x809c, 0x66e8,  0x5100, 0x5c00, 0x71a2, 0x51b8, 0x4e9e,  0xc6,
		0x6200, 0x70c4,  0x8022, 0x7d00, 0x439c, 0x63b8, 0x6ae0,  0x54c0,
		0x47e8, 0xe2,    0x5192, 0x6fc4, 0x4900, 0x60e8, 0x100ca, 0x14fe8,
		0x6000, 0x44e92, 0x6300, 0x57c4, 0xae,   0x4ecc, 0x62de,  0xc6,
		0xafae, 0x70c4,  0x9e,   0x4ec6, 0x639c, 0x5100, 0x4ecc,  0x74a2,
		0x9e,   0x54e8,  0x7100, 0x608a };
const long n = 9147811012615426336;
long
main ()
{
	if (i < 152)
	{
		char shifter;
		if (i % 2 == 0)
		{
			shifter = 8;
		}
		else
		{
			shifter = 1;
		}
		char adder1 = (b[i >> 1] >> shifter) & 64;

		char adder2 = (n >> (b[i >> 1] >> shifter)) & 63;

		char to_print = (char)adder1 + adder2;
		i++;
		main ();
		printf ("%c", to_print);
	}
	return 63;
}

        Needless to say, the return value doesn’t matter any more. So you can change it to 0 or 69 depending upon your preferences.

        • ulterno@lemmy.kde.socialEnglish
          1·
          12 days ago

          And more de-obf:

          #include <stdio.h>

const char addarr1[]
	= { 0x40, 0x40, 0x0,  0x40, 0x40, 0x40, 0x40, 0x0,  0x40, 0x40, 0x40, 0x40,
		0x0,  0x40, 0x40, 0x40, 0x40, 0x40, 0x40, 0x0,  0x40, 0x40, 0x40, 0x0,
		0x40, 0x0,  0x40, 0x40, 0x40, 0x40, 0x40, 0x0,  0x40, 0x40, 0x0,  0x40,
		0x40, 0x40, 0x0,  0x40, 0x40, 0x40, 0x40, 0x0,  0x40, 0x40, 0x0,  0x40,
		0x40, 0x40, 0x40, 0x40, 0x40, 0x0,  0x40, 0x40, 0x40, 0x40, 0x40, 0x40,
		0x40, 0x40, 0x40, 0x40, 0x0,  0x40, 0x0,  0x0,  0x40, 0x40, 0x0,  0x40,
		0x40, 0x0,  0x40, 0x40, 0x40, 0x40, 0x40, 0x40, 0x0,  0x40, 0x0,  0x40,
		0x40, 0x40, 0x40, 0x0,  0x40, 0x40, 0x40, 0x40, 0x40, 0x40, 0x0,  0x0,
		0x40, 0x0,  0x40, 0x40, 0x40, 0x0,  0x40, 0x0,  0x40, 0x40, 0x40, 0x40,
		0x0,  0x0,  0x40, 0x40, 0x40, 0x0,  0x40, 0x40, 0x40, 0x40, 0x0,  0x40,
		0x0,  0x40, 0x40, 0x0,  0x40, 0x40, 0x40, 0x40, 0x0,  0x40, 0x40, 0x0,
		0x40, 0x40, 0x40, 0x40, 0x0,  0x40, 0x0,  0x40, 0x40, 0x40, 0x40, 0x0,
		0x40, 0x40, 0x40, 0x0,  0x0,  0x0,  0x0,  0x0 };

const char addarr2[]
	= { 0x9,  0x26, 0x20, 0x39, 0x2f, 0x35, 0x32, 0x20, 0x2c, 0x2f, 0x36, 0x25,
		0x20, 0x2c, 0x25, 0x34, 0x34, 0x25, 0x32, 0x20, 0x29, 0x33, 0x2e, 0x27,
		0x34, 0x20, 0x27, 0x29, 0x36, 0x25, 0x2e, 0x20, 0x29, 0x2e, 0x20, 0x34,
		0x28, 0x25, 0x20, 0x26, 0x2f, 0x32, 0x2d, 0x20, 0x2f, 0x26, 0x20, 0x28,
		0x29, 0x27, 0x28, 0x2c, 0x39, 0x20, 0x2f, 0x22, 0x26, 0x35, 0x33, 0x23,
		0x21, 0x34, 0x25, 0x24, 0x20, 0x3,  0x2c, 0x20, 0x29, 0x33, 0x20, 0x29,
		0x34, 0x20, 0x32, 0x25, 0x21, 0x2c, 0x2c, 0x39, 0x20, 0x21, 0x20, 0x2c,
		0x2f, 0x36, 0x25, 0x20, 0x2c, 0x25, 0x34, 0x34, 0x25, 0x32, 0x3f, 0xa,
		0x9,  0x20, 0x24, 0x2f, 0x2e, 0x27, 0x34, 0x20, 0x2b, 0x2e, 0x2f, 0x37,
		0x2c, 0x20, 0x22, 0x35, 0x34, 0x20, 0x37, 0x28, 0x21, 0x34, 0x20, 0x9,
		0x20, 0x24, 0x2f, 0x20, 0x2b, 0x2e, 0x2f, 0x37, 0x20, 0x29, 0x33, 0x20,
		0x34, 0x28, 0x21, 0x34, 0x20, 0x9,  0x20, 0x2c, 0x2f, 0x36, 0x25, 0x20,
		0x39, 0x2f, 0x35, 0x21, 0x20, 0x3c, 0x33, 0xa };

int main ()
{
	for (int i = 0; i < 152; i++)
	{
		char adder1 = addarr1[i];

		char adder2 = addarr2[i];

		char to_print = (char)adder1 + adder2;

		printf ("%c", to_print);
	}
	return 63;
}

          I guess I should have kept the recursion and straightened it out in the next step, but now that it’s done…

          The next step will just have an array of the characters that would be printed, so I’ll leave it here.

    • 1stTime4MeInMCU@mander.xyzEnglish
      1·
      12 days ago

      If your love letter isn’t given in the form of highly obfuscated C, is it really a love letter? I don’t know, but what I do know is that I love you! <3