Let's Disect

Code:

/* * christmas.c * a program that prints out the twelve days of chrismas * in clear and obvious fashion * * adopted from the obfuscated version by Michael Bayne <[email protected]> * and Paul Phillips <[email protected]> */ #include <stdio.h> #define babble1 "@n'+,#'/*{}w+/w#cdnr/+,{}r/*de}+,/*{*+,/w\ {%+,/w#q#n+,/#{l+,/n{n+,/+#n+,/#;#q#n+,/+k#;*+,/'r :'d*'3,}\ {w+K w'K:'+}e#';dq#'l q#'+d'K#!/+k#;q#'r}eKK#}w'r}eKK\ {nl]'/#;#q#n'){)#}w'){){nl]'/+#n';d}rw' i;# ){nl]!/n{n#'; r{#w'r \ nc{nl]'/#{l,+'K {rw' iK{;[{nl]'/w#q#n'wk nw' iwk{KK{nl]!/w\ {%'l##w#' i; :{nl]'/*{q#'ld;r'}{nlwb!/*de}'c ;;{nl'-{}rw]'/+,}\ ##'*}#nc,',#nw]'/+kd'+e}+;#'rdq#w! nr'/ ') }+}{rl#'{n' ')#}'+}\ ##(!!/" #define babble2 "!ek;dc i@bK'(q)-[w]*%n+r3#l,{}:\nuwloca-\ O;m .vpbks,fxntdCeghiry" int main (int arg1, int arg2, char* arg3) { /* if (!0 < arg1) { */ if (arg1 > 1) { int rv; /* arg1 == 2, start printing day #arg2 */ /* if (arg1 < 3) { */ if (arg1 == 2) { int rv1, rv2; /* prints out 'On the ' */ /* arg3 ignored, results in call to main(0, -86, babble1) */ rv1 = main(-86, 0, arg3 + 1); /* prints out the ordinal date (ie. tenth) */ /* arg3 ignored, results in call to main(1-day, -87, babble1) */ rv2 = main(-87, 1 - arg2, rv1 + arg3); /* prints out 'day of Christmas my true love gave to me' */ /* arg3 ignored, results in call to main(-13, -79, babble1) */ main(-79, -13, arg3 + rv2); } /* else 1; */ /* here we stack up a bunch of calls to ourself in 'phrase printing' mode to print out each successive phrase (partridge in a pear tree, and, two turtle doves). each call prints out the successively previous phrase, and we keep recursing until we've made one call for each day */ if (arg1 < arg2) { main(arg1 + 1, arg2, arg3); } /* else 3; */ /* this call causes us to print out the (-27 + arg1) th phrase in the phrase buffer, there are 26 phrases, so arg1 varies from 2 to 13 in order to print out each of the day's' gifts */ /* arg3 ignored, results in call to main(-27 + arg1, -94, babble1) */ rv = main(-94, -27 + arg1, arg3); /* rv is always 16 (hence true). see below where we return 16 instead of advancing to the next day */ /* we don't execute this on the stacked up recursive calls made above (where we pass in arg1 values from 3 to the number of the current day), but only on the primary call (where we passed in 2), because this code causes us to write out the next day */ if (rv && (arg1 == 2)) { /* here's where we decide how many days to print */ if (arg2 < 13) { /* 2 means we're a main recursive call that does a new day arg2 + 1 increments the day "%s %d %d" is to confuse us (it's ignored) */ return main(2, arg2 + 1, "%s %d %d\n"); } else { /* this is the return value of the command -- echo $status after running this [:-)] */ return 9; } } else { /* this is the return value of the above described recursive calls, must be non-zero in original code because its return value is checked */ return 16; } } else { if (arg1 < 0) { if (arg1 < -72) { /* this is the stealthy switch of the first two arguments and replacement of argument 3 with a pointer to the word buffer */ return main(arg2, arg1, babble1); } else { /* if we're here, we want to translate and print out a character */ if (arg1 < -50) { /* if arg2 (the untranslated character we want to print) is equal to *arg3 (some character in the translation buffer */ if (arg2 == *arg3) { /* if so, we're all set, print out the translated version of the character (which is 31 characters up the translation table) */ return putchar(arg3[31]); } else { /* if not, call ourselves with the printout magic number, the same untranslated character and the next character up in the translation table */ return main(-65, arg2, arg3 + 1); } /* if we're here, we want to print out a phrase. we start out with arg3 pointing to babble1, and arg1 being -(the number of phrases to skip before printing out a phrase), we increment the pointer, incrementing arg1 each time we pass a slash, so when arg1 equals zero, we've skipped the appropriate number of slashes (phrases) and we end up down in the phrase printing part of the code */ } else { /* pushes babble1 pointer forward (skipping slashes) and incrementing arg1 (every time a slash is seen) until arg1 == 0. this positions a pointer in the babble2 buffer at the beginning of the -arg1 th (slash-seperated) string */ return main((*arg3 == '/') + arg1, arg2, arg3+1); } } } else { /* if (0 < arg1) { */ if (arg1 == 1) { /* this is the original call, since we are originally called with arg1 == 1 (argc) */ return main(2, 2, "%s"); } else { /* arg1 == 0 */ /* we've got a correctly positioned pointer into babble2 in arg3 */ if (*arg3 == '/') { /* we've hit the end of the phrase that we wanted to translate and print */ return 1; } else { /* -61 could be any number greater than -72 but less than -50 to cause us to go into printing out mode *arg2 is a pointer to the untranslated character that we want to print */ int rv = main(-61, *arg3, babble2); /* we call ourselves again with the next character in the word buffer, we'll end up right back here until we hit a '/' at which point, we'll return (1) all the way up */ return main(0, rv, arg3+1); } } } } }

i got this from:

http://www.cs.rit.edu/usr/local/pub/...09/christmas.c