Reminds me of the 'goes to' operator: for (int i = 42; i --> 0;) { foo(); }

lilyball · on March 11, 2012

That's really cute, but it's also buggy. The loop will start at i == 41.

jacquesm · on March 11, 2012

You are right about it starting on 41, but chances are that that is its intended behavior, with a bit of context that is more clear:

  #define ARRAYSIZE 42

  int array[ARRAYSIZE];

  void f() {
     for (i=ARRAYSIZE; i --> 0;) {
       printf("%d\n", array[i]);
     }
  }

If it were to start at 42 you'd be referencing the array outside of its bounds.

TazeTSchnitzel · on March 11, 2012

In C you're used to less-than-one loops though. for(i = 0; i < 5; i++) terminates on 4. And it's what you want if you're going through an array.

delinka · on March 11, 2012

The observant (yeah, newbies aren't so much, typically) will note that it really shouldn't be called "goes to" because it doesn't work the other way:

   for(int i=0; i --> 42;) { bar(); }