• ulterno@lemmy.kde.socialEnglish
    9·
    2 months 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
      4·
      2 months 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.