src/freedreno/ir3/ir3_group.c

   1 /*
   2  * Copyright (C) 2014 Rob Clark <robclark@freedesktop.org>
   3  *
   4  * Permission is hereby granted, free of charge, to any person obtaining a
   5  * copy of this software and associated documentation files (the "Software"),
   6  * to deal in the Software without restriction, including without limitation
   7  * the rights to use, copy, modify, merge, publish, distribute, sublicense,
   8  * and/or sell copies of the Software, and to permit persons to whom the
   9  * Software is furnished to do so, subject to the following conditions:
  10  *
  11  * The above copyright notice and this permission notice (including the next
  12  * paragraph) shall be included in all copies or substantial portions of the
  13  * Software.
  14  *
  15  * THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND, EXPRESS OR
  16  * IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF MERCHANTABILITY,
  17  * FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT.  IN NO EVENT SHALL
  18  * THE AUTHORS OR COPYRIGHT HOLDERS BE LIABLE FOR ANY CLAIM, DAMAGES OR OTHER
  19  * LIABILITY, WHETHER IN AN ACTION OF CONTRACT, TORT OR OTHERWISE, ARISING FROM,
  20  * OUT OF OR IN CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER DEALINGS IN THE
  21  * SOFTWARE.
  22  *
  23  * Authors:
  24  *    Rob Clark <robclark@freedesktop.org>
  25  */
  26
  27 #include "ir3.h"
  28
  29 /*
  30  * Find/group instruction neighbors:
  31  */
  32
  33 /* bleh.. we need to do the same group_n() thing for both inputs/outputs
  34  * (where we have a simple instr[] array), and fanin nodes (where we have
  35  * an extra indirection via reg->instr).
  36  */
  37 struct group_ops {
  38         struct ir3_instruction *(*get)(void *arr, int idx);
  39         void (*insert_mov)(void *arr, int idx, struct ir3_instruction *instr);
  40 };
  41
  42 static struct ir3_instruction *arr_get(void *arr, int idx)
  43 {
  44         return ((struct ir3_instruction **)arr)[idx];
  45 }
  46 static void arr_insert_mov_out(void *arr, int idx, struct ir3_instruction *instr)
  47 {
  48         ((struct ir3_instruction **)arr)[idx] =
  49                         ir3_MOV(instr->block, instr, TYPE_F32);
  50 }
  51 static void arr_insert_mov_in(void *arr, int idx, struct ir3_instruction *instr)
  52 {
  53         /* so, we can't insert a mov in front of a meta:in.. and the downstream
  54          * instruction already has a pointer to 'instr'.  So we cheat a bit and
  55          * morph the meta:in instruction into a mov and insert a new meta:in
  56          * in front.
  57          */
  58         struct ir3_instruction *in;
  59
  60         debug_assert(instr->regs_count == 1);
  61
  62         in = ir3_instr_create(instr->block, OPC_META_INPUT);
  63         in->inout.block = instr->block;
  64         ir3_reg_create(in, instr->regs[0]->num, 0);
  65
  66         /* create src reg for meta:in and fixup to now be a mov: */
  67         ir3_reg_create(instr, 0, IR3_REG_SSA)->instr = in;
  68         instr->opc = OPC_MOV;
  69         instr->cat1.src_type = TYPE_F32;
  70         instr->cat1.dst_type = TYPE_F32;
  71
  72         ((struct ir3_instruction **)arr)[idx] = in;
  73 }
  74 static struct group_ops arr_ops_out = { arr_get, arr_insert_mov_out };
  75 static struct group_ops arr_ops_in = { arr_get, arr_insert_mov_in };
  76
  77 static struct ir3_instruction *instr_get(void *arr, int idx)
  78 {
  79         return ssa(((struct ir3_instruction *)arr)->regs[idx+1]);
  80 }
  81 static void
  82 instr_insert_mov(void *arr, int idx, struct ir3_instruction *instr)
  83 {
  84         ((struct ir3_instruction *)arr)->regs[idx+1]->instr =
  85                         ir3_MOV(instr->block, instr, TYPE_F32);
  86 }
  87 static struct group_ops instr_ops = { instr_get, instr_insert_mov };
  88
  89 /* verify that cur != instr, but cur is also not in instr's neighbor-list: */
  90 static bool
  91 in_neighbor_list(struct ir3_instruction *instr, struct ir3_instruction *cur, int pos)
  92 {
  93         int idx = 0;
  94
  95         if (!instr)
  96                 return false;
  97
  98         if (instr == cur)
  99                 return true;
 100
 101         for (instr = ir3_neighbor_first(instr); instr; instr = instr->cp.right)
 102                 if ((idx++ != pos) && (instr == cur))
 103                         return true;
 104
 105         return false;
 106 }
 107
 108 static void
 109 group_n(struct group_ops *ops, void *arr, unsigned n)
 110 {
 111         unsigned i, j;
 112
 113         /* first pass, figure out what has conflicts and needs a mov
 114          * inserted.  Do this up front, before starting to setup
 115          * left/right neighbor pointers.  Trying to do it in a single
 116          * pass could result in a situation where we can't even setup
 117          * the mov's right neighbor ptr if the next instr also needs
 118          * a mov.
