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32 lines
519 B
C
32 lines
519 B
C
void abort (void);
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int x[10000000];
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void parloop (int N)
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{
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int i;
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for (i = 0; i < N; i++)
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x[i] = i + 1;
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for (i = 0; i < N; i++)
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x[i] = i + 3;
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for (i = 0; i < N; i++)
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{
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if (x[i] != i + 3)
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abort ();
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}
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}
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int main(void)
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{
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parloop(10000000);
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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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/* { dg-final { scan-tree-dump-times "2 loops carried no dependency" 1 "graphite" } } */
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/* { dg-final { scan-tree-dump-times "loopfn" 8 "optimized" } } */
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