Thread: Using an if in a for loop

Hi, everyone. I want to conditionally increment a value within a for loop, like this:

Code:

  for (attempt = 1; attempt <= uinumberofattempts || h->prefs.persevere;
       if (attempt) attempt++)

However, using a statement instead of an expression is not allowed. This is what I've come up with. Is this acceptable or hacky? The voids are necessary to suppress clang compiler warnings about unused values.

Code:

  for (attempt = 1; attempt <= uinumberofattempts || h->prefs.persevere;
       attempt ? (void) attempt++ : (void) 0)

Originally Posted by CommonTater

Code:

attempt = 1;
while ((attempt <= uinumberofattempts) || (h-> perfs.persevere))
  { // do your stuff here
     if ( whatevercondition )
       attempt++; }

using... if (attempt) attempt++ ... does nothing. It just increments the counter since attempt is never 0.

My idea is that attempt will reach 0 when it overflows. Then it stays at 0 and lower down I have switch (attempt) that will print out "We lost count" instead of "%d", attempt.

Originally Posted by CommonTater

Go back and look at my post... I did a little re-edit.

Working like that it will *never* overflow.

Ah, but I want it to count the attempts and print them to screen even if the 'persevere' option is set.

And the attempt++ thing - I want it to be in the control statement because there are several 'continue's within the loop. This way I know it's going to execute every iteration.

Ok ... well, have fun with that.

Behold my masterpiece:

Code:

  unsigned int attempt;

  for (attempt = 1; attempt <= uinumberofattempts || h->prefs.persevere;
       attempt ? (void) attempt++ : (void) 0)
    {
      switch (attempt)
	{
	case 0:
	      printf ( "We've lost count of the number of tries.\n") 
	  break;
	case 1:
	 printf ("Attempting to login to Wikipedia.\n");
	  break;
	default:
	printf ( "Attempting to login to Wikipedia (try #%d).\n",
	       attempt);
	  break;
	}
    }

And you win the award for the weirdest code I've ever seen.

Originally Posted by CommonTater

And you win the award for the weirdest code I've ever seen.

Thanks, Tater. I'm delighted that you have honoured me in this way.

Originally Posted by Richardcavell

Thanks, Tater. I'm delighted that you have honoured me in this way.

For the 4th month in a row....

Code:

 for (attempt = 1; attempt <= uinumberofattempts || h->prefs.persevere;
       attempt += attempt > 0)

And I assume 'attempt' would have a chance at getting set to zero somewhere in the loop.

Originally Posted by nonoob

Code:

 for (attempt = 1; attempt <= uinumberofattempts || h->prefs.persevere;
       attempt += attempt > 0)

And I assume 'attempt' would have a chance at getting set to zero somewhere in the loop.

That's the problem... from his previous posts, it would appear he intends to just let it overflow and hope it wraps around to 0.

Originally Posted by nonoob

Gimme a break. Nothing will happen. Check the assembly code. I know of no machine architecture that generates an interrupt on overflow.

Who said anything about interrupts?

What happens when you overflow a signed int? Does it go to 0, -1, -INT_MAX ??? I should think that's pretty much implementation specific behavior which may not be the same from one machine to the next. That is... you can't trust it.

Why not just have the loop stop at INT_MAX? It's one number before overflow, and that hardly matters, since we're talking about 2 billion login attempts to Wikipedia, which will probably blacklist you as spam/DoS attack long before then.