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!
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
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; }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
0or69depending upon your preferences.
If there was ever a time to replace this Drake format with the Geordi LaForge alternative, this was it.
On it.
return 69
; }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!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.
The code is even wink at u
