1 /* -*- mode: C; c-file-style: "k&r"; tab-width 4; indent-tabs-mode: t; -*- */
2 
3 /*
4  * Copyright (C) 2014 Rob Clark <robclark@freedesktop.org>
5  *
6  * Permission is hereby granted, free of charge, to any person obtaining a
7  * copy of this software and associated documentation files (the "Software"),
8  * to deal in the Software without restriction, including without limitation
9  * the rights to use, copy, modify, merge, publish, distribute, sublicense,
10  * and/or sell copies of the Software, and to permit persons to whom the
11  * Software is furnished to do so, subject to the following conditions:
12  *
13  * The above copyright notice and this permission notice (including the next
14  * paragraph) shall be included in all copies or substantial portions of the
15  * Software.
16  *
17  * THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND, EXPRESS OR
18  * IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF MERCHANTABILITY,
19  * FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT.  IN NO EVENT SHALL
20  * THE AUTHORS OR COPYRIGHT HOLDERS BE LIABLE FOR ANY CLAIM, DAMAGES OR OTHER
21  * LIABILITY, WHETHER IN AN ACTION OF CONTRACT, TORT OR OTHERWISE, ARISING FROM,
22  * OUT OF OR IN CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER DEALINGS IN THE
23  * SOFTWARE.
24  *
25  * Authors:
26  *    Rob Clark <robclark@freedesktop.org>
27  */
28 
29 #include "freedreno_util.h"
30 
31 #include "ir3.h"
32 
33 /*
34  * Find/group instruction neighbors:
35  */
36 
37 /* bleh.. we need to do the same group_n() thing for both inputs/outputs
38  * (where we have a simple instr[] array), and fanin nodes (where we have
39  * an extra indirection via reg->instr).
40  */
41 struct group_ops {
42 	struct ir3_instruction *(*get)(void *arr, int idx);
43 	void (*insert_mov)(void *arr, int idx, struct ir3_instruction *instr);
44 };
45 
arr_get(void * arr,int idx)46 static struct ir3_instruction *arr_get(void *arr, int idx)
47 {
48 	return ((struct ir3_instruction **)arr)[idx];
49 }
arr_insert_mov_out(void * arr,int idx,struct ir3_instruction * instr)50 static void arr_insert_mov_out(void *arr, int idx, struct ir3_instruction *instr)
51 {
52 	((struct ir3_instruction **)arr)[idx] =
53 			ir3_MOV(instr->block, instr, TYPE_F32);
54 }
arr_insert_mov_in(void * arr,int idx,struct ir3_instruction * instr)55 static void arr_insert_mov_in(void *arr, int idx, struct ir3_instruction *instr)
56 {
57 	/* so, we can't insert a mov in front of a meta:in.. and the downstream
58 	 * instruction already has a pointer to 'instr'.  So we cheat a bit and
59 	 * morph the meta:in instruction into a mov and insert a new meta:in
60 	 * in front.
61 	 */
62 	struct ir3_instruction *in;
63 
64 	debug_assert(instr->regs_count == 1);
65 
66 	in = ir3_instr_create(instr->block, OPC_META_INPUT);
67 	in->inout.block = instr->block;
68 	ir3_reg_create(in, instr->regs[0]->num, 0);
69 
70 	/* create src reg for meta:in and fixup to now be a mov: */
71 	ir3_reg_create(instr, 0, IR3_REG_SSA)->instr = in;
72 	instr->opc = OPC_MOV;
73 	instr->cat1.src_type = TYPE_F32;
74 	instr->cat1.dst_type = TYPE_F32;
75 
76 	((struct ir3_instruction **)arr)[idx] = in;
77 }
78 static struct group_ops arr_ops_out = { arr_get, arr_insert_mov_out };
79 static struct group_ops arr_ops_in = { arr_get, arr_insert_mov_in };
80 
instr_get(void * arr,int idx)81 static struct ir3_instruction *instr_get(void *arr, int idx)
82 {
83 	return ssa(((struct ir3_instruction *)arr)->regs[idx+1]);
84 }
85 static void
instr_insert_mov(void * arr,int idx,struct ir3_instruction * instr)86 instr_insert_mov(void *arr, int idx, struct ir3_instruction *instr)
87 {
88 	((struct ir3_instruction *)arr)->regs[idx+1]->instr =
89 			ir3_MOV(instr->block, instr, TYPE_F32);
90 }
91 static struct group_ops instr_ops = { instr_get, instr_insert_mov };
92 
93 /* verify that cur != instr, but cur is also not in instr's neighbor-list: */
94 static bool
in_neighbor_list(struct ir3_instruction * instr,struct ir3_instruction * cur,int pos)95 in_neighbor_list(struct ir3_instruction *instr, struct ir3_instruction *cur, int pos)
96 {
97 	int idx = 0;
98 
99 	if (!instr)
100 		return false;
101 
102 	if (instr == cur)
103 		return true;
104 
105 	for (instr = ir3_neighbor_first(instr); instr; instr = instr->cp.right)
106 		if ((idx++ != pos) && (instr == cur))
107 			return true;
108 
109 	return false;
110 }
111 
112 static void
group_n(struct group_ops * ops,void * arr,unsigned n)113 group_n(struct group_ops *ops, void *arr, unsigned n)
114 {
115 	unsigned i, j;
116 
117 	/* first pass, figure out what has conflicts and needs a mov
118 	 * inserted.  Do this up front, before starting to setup
119 	 * left/right neighbor pointers.  Trying to do it in a single
120 	 * pass could result in a situation where we can't even setup
121 	 * the mov's right neighbor ptr if the next instr also needs
122 	 * a mov.
123 	 */
124 restart:
125 	for (i = 0; i < n; i++) {
126 		struct ir3_instruction *instr = ops->get(arr, i);
127 		if (instr) {
128 			struct ir3_instruction *left = (i > 0) ? ops->get(arr, i - 1) : NULL;
129 			struct ir3_instruction *right = (i < (n-1)) ? ops->get(arr, i + 1) : NULL;
130 			bool conflict;
131 
132 			/* check for left/right neighbor conflicts: */
133 			conflict = conflicts(instr->cp.left, left) ||
134 				conflicts(instr->cp.right, right);
135 
136 			/* RA can't yet deal very well w/ group'd phi's: */
137 			if (instr->opc == OPC_META_PHI)
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
instr_find_neighbors(struct ir3_instruction * instr)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
pad_and_group_input(struct ir3_instruction ** input,unsigned n)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
find_neighbors(struct ir3 * ir)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 }
264 
265 void
ir3_group(struct ir3 * ir)266 ir3_group(struct ir3 *ir)
267 {
268 	ir3_clear_mark(ir);
269 	find_neighbors(ir);
270 }
271