On 04/05/2014 02:45 AM, Daniel Vetter wrote:
> This is C standard hair-splitting, but afaict
> - sum will be promoted to signed int in computation since
> uint8_t fits
> - signed overflow is undefined.
>
> No we need to add up an awful lot of bytes to actually make it
^^
Now
> overflow. But I guess the real risk is gcc spotting this and going
> bananas. Fix this by simply using unsigned in to force all computations
> to use the well-defined unsigned behaviour.
Seems reasonable... it also seems impossible (ha!) to break anything.
Reviewed-by: Ian Romanick <ian.d.romanick@xxxxxxxxx>
> Spotted by coverity.
>
> Cc: Russell King <rmk+kernel@xxxxxxxxxxxxxxxx>
> Cc: Rob Clark <robdclark@xxxxxxxxx>
> Cc: Jean-Francois Moine <moinejf@xxxxxxx>
> Signed-off-by: Daniel Vetter <daniel.vetter@xxxxxxxx>
> ---
> drivers/gpu/drm/i2c/tda998x_drv.c | 2 +-
> 1 file changed, 1 insertion(+), 1 deletion(-)
>
> diff --git a/drivers/gpu/drm/i2c/tda998x_drv.c b/drivers/gpu/drm/i2c/tda998x_drv.c
> index 48af5cac1902..ae2754760d77 100644
> --- a/drivers/gpu/drm/i2c/tda998x_drv.c
> +++ b/drivers/gpu/drm/i2c/tda998x_drv.c
> @@ -568,7 +568,7 @@ static irqreturn_t tda998x_irq_thread(int irq, void *data)
>
> static uint8_t tda998x_cksum(uint8_t *buf, size_t bytes)
> {
> - uint8_t sum = 0;
> + unsigned sum = 0;
>
> while (bytes--)
> sum += *buf++;
>
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