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分析一段H264视频数据

2013年08月08日 ⁄ 综合 ⁄ 共 6359字 ⁄ 字号 评论关闭

实时传送协议(Real-time Transport Protocol或简写RTP,也可以写成RTTP)是一个网络传输协议,它是由IETF的多媒体传输工作小组1996年在RFC 1889中公布的.

 

00 00 00 01 67 42 00 1E 99 A0 B1 31 00 00 00 01

 

H264的数据流分为两种,一种是NAL UNIT stream(RTP),一种是 bits stream,

两者可以互相转换。我们分析的这个是 bit stream,根据Annex B

00 00 00 01   67 42 00 1E 99 A0 B1 31   00 00 00 01是 一个NAL,在两个00 00 00 01之间

0110 0111 0100 0010 0000 0000 0001 1110 1001 1001 1010 0000 1011 0001 0011 0001

老毕的书154

forbidden_zero_bit(1 = 0

nal_ref_idc(2)= 11

nal_unit_type(5 = 0 0111seq_parameter_set_rbsp( )  所以 processSPS

 

profile_idc(8):    42:0100 0010

constraint_set0_flag(1):0

constraint_set1_flag(1):0

constraint_set2_flag(1):0

constraint_set3_flag(1):0

reserved_zero_4bits(4):0

level_idc(8):1E

seq_parameter_set_id(UE(V)):

     ue(v): unsigned integer Exp-Golomb-coded syntax element with the left bit first. The       parsing从语法上描述或分析  解析 剖析 process for this descriptor is specified in subclause 9.1

 

uvlC: 1001:根据Table9.1 , value= 0,只占1bit.

 

根据profile_idc忽略掉一部分。

 

log2_max_frame_num_minus4(ue(v)): 001 1001,len = 5,value= 5

 

pic_order_cnt_type(ue(v)):01 1010,len = 3,value = 2

根据pic_order_cnt_type忽略几个参数

num_ref_framesue):010,len = 3,value = 1

 

0000 1011 0001 0011 0001

gaps_in_frame_num_value_allowed_flag(1) = 0

pic_width_in_mbs_minus1(ue):000 1011 ,len = 7,value = 10;

pic_height_in_map_units_minus1(ue):0001 001,len = 7,value = 8

frame_mbs_only_flag(1) = 1

忽略1

direct_8x8_inference_flag(1):0

忽略

vui_parameters_present_flag(1):0

忽略

NALU结束

 

68 CE 38 80 00 00 00 01

 

0110 1000 == 68

forbidden_zero_bit(1 = 0

nal_ref_idc(2)= 11

nal_unit_type(5 =01000pic_parameter_set_rbsp( ),7.3.2.2

 

1100 C

pic_parameter_set_id (ue)=0   P156老毕

seq_parameter_set_id(ue)=0

entropy_coding_mode_flag(1) :   0,重要的flag,0 表示编码Exp-Golomb coded and CAVLC,1表 示 CABAC

pic_order_present_flag(1):0

 

1110 E

num_slice_groups_minus1(ue):0

忽略

num_ref_idx_l0_active_minus1(ue):0

num_ref_idx_l1_active_minus1(ue):0

weighted_pred_flag(1);0

 

0011 1000 1000 0000 ==38 80

weighted_bipred_idc(2):00

pic_init_qp_minus26 /* relative to 26 */(se):0

pic_init_qs_minus26 (se):0

chroma_qp_index_offset(se):0

deblocking_filter_control_present_flag(1);0

constrained_intra_pred_flag(1):0

redundant_pic_cnt_present_flag(1):0

忽略

NALU结束

 

65 88 80 21 71 27 1B 88…….3888*16 byte

65:0110 0101

forbidden_zero_bit(1 = 0

nal_ref_idc(2)= 11

nal_unit_type(5 =0 0101:slice_layer_without_partitioning_rbsp( ),IDR浈

 

难道从这里可以看出是什么数据吗

从169页的nal_unit_type表7.20

这里是5 说明是IDR图片中的片 IDR图像不能使用片分区

 

