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From: schmidt@bonnie.ics.uci.edu (Douglas C. Schmidt)
Newsgroups: comp.lang.c
Subject: Explanation, please!
Message-ID: <638@paris.ICS.UCI.EDU>
Date: 25 Aug 88 02:16:14 GMT
Sender: news@paris.ics.uci.edu
Reply-To: schmidt@bonnie.ics.uci.edu (Douglas C. Schmidt)
Organization: University of California, Irvine - Dept. of ICS
Lines: 36
References:

The following piece of wonderful obscurity comes from Stroustup's
C++ Programming Language book, page 100:

void send(int *to,int *from, int count) {
   int n = (count + 7) / 8;
   
   switch(count % 8) {
      case 0:  do { *to++ = *from++;
      case 7:       *to++ = *from++;
      case 6:       *to++ = *from++;
      case 5:       *to++ = *from++;
      case 4:       *to++ = *from++;
      case 3:       *to++ = *from++;
      case 2:       *to++ = *from++;
      case 1:       *to++ = *from++;
               } while (--n > 0);
   }

}

Now, much to my surprise, this is not only valid C++, it is also valid C!
Could some one please explain to me why this is so?  It seems like
the case 7-1 labels are actually nested inside the do {} while loop,
and thus not in the scope of the switch (should a break statement exit
both the switch and the loop, or just one?!?!).

Finally, Stroustrup asks the rhetorical question ``why would anyone
want to write something like this.''  Any guesses?!

thanks,

   Doug Schmidt
--
Douglas Schmidt
"If our behavior is strict, we do not need fun." -Zippy th' Pinhead
"If our behavior is struct, we do not need defun." -Anon