1 // Copyright 2009 The Go Authors. All rights reserved. 2 // Use of this source code is governed by a BSD-style 3 // license that can be found in the LICENSE file. 4 5 package jpeg 6 7 // This is a Go translation of idct.c from 8 // 9 // http://standards.iso.org/ittf/PubliclyAvailableStandards/ISO_IEC_13818-4_2004_Conformance_Testing/Video/verifier/mpeg2decode_960109.tar.gz 10 // 11 // which carries the following notice: 12 13 /* Copyright (C) 1996, MPEG Software Simulation Group. All Rights Reserved. */ 14 15 /* 16 * Disclaimer of Warranty 17 * 18 * These software programs are available to the user without any license fee or 19 * royalty on an "as is" basis. The MPEG Software Simulation Group disclaims 20 * any and all warranties, whether express, implied, or statuary, including any 21 * implied warranties or merchantability or of fitness for a particular 22 * purpose. In no event shall the copyright-holder be liable for any 23 * incidental, punitive, or consequential damages of any kind whatsoever 24 * arising from the use of these programs. 25 * 26 * This disclaimer of warranty extends to the user of these programs and user's 27 * customers, employees, agents, transferees, successors, and assigns. 28 * 29 * The MPEG Software Simulation Group does not represent or warrant that the 30 * programs furnished hereunder are free of infringement of any third-party 31 * patents. 32 * 33 * Commercial implementations of MPEG-1 and MPEG-2 video, including shareware, 34 * are subject to royalty fees to patent holders. Many of these patents are 35 * general enough such that they are unavoidable regardless of implementation 36 * design. 37 * 38 */ 39 40 const blockSize = 64 // A DCT block is 8x8. 41 42 type block [blockSize]int32 43 44 const ( 45 w1 = 2841 // 2048*sqrt(2)*cos(1*pi/16) 46 w2 = 2676 // 2048*sqrt(2)*cos(2*pi/16) 47 w3 = 2408 // 2048*sqrt(2)*cos(3*pi/16) 48 w5 = 1609 // 2048*sqrt(2)*cos(5*pi/16) 49 w6 = 1108 // 2048*sqrt(2)*cos(6*pi/16) 50 w7 = 565 // 2048*sqrt(2)*cos(7*pi/16) 51 52 w1pw7 = w1 + w7 53 w1mw7 = w1 - w7 54 w2pw6 = w2 + w6 55 w2mw6 = w2 - w6 56 w3pw5 = w3 + w5 57 w3mw5 = w3 - w5 58 59 r2 = 181 // 256/sqrt(2) 60 ) 61 62 // idct performs a 2-D Inverse Discrete Cosine Transformation. 63 // 64 // The input coefficients should already have been multiplied by the 65 // appropriate quantization table. We use fixed-point computation, with the 66 // number of bits for the fractional component varying over the intermediate 67 // stages. 68 // 69 // For more on the actual algorithm, see Z. Wang, "Fast algorithms for the 70 // discrete W transform and for the discrete Fourier transform", IEEE Trans. on 71 // ASSP, Vol. ASSP- 32, pp. 803-816, Aug. 1984. 