Slice  88

Slice_Header:    片头 P158

first_mb_in_slice(ue):   0

slice_type(ue):         000 1000 = 7

pic_parameter_set_id(ue) =    0

80 21:000 0000 0010 0001

frame_num(u(v)):

    frame_num is used as an identifier for pictures and shall be represented by log2_max_frame_num_minus4 + 4 bits, 9 bits = 0

忽略

if( nal_unit_type = = 5 ) //IDR frame

idr_pic_id(u(e)):0

忽略N

 

P159 参考帧队列重排列 

ref_pic_list_reordering( )     见7。3。3。1忽略,I slice,SI slice,B slice

nal_ref_idc  =11 所以dec_ref_pic_marking( )

       nal_unit_type = 5,所以

       no_output_of_prior_pics_flag(1):0

long_term_reference_flag(1):    0

忽略

 

。。71 27

001 0111 0001 0010 0111

slice_qp_deltasev)):001 01 4-2

忽略

 

slice_data( ):7.3.4   表7.13  p161

对I-Slice:忽略N

进入

if(moreDataFlag)//p162

{

  if( MbaffFrameFlag && (CurrMbAddr % 2 = = 0 ||( CurrMbAddr % 2 = = 1 && prevMbSkipped )))

  mb_field_decoding_flag

  macroblock_layer()

}

mb_field_decoding_flag  忽略

macroblock_layer()//p163

mb_type(ue(v)):     0

mb_pred(mb_type)

prev_intra4x4_pred_mode_flag[ luma4x4BlkIdx ] (1bit, 对babac是ae(v)):    1

 

1 27:0001 0010 0111

prev_intra4x4_pred_mode_flag[ 1 ] : 0001,0,001

0010 0111

prev_intra4x4_pred_mode_flag[ 2 ] : 0010,0,010

prev_intra4x4_pred_mode_flag[ 3 ] : 0111,0,111

 

……16

1b 88 00 3e cf.

intra_chroma_pred_modeue(v) : 最后的一个1bit:0

 

接下来是macroblock_layer的coded_block_pattern和run level,即系数

c0 06 ad a0 18

1100 0000 0000 0110 1010 0000 0001 1000

coded_block_pattern(me(v):0,

根据Table 9? 4,= 47,0x2f

mb_qp_delta(se(v):):0 len =1

 

residual()见7.3.5.3 残差语句 p164

residual_block( LumaLevel[ i8x8 * 4 + i4x4 ], 16 )

coeff_token(ce(v)): 00 0000 0000 0110 1

       nc = 0(left block and top block 相关的)

len: {      // 0702

      { 1, 6, 8, 9,10,11,13,13,13,14,14,15,15,16,16,16,16},

      { 0, 2, 6, 8, 9,10,11,13,13,14,14,15,15,15,16,16,16},

      { 0, 0, 3, 7, 8, 9,10,11,13,13,14,14,15,15,16,16,16},

      { 0, 0, 0, 5, 6, 7, 8, 9,10,11,13,14,14,15,15,16,16},

    },                                                

    {                                                  

      { 2, 6, 6, 7, 8, 8, 9,11,11,12,12,12,13,13,13,14,14},

      { 0, 2, 5, 6, 6, 7, 8, 9,11,11,12,12,13,13,14,14,14},

      { 0, 0, 3, 6, 6, 7, 8, 9,11,11,12,12,13,13,13,14,14},

      { 0, 0, 0, 4, 4, 5, 6, 6, 7, 9,11,11,12,13,13,13,14},

    },                                                

    {                                                 

      { 4, 6, 6, 6, 7, 7, 7, 7, 8, 8, 9, 9, 9,10,10,10,10},

      { 0, 4, 5, 5, 5, 5, 6, 6, 7, 8, 8, 9, 9, 9,10,10,10},

      { 0, 0, 4, 5, 5, 5, 6, 6, 7, 7, 8, 8, 9, 9,10,10,10},

      { 0, 0, 0, 4, 4, 4, 4, 4, 5, 6, 7, 8, 8, 9,10,10,10},

    },

code:

