[drivers/iio] Question about `iio_gts_build_avail_time_table`

[Chenyuan Yang]

Dear Linux Developers for IIO Driver,

We are curious about the functionality of
`iio_gts_build_avail_time_table`
(https://elixir.bootlin.com/linux/latest/source/drivers/iio/industrialio-gts-helper.c#L363)

```
static int iio_gts_build_avail_time_table(struct iio_gts *gts)
{
  int *times, i, j, idx = 0, *int_micro_times;

  if (!gts->num_itime)
    return 0;

  times = kcalloc(gts->num_itime, sizeof(int), GFP_KERNEL);
  if (!times)
    return -ENOMEM;

  /* Sort times from all tables to one and remove duplicates */
  for (i = gts->num_itime - 1; i >= 0; i--) {
    int new = gts->itime_table[i].time_us;

    if (times[idx] < new) {
      times[idx++] = new;
      continue;
    }

    for (j = 0; j <= idx; j++) {
      if (times[j] > new) {
        memmove(&times[j + 1], &times[j],
                                (idx - j) * sizeof(int));
        times[j] = new;
        idx++;
      }
    }
  ...
}
```

For this function, we are trying to understand how it works but not
sure how this sorting is done.

1. When the gts->itime_table[i].time_us is zero, e.g., the time
sequence is `3, 0, 1`, the inner for-loop will not terminate and do
out-of-bound writes. This is because once `times[j] > new`, the value
`new` will be added in the current position and the `times[j]` will be
moved to `j+1` position, which makes the if-condition always hold.
Meanwhile, idx will be added one, making the loop keep running without
termination and out-of-bound write.
2. If none of the gts->itime_table[i].time_us is zero, the elements
will just be copied without being sorted as described in the comment
"Sort times from all tables to one and remove duplicates".

Please correct us if we miss some key prerequisites for this function
or the data structure.
Thanks in advance!

A possible patch based on our understanding is attached.

Best,
Chenyuan

Attachment: iio.patch
Description: Binary data