> I did consider that, but I wrote "in general" for a reason. It works very specifically in the case of "add one" or "subtract one", but it doesn't work with anything more complicated, like chasing pointers or adding/subtracting more than one at a time.

You're reminding me of the book "Modern Compiler Design." The author goes over how to compile a general Pascal-style for-loop correctly, accounting for increasing or decreasing ranges, differing step sizes, and accounting for overflow. It was written using just goto statements, so I adapted a version of it to C. Just replace "intN_t" with an actual integer size. It works by calculating the number of times the loop will run. If "from" is equal to "to," it's still going to run at least once. Again, this is not mine, just adapted from the author's code (Dick Grune's).

  // enumerate: print out numbers from "from" to "to", inclusive, with step size "by"
  void enumerate(intN_t from, intN_t to, intN_t by) {
      uintN_t loop_count;
      intN_t i;
      if (by > 0) {
          if (from > to) return;
          loop_count = (to - from) / by + 1;
      } else if (by < 0) {
          if (from < to) return;
          loop_count = (from - to) / -by + 1;
      } else /* (by == 0) */ {
          // Maybe some kind of error
          return;
      }
      for (i = from; ; i += by) {
          printf("%d\n", i);
          if (--loop_count == 0) break;
      }
  }

You can see it's more complicated than the idiomatic C for-loop, haha. But that's just a general solution. Like you guys noted, it could be simpler for step sizes of 1.