this post was submitted on 07 Sep 2024
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The best love letters are written in C (apologies to the Rust fans) (fedia.io)
submitted 1 week ago by [email protected] to c/[email protected]
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[–] [email protected] 65 points 1 week ago (2 children)

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!

[–] [email protected] 44 points 1 week ago* (last edited 1 week 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

[–] [email protected] 8 points 1 week ago* (last edited 1 week ago) (2 children)

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;
}
[–] [email protected] 9 points 1 week ago (1 children)

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.

[–] [email protected] 5 points 1 week 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.

[–] [email protected] 1 points 1 week ago (1 children)

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%

[–] [email protected] 3 points 1 week ago* (last edited 1 week ago)

Whoops! When I looked at the second time that the shift value is calculated, I wondered if it would be inverted from the first time, but for some reason I decided that it wouldn't be. But looking at it again it's clear now that (1 - i) = (-i + 1) = ((~i + 1) + 1), making bit 0 the inverse. Then I wondered why there wasn't more corruption and realized that the author's compiler must perform postfix increments and decrements immediately after the variable is used, so the initial shift is also inverted. That's why the character pairs are flipped, but they still decode correctly otherwise. I hope this version works better:

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;
}

EDIT: I just got a chance to compile it and it does work.

[–] [email protected] 33 points 1 week ago (1 children)

If there was ever a time to replace this Drake format with the Geordi LaForge alternative, this was it.

[–] [email protected] 69 points 1 week ago

On it.

[–] [email protected] 21 points 1 week ago* (last edited 1 week ago)

return 69
; }

[–] [email protected] 15 points 1 week ago (1 children)

As a rust fan I can say:

Don't be sorry, cause the code is in C!
Be sorry for what you did in C!

[–] [email protected] 0 points 1 week ago* (last edited 1 week ago) (1 children)

I mean, of course it wouldn't be fair to expect everyone to be migrated over yet, but at some point it's going to be an obsolete language. Memory unsafety is a pretty nasty quirk; just one that was previously unavoidable, as far as I know.

[–] [email protected] 6 points 1 week ago (1 children)

at some point it’s going to be an obsolete language.

Yeah, COBOL went the way of the dodo too.

[–] [email protected] 1 points 1 week ago* (last edited 1 week ago) (1 children)

Exactly. COBOL still gets used in legacy stuff, but at this point you'd have to be either insane or a historical re-enactor to build something new in it.

I've used C more than anything else, for reference.

[–] [email protected] 1 points 1 week ago (1 children)

COBOL is on my to-learn list...

[–] [email protected] 1 points 1 week ago

I really want to build and learn to use a medieval-style pole lathe.

[–] [email protected] 7 points 1 week ago

The code is even wink at u