1 /*
2 * Copyright (c) 2003-2004, Mark Borgerding. All rights reserved.
3 * This file is part of KISS FFT - https://github.com/mborgerding/kissfft
4 *
5 * SPDX-License-Identifier: BSD-3-Clause
6 * See COPYING file for more information.
7 */
8
9 #include "kiss_fftnd.h"
10 #include "_kiss_fft_guts.h"
11
12 struct kiss_fftnd_state{
13 int dimprod; /* dimsum would be mighty tasty right now */
14 int ndims;
15 int *dims;
16 kiss_fft_cfg *states; /* cfg states for each dimension */
17 kiss_fft_cpx * tmpbuf; /*buffer capable of hold the entire input */
18 };
19
kiss_fftnd_alloc(const int * dims,int ndims,int inverse_fft,void * mem,size_t * lenmem)20 kiss_fftnd_cfg kiss_fftnd_alloc(const int *dims,int ndims,int inverse_fft,void*mem,size_t*lenmem)
21 {
22 kiss_fftnd_cfg st = NULL;
23 int i;
24 int dimprod=1;
25 size_t memneeded = sizeof(struct kiss_fftnd_state);
26 char * ptr;
27
28 for (i=0;i<ndims;++i) {
29 size_t sublen=0;
30 kiss_fft_alloc (dims[i], inverse_fft, NULL, &sublen);
31 memneeded += sublen; /* st->states[i] */
32 dimprod *= dims[i];
33 }
34 memneeded += sizeof(int) * ndims;/* st->dims */
35 memneeded += sizeof(void*) * ndims;/* st->states */
36 memneeded += sizeof(kiss_fft_cpx) * dimprod; /* st->tmpbuf */
37
38 if (lenmem == NULL) {/* allocate for the caller*/
39 st = (kiss_fftnd_cfg) malloc (memneeded);
40 } else { /* initialize supplied buffer if big enough */
41 if (*lenmem >= memneeded)
42 st = (kiss_fftnd_cfg) mem;
43 *lenmem = memneeded; /*tell caller how big struct is (or would be) */
44 }
45 if (!st)
46 return NULL; /*malloc failed or buffer too small */
47
48 st->dimprod = dimprod;
49 st->ndims = ndims;
50 ptr=(char*)(st+1);
51
52 st->states = (kiss_fft_cfg *)ptr;
53 ptr += sizeof(void*) * ndims;
54
55 st->dims = (int*)ptr;
56 ptr += sizeof(int) * ndims;
57
58 st->tmpbuf = (kiss_fft_cpx*)ptr;
59 ptr += sizeof(kiss_fft_cpx) * dimprod;
60
61 for (i=0;i<ndims;++i) {
62 size_t len;
63 st->dims[i] = dims[i];
64 kiss_fft_alloc (st->dims[i], inverse_fft, NULL, &len);
65 st->states[i] = kiss_fft_alloc (st->dims[i], inverse_fft, ptr,&len);
66 ptr += len;
67 }
68 /*
69 Hi there!
70
71 If you're looking at this particular code, it probably means you've got a brain-dead bounds checker
72 that thinks the above code overwrites the end of the array.
73
74 It doesn't.
75
76 -- Mark
77
78 P.S.
79 The below code might give you some warm fuzzies and help convince you.
80 */
81 if ( ptr - (char*)st != (int)memneeded ) {
82 fprintf(stderr,
83 "################################################################################\n"
84 "Internal error! Memory allocation miscalculation\n"
85 "################################################################################\n"
86 );
87 }
88 return st;
89 }
90
91 /*
92 This works by tackling one dimension at a time.
93
94 In effect,
95 Each stage starts out by reshaping the matrix into a DixSi 2d matrix.
96 A Di-sized fft is taken of each column, transposing the matrix as it goes.
97
98 Here's a 3-d example:
99 Take a 2x3x4 matrix, laid out in memory as a contiguous buffer
100 [ [ [ a b c d ] [ e f g h ] [ i j k l ] ]
101 [ [ m n o p ] [ q r s t ] [ u v w x ] ] ]
102
103 Stage 0 ( D=2): treat the buffer as a 2x12 matrix
104 [ [a b ... k l]
105 [m n ... w x] ]
106
107 FFT each column with size 2.
108 Transpose the matrix at the same time using kiss_fft_stride.
109
110 [ [ a+m a-m ]
111 [ b+n b-n]
112 ...
