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#define N 1500
int foo(void)
{
int i, j;
int x[N][N], y[N];
for (i = 0; i < N; i++)
{
y[i] = i;
for (j = 0; j < N; j++)
{
if (j > 500)
{
x[i][j] = i + j + 3;
y[j] = i*j + 10;
}
else
x[i][j] = x[i][j]*3;
}
}
return x[2][5]*y[8];
}
int main(void)
{
foo();
return 0;
}
/* Check that parallel code generation part make the right answer. */
/* { dg-final { scan-tree-dump-times "2 loops carried no dependency" 1 "graphite" } } */
/* { dg-final { cleanup-tree-dump "graphite" } } */
/* { dg-final { scan-tree-dump-times "loopfn.0" 5 "optimized" } } */
/* { dg-final { scan-tree-dump-times "loopfn.1" 5 "optimized" } } */
/* { dg-final { cleanup-tree-dump "parloops" } } */
/* { dg-final { cleanup-tree-dump "optimized" } } */
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