      { 1, 5, 7, 7, 7, 7,15,11, 8,15,11,15,11,15,11, 7,4},

      { 0, 1, 4, 6, 6, 6, 6,14,10,14,10,14,10, 1,14,10,6},

      { 0, 0, 1, 5, 5, 5, 5, 5,13, 9,13, 9,13, 9,13, 9,5},

      { 0, 0, 0, 3, 3, 4, 4, 4, 4, 4,12,12, 8,12, 8,12,8},

    },

    {

      { 3,11, 7, 7, 7, 4, 7,15,11,15,11, 8,15,11, 7, 9,7},

      { 0, 2, 7,10, 6, 6, 6, 6,14,10,14,10,14,10,11, 8,6},

      { 0, 0, 3, 9, 5, 5, 5, 5,13, 9,13, 9,13, 9, 6,10,5},

      { 0, 0, 0, 5, 4, 6, 8, 4, 4, 4,12, 8,12,12, 8, 1,4},

    },

    {

      {15,15,11, 8,15,11, 9, 8,15,11,15,11, 8,13, 9, 5,1},

      { 0,14,15,12,10, 8,14,10,14,14,10,14,10, 7,12, 8,4},

      { 0, 0,13,14,11, 9,13, 9,13,10,13, 9,13, 9,11, 7,3},

      { 0, 0, 0,12,11,10, 9, 8,13,12,12,12, 8,12,10, 6,2},

    },

根据表查的:

code = 13,len = 15,i= 12,j=2

       所以numcoeff = 12,numtrailingones = 2

010 0000 0001 1000: totalzeros:

根据numcoeff

 int lentab[TOTRUN_NUM][16] =

 {

   

    { 1,3,3,4,4,5,5,6,6,7,7,8,8,9,9,9}, 

    { 3,3,3,3,3,4,4,4,4,5,5,6,6,6,6}, 

    { 4,3,3,3,4,4,3,3,4,5,5,6,5,6}, 

    { 5,3,4,4,3,3,3,4,3,4,5,5,5}, 

    { 4,4,4,3,3,3,3,3,4,5,4,5}, 

    { 6,5,3,3,3,3,3,3,4,3,6}, 

    { 6,5,3,3,3,2,3,4,3,6}, 

    { 6,4,5,3,2,2,3,3,6}, 

    { 6,6,4,2,2,3,2,5}, 

    { 5,5,3,2,2,2,4}, 

    { 4,4,3,3,1,3}, 

    { 4,4,2,1,3}, numcoeff开始

    { 3,3,1,2}, 

    { 2,2,1}, 

    { 1,1}, 

 };

 

 int codtab[TOTRUN_NUM][16] =

 {

    {1,3,2,3,2,3,2,3,2,3,2,3,2,3,2,1},

    {7,6,5,4,3,5,4,3,2,3,2,3,2,1,0},

    {5,7,6,5,4,3,4,3,2,3,2,1,1,0},

    {3,7,5,4,6,5,4,3,3,2,2,1,0},

    {5,4,3,7,6,5,4,3,2,1,1,0},

    {1,1,7,6,5,4,3,2,1,1,0},

    {1,1,5,4,3,3,2,1,1,0},

    {1,1,1,3,3,2,2,1,0},

    {1,0,1,3,2,1,1,1,},

    {1,0,1,3,2,1,1,},

    {0,1,1,2,1,3},

    {0,1,1,1,1}, numcoeff开始

    {0,1,1,1},

    {0,1,1},

    {0,1}, 

 };

Code = 1,len = 2,i=2,j = 0, totzeros = 2

Read run: 0 0000 0001 1000根据totzeros = 2

 int lentab[TOTRUN_NUM][16] =

 {

    {1,1},

    {1,2,2},

    {2,2,2,2},

    {2,2,2,3,3},

    {2,2,3,3,3,3},

    {2,3,3,3,3,3,3},

    {3,3,3,3,3,3,3,4,5,6,7,8,9,10,11},

 };

 

 int codtab[TOTRUN_NUM][16] =

 {

    {1,0},

    {1,1,0},

    {3,2,1,0},

    {3,2,1,1,0},

    {3,2,3,2,1,0},

    {3,0,1,3,2,5,4},

    {7,6,5,4,3,2,1,1,1,1,1,1,1,1,1},

Code = 1,len =1,I = 0,j = 0

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