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iio: ensure scan index is unique at device register

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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>
This commit is contained in:
Vlad Dogaru 2014-12-29 11:37:48 +02:00 committed by Jonathan Cameron
parent c22d2672c8
commit 8f5d8727a7
1 changed files with 27 additions and 0 deletions
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27
drivers/iio/industrialio-core.c
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@ -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);