Explanation, someone??

[Trent Tobler]

>From article <33641 at usc.edu>, by burzin at skat.usc.edu (Burzin N. Engineer):
> Hi,
> 	I just gave an interview which had a C quiz and there was this program
> that really confused me. I have compiled the program and can still not figure
> out what is going on. Any reference or help will be appreciated.
> ---
> char *c[] = {
>   "ENTER",
>   "NEW",
>   "POINT",
>   "FIRST"
>   };
> char **cp[] = { c+3, c+2, c+1, c};

translation to something humans can grasp (not really valid C code):

cp[] = { &"FIRST", &"POINT", &"NEW", &"ENTER" };

> char ***cpp = cp;
> main()
> {
>   printf("%s", *++cpp);

This statement may cause problems, because *++cpp is not a (char *),
it is a (char **), so whatever the address of "POINT" is gets interpreted
as a array of characters.  Depends on the compiler/machine what happens.
Anyway, if the machine doesn't trap it, it does have the effect of setting
cpp to &cp[1].

>   printf("%s", *--*++cpp+3);

  cpp now points to &cp[2].  cp[2] gets decremented (set to c+0, or &"ENTER"),
  which we dereference to "ENTER", and then add 3 to get "ER", which is
  printed.

>   printf("%s", *cpp[-2]+3);

  cpp[-2] is cp[0], equal to &"FIRST", which is dereferenced to give "FIRST",
  we add 3 and get "ST", which is printed.

>   printf("%s\n", cpp[-1][-1]+1 );

  cpp[-1] is cp[1], equal to c + 2 (&"POINT"), which we dereference after
  subtracting one giveing c[1], or "NEW".  Add one to get "EW", and print
  it, followed by a line-feed.

> }
> 
> When run it produces:
> ERSTEW

Yes, that is the result, if we ignore what happens in the first printf
statement.


> =========================================
> Another one is :
> char input[] = "SSSWILTECH1\1\11W\1WALLMP1";
> main()
> {
>   int i, c ;
> 
> for (i=2; (c=input[i]) != '\0'; i++){
>   switch(c) {
>   case 'a': putchar('i'); continue;
>   case '1': break;
>   case 1: while((c = input[++i]) != '\1' && c!='0' );
>   case 9: putchar('S');
>   case 'E':
>   case 'L': continue;
>   default: putchar(c);
>            continue;
>   }
>   putchar(' ');
> }
> putchar('\n');
> }

Because this has a loop, I will roll it out, and show only the important
parts:
__i__  __c__   __switch_action__
  2     'S'     prints an 'S', restarts loop
  3     'W'     prints a 'W', restarts loop
  4     'I'     prints an 'I', restarts loop
  5     'L'     restarts loop
  6     'T'     prints a 'T', restarts loop
  7     'E'     restarts loop
  8     'C'     prints a 'C', restarts loop
  9     'H'     prints an 'H', restarts loop
  10    '1'     breaks out of case, and prints a ' '.
  11    1       begin a while loop
        9          i gets incremented; test succeeds.
        'W'        i gets incremented; test succeeds
        1          i gets incremented; test fails
                prints an 'S'. restarts loop.
  15    'W'     prints a 'W', restarts loop.
  16    'A'     print an 'A', restart loop.
  17    'L'     restart loop
  18    'L'     restart loop
  19    'M'     prints an 'M', restarts loop.
  20    'P'     prints a 'P', restarts loop.
  21    '1'     breaks out of case, and prints a ' '.
  22    0
 at this point, the for loop test fails, and we drop out of the loop, print
 a line feed, and exit.

> 
> which produces:
> SWITCH SWAMP

Yep, that's what I got too.

> 
> ...and to think I thought I knew C.
> I would sincerely appreciate some explanation. Thank You.

Yes, these are quite obscure.  I would wonder why they gave such a test.
Even very experience programmers could get caught up in the complexity and
flounder around.  Also, were you told that the first one will print
what it did, because it is very dependant on the compiler and machine, and
could very possibly cause an illegal reference.

--
  Trent Tobler - ttobler at csulx.weber.edu