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/* { dg-do compile } */

/* { dg-do compile } */

/* { dg-options "-O2 -funsafe-loop-optimizations -Wunsafe-loop-optimizations" } */

/* { dg-options "-O2 -funsafe-loop-optimizations -Wunsafe-loop-optimizations" } */

extern void g(void);

extern void g(void);

void

void

f (unsigned n)

f (unsigned n)

{

{

  unsigned k;

  unsigned k;

  for(k = 0;k <= n;k++) /* { dg-warning "assuming.*not infinite" } */

  for(k = 0;k <= n;k++) /* { dg-warning "assuming.*not infinite" } */

    g();

    g();

  for(k = 5;k <= n;k += 4) /* { dg-warning "assuming.*not overflow" } */

  for(k = 5;k <= n;k += 4) /* { dg-warning "assuming.*not overflow" } */

    g();

    g();

  /* We used to get warning for this loop.  However, since then # of iterations

  /* We used to get warning for this loop.  However, since then # of iterations

     analysis improved, and we can now prove that this loop does not verflow.

     analysis improved, and we can now prove that this loop does not verflow.

     This is because the only case when it would overflow is if n = ~0 (since

     This is because the only case when it would overflow is if n = ~0 (since

     ~0 is divisible by 5), and this cannot be the case, since when we got

     ~0 is divisible by 5), and this cannot be the case, since when we got

     here, the previous loop exited, thus there exists k > n.  */

     here, the previous loop exited, thus there exists k > n.  */

  for(k = 5;k <= n;k += 5)

  for(k = 5;k <= n;k += 5)

    g();

    g();

  for(k = 4;k <= n;k += 5) /* { dg-warning "assuming.*not overflow" } */

  for(k = 4;k <= n;k += 5) /* { dg-warning "assuming.*not overflow" } */

    g();

    g();

  for(k = 15;k >= n;k--) /* { dg-warning "assuming.*not infinite" } */

  for(k = 15;k >= n;k--) /* { dg-warning "assuming.*not infinite" } */

    g();

    g();

}

}