Lines Matching +full:8 +full:a
17 * - avoid l2 bank-conflicts by not storing into the same 16-byte bank within a single
21 * First, note that L1 has a line-size of 64 bytes and L2 a line-size of 128 bytes.
22 * To avoid secondary misses in L2, we prefetch both source and destination with a line-size
28 * We use a software-pipelined loop to control the overall operation. The pipeline
32 * the prefetches. The four relevant points in the pipelined are called A, B, C, D:
33 * p[A] is TRUE if a source-line should be prefetched, p[B] is TRUE if a destination-line
35 * into L1D and p[D] is TRUE if a cacheline needs to be copied.
40 * As a secondary optimization, the first 2*PREFETCH_DIST iterations are implemented
67 #define PREFETCH_DIST 8 // McKinley sustains 16 outstanding L2 misses (8 ld, 8 st)
95 #define A 0 macro
122 add src_pre_l2 = 8*8, in1
123 add dst_pre_l2 = 8*8, in0
124 add src0 = 8, in1 // first t1 src
125 add src1 = 3*8, in1 // first t3 src
126 add dst0 = 8, in0 // first t1 dst
127 add dst1 = 3*8, in0 // first t3 dst
134 (p[A]) ld8 v[A] = [src_pre_mem], 128 // M0
139 mov ar.lc = t1 // with 64KB pages, t1 is too big to fit in 8 bits!
143 (p[D]) ld8 t2 = [src0], 3*8 // M0
144 (p[D]) ld8 t4 = [src1], 3*8 // M1
148 (p[A]) ld8 v[A] = [src_pre_mem], 128 // M0 prefetch src from memory
150 (p[D]) st8 [dst0] = t1, 8 // M2
151 (p[D]) st8 [dst1] = t3, 8 // M3
153 (p[D]) ld8 t5 = [src0], 8
154 (p[D]) ld8 t7 = [src1], 3*8
155 (p[D]) st8 [dst0] = t2, 3*8
156 (p[D]) st8 [dst1] = t4, 3*8
158 (p[D]) ld8 t6 = [src0], 3*8
159 (p[D]) ld8 t10 = [src1], 8
160 (p[D]) st8 [dst0] = t5, 8
161 (p[D]) st8 [dst1] = t7, 3*8
163 (p[D]) ld8 t9 = [src0], 3*8
164 (p[D]) ld8 t11 = [src1], 3*8
165 (p[D]) st8 [dst0] = t6, 3*8
166 (p[D]) st8 [dst1] = t10, 8
168 (p[D]) ld8 t12 = [src0], 8
169 (p[D]) ld8 t14 = [src1], 8
170 (p[D]) st8 [dst0] = t9, 3*8
171 (p[D]) st8 [dst1] = t11, 3*8
173 (p[D]) ld8 t13 = [src0], 4*8
174 (p[D]) ld8 t15 = [src1], 4*8
175 (p[D]) st8 [dst0] = t12, 8
176 (p[D]) st8 [dst1] = t14, 8
178 (p[D-1])ld8 t1 = [src0], 8
179 (p[D-1])ld8 t3 = [src1], 8
180 (p[D]) st8 [dst0] = t13, 4*8
181 (p[D]) st8 [dst1] = t15, 4*8