Ioccc best one liner

This is a brief analysis of korn.c, winner of the 1987 IOCCC. Why? Because. And because many people who shot some explanations have many flaws and conceptual mistakes.

#include <stdio.h>
/*
* Review of one of the best IOCCC one liner winners.
* See International Obfuscated C Code Contest (http://www.ioccc.org)
* ...found other reviews with conceptual mistakes, so I wrote mine...
*
* Winners:  http://www.ioccc.org/years.html#1987
* The code: http://www.ioccc.org/1987/korn.c
* Hints:    http://www.ioccc.org/1987/korn.hint
*
* Rodolfo Alcazar Portillo <rodolfoap@gmail.com>
*/

main() {
/* a simple text */
printf("%s\n","unix");

/* or... */
printf("%six\n","un");

/* adding an unnecesary C text terminator does not change output */
printf("%six\n\0","un");

/* replace \n==\012 which is a printable form of an octal value */
printf("%six\012\0","un");

/* the additional X won't print due to +1 offset */
printf("X%six\012\0"+1,"un");

/* nor any other character */
printf("\021%six\012\0"+1,"un");

/* rephrasing syntax as a text array */
printf(&"\021%six\012\0"[1],"un");

/* And the preprocessor said "Let unix==1". Verification
*      $ cpp -dM /dev/null | grep unix
*   #define __unix__ 1
*   #define __unix 1
*   #define unix 1
*
* Therefore it won't run on Windows unless compiled with cygwin or defined in preprocessor as:
* #define unix 1
*
* ...and therefore redefining the unix variable would raise an error.
*/
printf(&"\021%six\012\0"[unix],"un");

/* &text[n]==&n[text] is basic pointer arithmetics, or addition communitivity */
printf(&unix["\021%six\012\0"],"un");

/* 'a'==0x61 then ('a'-0x61)==0 */
printf(&unix["\021%six\012\0"],"un"+'a'-0x61);

/* Then add any char -X- and a offset of 1 (=='a'-0x60) */
printf(&unix["\021%six\012\0"],"Xun"+'a'-0x60);

/* use a double-quoted string, and its offset, which gives an integer value: 'a'=="a"[0] */
printf(&unix["\021%six\012\0"],"Xun"+"a"[0]-0x60);

/* or the same, with an arbitrary text and offset */
printf(&unix["\021%six\012\0"],"Xun"+"XaXX"[1]-0x60);

/* again, replace 1 with the predefined preprocessor unix==1 value...  */
printf(&unix["\021%six\012\0"],"Xun"+"XaXX"[unix]-0x60);

/* rephrasing, array pointer arithmetics, addition is communitive */
printf(&unix["\021%six\012\0"],"Xun"+(unix)["XaXX"]-0x60);

/* ditto */
printf(&unix["\021%six\012\0"],(unix)["XaXX"]+"Xun"-0x60);

/* at last, change ignored chars to something meaningful */
printf(&unix["\021%six\012\0"],(unix)["have"]+"fun"-0x60);
}