113 [ k+w k-w ]
114 [ l+x l-x ] ]
115
116 Note fft([x y]) == [x+y x-y]
117
118 Stage 1 ( D=3) treats the buffer (the output of stage D=2) as an 3x8 matrix,
119 [ [ a+m a-m b+n b-n c+o c-o d+p d-p ]
120 [ e+q e-q f+r f-r g+s g-s h+t h-t ]
121 [ i+u i-u j+v j-v k+w k-w l+x l-x ] ]
122
123 And perform FFTs (size=3) on each of the columns as above, transposing
124 the matrix as it goes. The output of stage 1 is
125 (Legend: ap = [ a+m e+q i+u ]
126 am = [ a-m e-q i-u ] )
127
128 [ [ sum(ap) fft(ap)[0] fft(ap)[1] ]
129 [ sum(am) fft(am)[0] fft(am)[1] ]
130 [ sum(bp) fft(bp)[0] fft(bp)[1] ]
131 [ sum(bm) fft(bm)[0] fft(bm)[1] ]
132 [ sum(cp) fft(cp)[0] fft(cp)[1] ]
133 [ sum(cm) fft(cm)[0] fft(cm)[1] ]
134 [ sum(dp) fft(dp)[0] fft(dp)[1] ]
135 [ sum(dm) fft(dm)[0] fft(dm)[1] ] ]
136
137 Stage 2 ( D=4) treats this buffer as a 4*6 matrix,
138 [ [ sum(ap) fft(ap)[0] fft(ap)[1] sum(am) fft(am)[0] fft(am)[1] ]
139 [ sum(bp) fft(bp)[0] fft(bp)[1] sum(bm) fft(bm)[0] fft(bm)[1] ]
140 [ sum(cp) fft(cp)[0] fft(cp)[1] sum(cm) fft(cm)[0] fft(cm)[1] ]
141 [ sum(dp) fft(dp)[0] fft(dp)[1] sum(dm) fft(dm)[0] fft(dm)[1] ] ]
142
143 Then FFTs each column, transposing as it goes.
144
145 The resulting matrix is the 3d FFT of the 2x3x4 input matrix.
146
147 Note as a sanity check that the first element of the final
148 stage's output (DC term) is
149 sum( [ sum(ap) sum(bp) sum(cp) sum(dp) ] )
150 , i.e. the summation of all 24 input elements.
151
152 */
kiss_fftnd(kiss_fftnd_cfg st,const kiss_fft_cpx * fin,kiss_fft_cpx * fout)153 void kiss_fftnd(kiss_fftnd_cfg st,const kiss_fft_cpx *fin,kiss_fft_cpx *fout)
154 {
155 int i,k;
156 const kiss_fft_cpx * bufin=fin;
157 kiss_fft_cpx * bufout;
158
159 /*arrange it so the last bufout == fout*/
160 if ( st->ndims & 1 ) {
161 bufout = fout;
162 if (fin==fout) {
163 memcpy( st->tmpbuf, fin, sizeof(kiss_fft_cpx) * st->dimprod );
164 bufin = st->tmpbuf;
165 }
166 }else
167 bufout = st->tmpbuf;
168
169 for ( k=0; k < st->ndims; ++k) {
170 int curdim = st->dims[k];
171 int stride = st->dimprod / curdim;
172
173 for ( i=0 ; i<stride ; ++i )
174 kiss_fft_stride( st->states[k], bufin+i , bufout+i*curdim, stride );
175
176 /*toggle back and forth between the two buffers*/
177 if (bufout == st->tmpbuf){
178 bufout = fout;
179 bufin = st->tmpbuf;
180 }else{
181 bufout = st->tmpbuf;
182 bufin = fout;
183 }
184 }
185 }
186