Re: [PATCH] ASoC: Intel: Skylake: Fix available clock counter incrementation

["Amadeusz Sławiński" <amadeuszx.slawinski@linux.intel.com>]

On 2/24/2020 5:18 PM, Pierre-Louis Bossart wrote:
> 
> 
> On 2/24/20 6:52 AM, Amadeusz Sławiński wrote:
>> Incrementation of avail_clk_cnt was incorrectly moved to error path. Put
>> it back to success path.
>>
>> Fixes: 6ee927f2f01466 ('ASoC: Intel: Skylake: Fix NULL ptr dereference 
>> when unloading clk dev')
>> Signed-off-by: Amadeusz Sławiński <amadeuszx.slawinski@linux.intel.com>
>> ---
>>   sound/soc/intel/skylake/skl-ssp-clk.c | 4 +++-
>>   1 file changed, 3 insertions(+), 1 deletion(-)
>>
>> diff --git a/sound/soc/intel/skylake/skl-ssp-clk.c 
>> b/sound/soc/intel/skylake/skl-ssp-clk.c
>> index 1c0e5226cb5b..bd43885f3805 100644
>> --- a/sound/soc/intel/skylake/skl-ssp-clk.c
>> +++ b/sound/soc/intel/skylake/skl-ssp-clk.c
>> @@ -384,9 +384,11 @@ static int skl_clk_dev_probe(struct 
>> platform_device *pdev)
>>                   &clks[i], clk_pdata, i);
>>           if (IS_ERR(data->clk[data->avail_clk_cnt])) {
>> -            ret = PTR_ERR(data->clk[data->avail_clk_cnt++]);
>> +            ret = PTR_ERR(data->clk[data->avail_clk_cnt]);
> 
> Are you sure?
> 
> If you start with avail_clk_cnt set to zero, the error handling will 
> decrement and access offset -1
> 

Yes, I'm sure as far as I know c it will first check the value and then 
decrement it, so it will be 0 while doing the "while" check and it won't 
enter the loop.

You can double check with simplified usecase:
#include <stdio.h>
int main() {
int i = 0;
while(i--)
printf("do something with i, while i = %d\n", i);
}

which seems to work fine to me, by not entering the loop.

Use case is as following:
we start with avail_clk_cnt = 0;
register clock at index 0; increment avail_clk_cnt to 1;
register clock at index 1; increment avail_clk_cnt to 2;
register clock at index 2; increment avail_clk_cnt to 3

now let's assume that there is no more clocks to register
so we do our stuff and then we need to free clocks

so we enter loop
3 evaluates to true, so we decrement it and release clock at index 2
2 evaluates to true, so we decrement it and release clock at index 1
1 evaluates to true, so we decrement it and release clock at index 0
0 evaluates to false, so wo don't enter loop

similar thing happens if we fail to register clock and do error handling

Amadeusz