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