2012-03-27 23:13:14 +00:00
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void abort (void);
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void parloop (int N)
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{
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int i, j;
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int x[500][500];
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for (i = 0; i < N; i++)
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for (j = 0; j < N; j++)
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x[i][j] = i + j + 3;
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for (i = 0; i < N; i++)
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for (j = 0; j < N; j++)
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if (x[i][j] != i + j + 3)
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abort ();
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}
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int main(void)
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{
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parloop(500);
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return 0;
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}
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/* Check that parallel code generation part make the right answer. */
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2014-09-21 17:33:12 +00:00
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/* { dg-final { scan-tree-dump-times "2 loops carried no dependency" 1 "graphite" } } */
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/* { dg-final { scan-tree-dump-times "loopfn" 4 "optimized" } } */
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