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/* { dg-do compile { target { i?86-*-* x86_64-*-* } } } */
/* { dg-options "-O2 -fno-tree-vectorize -fprefetch-loop-arrays -march=amdfam10 --param simultaneous-prefetches=100 --param min-insn-to-prefetch-ratio=6 -fdump-tree-aprefetch-details" } */
#define N 1000
#define K 900
double a[N][N];
double test(void)
{
unsigned i, j;
double sum = 0;
/* Here, we should use non-temporal prefetch instruction. */
for (i = 0; i < K; i++)
for (j = 0; j < K; j++)
sum += a[i][j];
/* Here, we should not use non-temporal prefetch instruction, since the
value of a[i+10][j] is reused in L2 cache. */
for (i = 0; i < K; i++)
for (j = 0; j < K; j++)
sum += a[i][j] * a[i + 10][j];
/* Here, we should use non-temporal prefetch instruction, since the
value of a[i+100][j] is too far to be reused in L2 cache. */
for (i = 0; i < K; i++)
for (j = 0; j < K; j++)
sum += a[i][j] * a[i + 100][j];
/* Here, temporal prefetches should be used, since the volume of the
memory accesses is smaller than L2 cache. */
for (i = 0; i < 100; i++)
for (j = 0; j < 100; j++)
sum += a[i][j] * a[i + 100][j];
/* Temporal prefetches should be used here (even though the accesses to
a[j][i] are independent, the same cache line is almost always hit
every N iterations). */
for (i = 0; i < N; i++)
for (j = 0; j < N; j++)
sum += a[j][i];
return sum;
}
/* { dg-final { scan-tree-dump-times "Issued prefetch" 5 "aprefetch" } } */
/* { dg-final { scan-tree-dump-times "Issued nontemporal prefetch" 3 "aprefetch" } } */
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