72 func idct(src *block) { 73 // Horizontal 1-D IDCT. 74 for y := 0; y < 8; y++ { 75 y8 := y * 8 76 // If all the AC components are zero, then the IDCT is trivial. 77 if src[y8+1] == 0 && src[y8+2] == 0 && src[y8+3] == 0 && 78 src[y8+4] == 0 && src[y8+5] == 0 && src[y8+6] == 0 && src[y8+7] == 0 { 79 dc := src[y8+0] << 3 80 src[y8+0] = dc 81 src[y8+1] = dc 82 src[y8+2] = dc 83 src[y8+3] = dc 84 src[y8+4] = dc 85 src[y8+5] = dc 86 src[y8+6] = dc 87 src[y8+7] = dc 88 continue 89 } 90 91 // Prescale. 92 x0 := (src[y8+0] << 11) + 128 93 x1 := src[y8+4] << 11 94 x2 := src[y8+6] 95 x3 := src[y8+2] 96 x4 := src[y8+1] 97 x5 := src[y8+7] 98 x6 := src[y8+5] 99 x7 := src[y8+3] 100 101 // Stage 1. 102 x8 := w7 * (x4 + x5) 103 x4 = x8 + w1mw7*x4 104 x5 = x8 - w1pw7*x5 105 x8 = w3 * (x6 + x7) 106 x6 = x8 - w3mw5*x6 107 x7 = x8 - w3pw5*x7 108 109 // Stage 2. 110 x8 = x0 + x1 111 x0 -= x1 112 x1 = w6 * (x3 + x2) 113 x2 = x1 - w2pw6*x2 114 x3 = x1 + w2mw6*x3 115 x1 = x4 + x6 116 x4 -= x6 117 x6 = x5 + x7 118 x5 -= x7 119 120 // Stage 3. 121 x7 = x8 + x3 122 x8 -= x3 123 x3 = x0 + x2 124 x0 -= x2 125 x2 = (r2*(x4+x5) + 128) >> 8 126 x4 = (r2*(x4-x5) + 128) >> 8 127 128 // Stage 4. 129 src[y8+0] = (x7 + x1) >> 8 130 src[y8+1] = (x3 + x2) >> 8 131 src[y8+2] = (x0 + x4) >> 8 132 src[y8+3] = (x8 + x6) >> 8 133 src[y8+4] = (x8 - x6) >> 8 134 src[y8+5] = (x0 - x4) >> 8 135 src[y8+6] = (x3 - x2) >> 8 136 src[y8+7] = (x7 - x1) >> 8 137 } 138 139 // Vertical 1-D IDCT. 140 for x := 0; x < 8; x++ { 141 // Similar to the horizontal 1-D IDCT case, if all the AC components are zero, then the IDCT is trivial. 142 // However, after performing the horizontal 1-D IDCT, there are typically non-zero AC components, so 143 // we do not bother to check for the all-zero case. 144 145 // Prescale. 146 y0 := (src[8*0+x] << 8) + 8192 147 y1 := src[8*4+x] << 8 148 y2 := src[8*6+x] 149 y3 := src[8*2+x] 150 y4 := src[8*1+x] 151 y5 := src[8*7+x] 152 y6 := src[8*5+x] 153 y7 := src[8*3+x] 154 155 // Stage 1. 156 y8 := w7*(y4+y5) + 4 157 y4 = (y8 + w1mw7*y4) >> 3 158 y5 = (y8 - w1pw7*y5) >> 3 159 y8 = w3*(y6+y7) + 4 160 y6 = (y8 - w3mw5*y6) >> 3 161 y7 = (y8 - w3pw5*y7) >> 3 162 163 // Stage 2. 164 y8 = y0 + y1 165 y0 -= y1 166 y1 = w6*(y3+y2) + 4 167 y2 = (y1 - w2pw6*y2) >> 3 168 y3 = (y1 + w2mw6*y3) >> 3 169 y1 = y4 + y6 170 y4 -= y6 171 y6 = y5 + y7 172 y5 -= y7 173 174 // Stage 3. 175 y7 = y8 + y3 176 y8 -= y3 177 y3 = y0 + y2 178 y0 -= y2 179 y2 = (r2*(y4+y5) + 128) >> 8 180 y4 = (r2*(y4-y5) + 128) >> 8 181 182 // Stage 4. 183 src[8*0+x] = (y7 + y1) >> 14 184 src[8*1+x] = (y3 + y2) >> 14 185 src[8*2+x] = (y0 + y4) >> 14 186 src[8*3+x] = (y8 + y6) >> 14 187 src[8*4+x] = (y8 - y6) >> 14 188 src[8*5+x] = (y0 - y4) >> 14 189 src[8*6+x] = (y3 - y2) >> 14 190 src[8*7+x] = (y7 - y1) >> 14 191 } 192 } 193
