ulterno

I am starting to understand why stealth games portray guards as easily head-flippable.
It's because they already have it halfway there for the player.

Hey, if the grid is fine enough...

When Google Navigation tells you to "head North-East".

I am a fool who bought an Android phone without a magnetic compass.

I tried the rubber band. I tried the clip. Neither work.
Only the fridge does. And that works well enough. I either tuck it, or I take it all out and keep it on a tray. Open. If I keep it for long enough to make it dehydrated, it's my fault.

OIC. Good to know in case I ever have to work on some old CentOS 5 box lying around ever again.
It also looks kinda proper, using that instead of the @, so when making shell scripts, I might want to prefer this.

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.

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.

Some kind of Caesar cipher you made?

fIy uo rolevl teet rsi'n tigev nnit ehf ro mfoh gilh yboufcstadeC ,sii terlayla l vo eelttre ? Iod'n tnkwo ,ub thwtaI d onkwoi shttaI l vo eoy!u< 3%

And people say pointers are hard.

What if it's optikal?

You get all Lemmy results! Yaay!

Sorry, that's really all there's to it.

Try searching Google for "Saganumenousness"