On 03/01/2020 22:59, Derek Buitenhuis wrote:
On 02/01/2020 23:09, Michael Niedermayer wrote:
I think if entry 128 is 0 then the whole table must be 0.
If that is the case, checking the entry 128 of table 4 and 5 would be enough
and caching the entry comparission is maybe not needed.
Is this guaranteed somehow? It isn't mentioned in the spec.
Checking better the spec about how the quant table is stored in
bitstream, we may have missed in our past discussion that it is actually
not stored as 256 independent values, "QuantizationTable(i, j, scale)"
permits only increasing numbers from index 0 to 127 (so 128, as index
128 is the negative of index 127), so it seems that by design of how the
quant table is stored it is impossible to have 0 at index 128 without
having 0 at indexes 0-127 (one value of "len - 1", 127, seems to be the
only way to have 0 at index 128).
The need to check the 5th quant table is still valid.
Jérôme
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