re PR tree-optimization/33291 (a+=2; a+=2 not simplified to a+=4; with -O3 ...
[gcc.git] / gcc / testsuite / gcc.dg / tree-ssa / prefetch-6.c
1 /* { dg-do compile { target i?86-*-* x86_64-*-* } } */
2 /* { dg-require-effective-target ilp32 } */
3 /* { dg-options "-O2 -fprefetch-loop-arrays -march=athlon -msse2 -mfpmath=sse --param simultaneous-prefetches=100 -fdump-tree-aprefetch-details" } */
4
5 #define N 1000
6 #define K 900
7
8 double a[N][N];
9
10 double test(void)
11 {
12 unsigned i, j;
13 double sum = 0;
14
15 /* Here, we should use non-temporal prefetch instruction. */
16 for (i = 0; i < K; i++)
17 for (j = 0; j < K; j++)
18 sum += a[i][j];
19
20 /* Here, we should not use non-temporal prefetch instruction, since the
21 value of a[i+10][j] is reused in L2 cache. */
22 for (i = 0; i < K; i++)
23 for (j = 0; j < K; j++)
24 sum += a[i][j] * a[i + 10][j];
25
26 /* Here, we should use non-temporal prefetch instruction, since the
27 value of a[i+100][j] is too far to be reused in L2 cache. */
28 for (i = 0; i < K; i++)
29 for (j = 0; j < K; j++)
30 sum += a[i][j] * a[i + 100][j];
31
32 /* Here, temporal prefetches should be used, since the volume of the
33 memory accesses is smaller than L2 cache. */
34 for (i = 0; i < 100; i++)
35 for (j = 0; j < 100; j++)
36 sum += a[i][j] * a[i + 100][j];
37
38 /* Temporal prefetches should be used here (even though the accesses to
39 a[j][i] are independent, the same cache line is almost always hit
40 every N iterations). */
41 for (i = 0; i < N; i++)
42 for (j = 0; j < N; j++)
43 sum += a[j][i];
44
45 return sum;
46 }
47
48 /* { dg-final { scan-tree-dump-times "Issued prefetch" 5 "aprefetch" } } */
49 /* { dg-final { scan-tree-dump-times "Issued nontemporal prefetch" 3 "aprefetch" } } */
50
51 /* { dg-final { scan-assembler-times "prefetcht" 5 } } */
52 /* { dg-final { scan-assembler-times "prefetchnta" 3 } } */
53
54 /* { dg-final { cleanup-tree-dump "aprefetch" } } */