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68 lines
2.4 KiB
68 lines
2.4 KiB
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10 years ago
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/*****************************************************************************
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* startcode_helper.h: Startcodes helpers
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*****************************************************************************
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* Copyright (C) 2016 VideoLAN Authors
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*
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* This program is free software; you can redistribute it and/or modify it
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* under the terms of the GNU Lesser General Public License as published by
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* the Free Software Foundation; either version 2.1 of the License, or
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* (at your option) any later version.
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*
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* This program is distributed in the hope that it will be useful,
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* but WITHOUT ANY WARRANTY; without even the implied warranty of
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* MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
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* GNU Lesser General Public License for more details.
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*
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* You should have received a copy of the GNU Lesser General Public License
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* along with this program; if not, write to the Free Software Foundation,
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* Inc., 51 Franklin Street, Fifth Floor, Boston MA 02110-1301, USA.
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*****************************************************************************/
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#ifndef _STARTCODE_HELPER_H
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#define _STARTCODE_HELPER_H 1
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/* Looks up efficiently for an AnnexB startcode 0x00 0x00 0x01
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* by using a 4 times faster trick than single byte lookup.
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*
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* That code is adapted from libav's ff_avc_find_startcode_internal
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* and i believe the trick originated from
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* https://graphics.stanford.edu/~seander/bithacks.html#ZeroInWord
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*/
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static inline const uint8_t * startcode_FindAnnexB( const uint8_t *p, const uint8_t *end )
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{
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const uint8_t *a = p + 4 - ((intptr_t)p & 3);
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for (end -= 3; p < a && p < end; p++) {
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if (p[0] == 0 && p[1] == 0 && p[2] == 1)
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return p;
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}
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for (end -= 3; p < end; p += 4) {
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uint32_t x = *(const uint32_t*)p;
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if ((x - 0x01010101) & (~x) & 0x80808080)
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{
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/* matching DW isn't faster */
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if (p[1] == 0) {
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if (p[0] == 0 && p[2] == 1)
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return p;
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if (p[2] == 0 && p[3] == 1)
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return p+1;
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}
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if (p[3] == 0) {
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if (p[2] == 0 && p[4] == 1)
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return p+2;
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if (p[4] == 0 && p[5] == 1)
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return p+3;
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}
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}
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}
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for (end += 3; p < end; p++) {
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if (p[0] == 0 && p[1] == 0 && p[2] == 1)
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return p;
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}
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return NULL;
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}
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#endif
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