 119          */
 120 restart:
 121         for (i = 0; i < n; i++) {
 122                 struct ir3_instruction *instr = ops->get(arr, i);
 123                 if (instr) {
 124                         struct ir3_instruction *left = (i > 0) ? ops->get(arr, i - 1) : NULL;
 125                         struct ir3_instruction *right = (i < (n-1)) ? ops->get(arr, i + 1) : NULL;
 126                         bool conflict;
 127
 128                         /* check for left/right neighbor conflicts: */
 129                         conflict = conflicts(instr->cp.left, left) ||
 130                                 conflicts(instr->cp.right, right);
 131
 132                         /* Mixing array elements and higher register classes
 133                          * (ie. groups) doesn't really work out in RA.  See:
 134                          *
 135                          * https://trello.com/c/DqeDkeVf/156-bug-with-stk-70frag
 136                          */
 137                         if (instr->regs[0]->flags & IR3_REG_ARRAY)
 138                                 conflict = true;
 139
 140                         /* we also can't have an instr twice in the group: */
 141                         for (j = i + 1; (j < n) && !conflict; j++)
 142                                 if (in_neighbor_list(ops->get(arr, j), instr, i))
 143                                         conflict = true;
 144
 145                         if (conflict) {
 146                                 ops->insert_mov(arr, i, instr);
 147                                 /* inserting the mov may have caused a conflict
 148                                  * against the previous:
 149                                  */
 150                                 goto restart;
 151                         }
 152                 }
 153         }
 154
 155         /* second pass, now that we've inserted mov's, fixup left/right
 156          * neighbors.  This is guaranteed to succeed, since by definition
 157          * the newly inserted mov's cannot conflict with anything.
 158          */
 159         for (i = 0; i < n; i++) {
 160                 struct ir3_instruction *instr = ops->get(arr, i);
 161                 if (instr) {
 162                         struct ir3_instruction *left = (i > 0) ? ops->get(arr, i - 1) : NULL;
 163                         struct ir3_instruction *right = (i < (n-1)) ? ops->get(arr, i + 1) : NULL;
 164
 165                         debug_assert(!conflicts(instr->cp.left, left));
 166                         if (left) {
 167                                 instr->cp.left_cnt++;
 168                                 instr->cp.left = left;
 169                         }
 170
 171                         debug_assert(!conflicts(instr->cp.right, right));
 172                         if (right) {
 173                                 instr->cp.right_cnt++;
 174                                 instr->cp.right = right;
 175                         }
 176                 }
 177         }
 178 }
 179
 180 static void
 181 instr_find_neighbors(struct ir3_instruction *instr)
 182 {
 183         struct ir3_instruction *src;
 184
 185         if (ir3_instr_check_mark(instr))
 186                 return;
 187
 188         if (instr->opc == OPC_META_FI)
 189                 group_n(&instr_ops, instr, instr->regs_count - 1);
 190
 191         foreach_ssa_src(src, instr)
 192                 instr_find_neighbors(src);
 193 }
 194
 195 /* a bit of sadness.. we can't have "holes" in inputs from PoV of
 196  * register assignment, they still need to be grouped together.  So
 197  * we need to insert dummy/padding instruction for grouping, and
 198  * then take it back out again before anyone notices.
 199  */
 200 static void
 201 pad_and_group_input(struct ir3_instruction **input, unsigned n)
 202 {
 203         int i, mask = 0;
 204         struct ir3_block *block = NULL;
 205
 206         for (i = n - 1; i >= 0; i--) {
 207                 struct ir3_instruction *instr = input[i];
 208                 if (instr) {
 209                         block = instr->block;
 210                 } else if (block) {
 211                         instr = ir3_NOP(block);
 212                         ir3_reg_create(instr, 0, IR3_REG_SSA);    /* dummy dst */
 213                         input[i] = instr;
 214                         mask |= (1 << i);
 215                 }
 216         }
 217
 218         group_n(&arr_ops_in, input, n);
 219
 220         for (i = 0; i < n; i++) {
 221                 if (mask & (1 << i))
 222                         input[i] = NULL;
 223         }
 224 }
 225
 226 static void
 227 find_neighbors(struct ir3 *ir)
 228 {
 229         unsigned i;
 230
 231         /* shader inputs/outputs themselves must be contiguous as well:
 232          *
 233          * NOTE: group inputs first, since we only insert mov's
 234          * *before* the conflicted instr (and that would go badly
 235          * for inputs).  By doing inputs first, we should never
 236          * have a conflict on inputs.. pushing any conflict to
 237          * resolve to the outputs, for stuff like:
 238          *
 239          *     MOV OUT[n], IN[m].wzyx
 240          *
 241          * NOTE: we assume here inputs/outputs are grouped in vec4.
 242          * This logic won't quite cut it if we don't align smaller
 243          * on vec4 boundaries
 244          */
 245         for (i = 0; i < ir->ninputs; i += 4)
 246                 pad_and_group_input(&ir->inputs[i], 4);
 247         for (i = 0; i < ir->noutputs; i += 4)
 248                 group_n(&arr_ops_out, &ir->outputs[i], 4);
 249
 250         for (i = 0; i < ir->noutputs; i++) {
 251                 if (ir->outputs[i]) {
 252                         struct ir3_instruction *instr = ir->outputs[i];
 253                         instr_find_neighbors(instr);
 254                 }
 255         }
 256
 257         list_for_each_entry (struct ir3_block, block, &ir->block_list, node) {
 258                 for (i = 0; i < block->keeps_count; i++) {
 259                         struct ir3_instruction *instr = block->keeps[i];
 260                         instr_find_neighbors(instr);
 261                 }
 262
 263                 /* We also need to account for if-condition: */
 264                 if (block->condition)
 265                         instr_find_neighbors(block->condition);
 266         }
 267 }
 268
 269 void
 270 ir3_group(struct ir3 *ir)
 271 {
 272         ir3_clear_mark(ir);
 273         find_neighbors(ir);
 274 }