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iio: ensure scan index is unique at device register

Having two or more channels with the same positive scan_index field
makes no sense if the device supports buffering.  Prevent this situation
by failing to register such a device.

Signed-off-by: Vlad Dogaru <vlad.dogaru@intel.com>
Reviewed-by: Lars-Peter Clausen <lars@metafoo.de>
Signed-off-by: Jonathan Cameron <jic23@kernel.org>
Vlad Dogaru 10 years ago
parent
commit
8f5d8727a7
1 changed files with 27 additions and 0 deletions
  1. 27 0
      drivers/iio/industrialio-core.c

+ 27 - 0
drivers/iio/industrialio-core.c

@@ -1134,6 +1134,29 @@ static const struct file_operations iio_buffer_fileops = {
 	.compat_ioctl = iio_ioctl,
 };
 
+static int iio_check_unique_scan_index(struct iio_dev *indio_dev)
+{
+	int i, j;
+	const struct iio_chan_spec *channels = indio_dev->channels;
+
+	if (!(indio_dev->modes & INDIO_ALL_BUFFER_MODES))
+		return 0;
+
+	for (i = 0; i < indio_dev->num_channels - 1; i++) {
+		if (channels[i].scan_index < 0)
+			continue;
+		for (j = i + 1; j < indio_dev->num_channels; j++)
+			if (channels[i].scan_index == channels[j].scan_index) {
+				dev_err(&indio_dev->dev,
+					"Duplicate scan index %d\n",
+					channels[i].scan_index);
+				return -EINVAL;
+			}
+	}
+
+	return 0;
+}
+
 static const struct iio_buffer_setup_ops noop_ring_setup_ops;
 
 /**
@@ -1148,6 +1171,10 @@ int iio_device_register(struct iio_dev *indio_dev)
 	if (!indio_dev->dev.of_node && indio_dev->dev.parent)
 		indio_dev->dev.of_node = indio_dev->dev.parent->of_node;
 
+	ret = iio_check_unique_scan_index(indio_dev);
+	if (ret < 0)
+		return ret;
+
 	/* configure elements for the chrdev */
 	indio_dev->dev.devt = MKDEV(MAJOR(iio_devt), indio_dev->id);