aan 演算法 c 語言示例 (轉)

aan 演算法 c 語言示例 (轉)[@more@]

static int SpreSC[64]
={
  16384, 22725, 21407, 19266,  16384, 12873, 8867,  4520,
  22725, 31521, 29692, 26722,  22725, 17855, 12299, 6270,
  21407, 29692, 27969, 25172,  21407, 16819, 11585, 5906,
  19266, 26722, 25172, 22654,  19266, 15137, 10426, 5315,
  16384, 22725, 21407, 19266,  16384, 12873, 8867,  4520,
  12873, 17855, 16819, 15137,  25746, 20228, 13933, 7103,
  17734, 24598, 23170, 20853,  17734, 13933, 9597,  4892,
  18081, 25080, 23624, 21261,  18081, 14206, 9785,  4988
};

/*
  use this table to multiply the block[j] in the decode_macroblock
 stage;
 after multiply,shift to right 13 bits; (17-4)

*/

#define C2  30274  // 2 C(PI/8) 1.84776
#define C4  23171  // SQRT(2)  1.41421
#define C6  12540  // 2 SIN (PI/8)  0.76537
#define C6pC2  42814    2.61313
#define C6sC2  (-17734) //  -1.08239
#define RC0  14
#define RC1  18
#define ROUND  (1<

//  do 80 times of multiply; fastidct :176;

#define idct_unit(a,b) {x4 = x5 - x3; 
   t1 = x5 + x3; 
   x3 = (x2+x6)<   x2 = (x2-x6)*C4-x3; 
    t0 = x1 + x7; 
    x6=  x1 - x7; 
    x7=  (t0 + t1)<    x5 = (t0 - t1)*C4; 
   t0=C6*(x4+x6); 
    x4=C6sC2*x4-t0; 
    x6=C6pC2*x6-t0; 
   t0=x6-x7; 
    x1=(a-b)<    t1=t0-x5; 
    x6=(a+b)<   x0=x4-t1; 
    x4=x3+x6; 
    x6-=x3; 
    x3=x1+x2; 
    x5=x1-x2;  }

static void IDCTaan_c(short* block)
{
 int x0,x1,x2,x3,x4,x5,x6,x7;
 int t0,t1;
 short *blk;
// Field one
 blk  =  block;
 for(int i=0; i<8; i++)
 {
 if (!(  (x1 = blk[1]) 
    | (x2 = blk[2])
  | (x3 = blk[3])
  | (x4 = blk[4])
  | (x5 = blk[5])
  | (x6 = blk[6])
  | (x7 = blk[7])))
 {
  // blk[0]=
  blk[1]=blk[2]=blk[3]=blk[4]
    =blk[5]=blk[6]=blk[7]=(blk[0]);
 }
  else
 {

 idct_unit(blk[0],blk[4])

  blk[0]=(x4+x7)>>RC0 ;
  blk[7]=(x4-x7)>>RC0 ;
  blk[1]=(x3+t0)>>RC0 ;
  blk[6]=(x3-t0)>>RC0 ;
  blk[2]=(x5-t1)>>RC0 ;
  blk[5]=(x5+t1)>>RC0 ;
  blk[3]=(x6-x0)>>RC0 ;
  blk[4]=(x6+x0)>>RC0 ;
 }
 blk+=8;
 }

two
  blk=block;
 for ( i=0; i<8; i++)
 {
 if (!((x1 = blk[8])
 | (x2 = blk[16])
 | (x3 = blk[24])
 | (x4 = blk[32])
 | (x5 = blk[40]) 
 | (x6 = blk[48])
 | (x7 = blk[56])))
 {
  blk[0]=blk[8]=blk[16]=blk[24]
 =blk[32]=blk[40]=blk[48]=blk[56]=
  iclp[blk[0]>>4];
 }
 else
 {
 idct_unit(blk[0*8],blk[4*8])

 blk[0*8]=iclp[(x4+x7+ROUND)>>RC1];
   blk[7*8]=iclp[(x4-x7+ROUND)>>RC1];
   blk[1*8]=iclp[(x3+t0+ROUND)>>RC1];
   blk[6*8]=iclp[(x3-t0+ROUND)>>RC1];
   blk[2*8]=iclp[(x5-t1+ROUND)>>RC1];
   blk[5*8]=iclp[(x5+t1+ROUND)>>RC1];
   blk[3*8]=iclp[(x6-x0+ROUND)>>RC1];
   blk[4*8]=iclp[(x6+x0+ROUND)>>RC1];
 }
  blk++;
 }
}