[FFmpeg-devel] [PATCH] ALS decoder

Thilo Borgmann thilo.borgmann
Tue Aug 25 18:56:46 CEST 2009


>>>> +            // damaged block, write zero for the rest of the frame
>>>> +            while (b < ctx->num_blocks) {
>>>> +                memset(raw_sample, 0, div_blocks[b]);
>>>> +                raw_sample += div_blocks[b];
>>>> +                b++;
>>>> +            }
>>>> +            return -1;
>>> [...]
>>>> +            // damaged block, write zero for the rest of the frame
>>>> +            while (b < ctx->num_blocks) {
>>>> +                memset(raw_samples_L, 0, div_blocks[b]);
>>>> +                memset(raw_samples_R, 0, div_blocks[b]);
>>>> +                raw_samples_L += div_blocks[b];
>>>> +                raw_samples_R += div_blocks[b];
>>>> +                b++;
>>>> +            }
>>> [...]
>>>> +                // damaged block, write zero for the rest of the frame
>>>> +                while (b < ctx->num_blocks) {
>>>> +                    memset(raw_samples_L, 0, div_blocks[b]);
>>>> +                    raw_samples_L += div_blocks[b];
>>>> +                    b++;
>>>> +                }
>>> cant these be factored/combined?
>> They could be factored into decode_frame() but several parameters would
>> have to be passed somehow to know wich samples of which channel could be
>> decoded correctly already.
>> Alternative 1) loose all samples of the current RA unit.
>> Alternative 2) infunction {while() memset;} but this would add a second
>> loop for the code block in the middle.
>> Which way to go?
> 
> Hm, neither?
> get_remain_count():
>   while (b < ctx->num_blocks) sum += div_blocks[b];
>   return sum;
> 
> count = get_remain_count();
> memset(raw_samples_L, 0, count);
> raw_samples_L += count;
> memset(raw_samples_R, 0, count);
> raw_samples_R += count;
> 
> Though personally I'd suggest expanding the memset+increment
> into a bytestream_fill_byte() that does both.
I read this too fast for revision 9... will be a count function + memset
in revision 10 - it is zero function incl. count in rev. 9, which counts
twice...

Thanks!

-Thilo



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