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[Bug tree-optimization/52112] New: Vectorizer fails when using CRTP


http://gcc.gnu.org/bugzilla/show_bug.cgi?id=52112

             Bug #: 52112
           Summary: Vectorizer fails when using CRTP
    Classification: Unclassified
           Product: gcc
           Version: 4.6.1
            Status: UNCONFIRMED
          Severity: normal
          Priority: P3
         Component: tree-optimization
        AssignedTo: unassigned@gcc.gnu.org
        ReportedBy: ddesics@gmail.com


Created attachment 26565
  --> http://gcc.gnu.org/bugzilla/attachment.cgi?id=26565
The test case code.

Using the CRTP along with static_cast pointers prevents auto-vectorization.
If the code below is compiled with -ftree-vectorizer-verbose=7, the CRTP method
fails to vectorize with "not vectorized: control flow in loop."  However, if
compiled with -O3 -S -fno-tree-vectorize, both methods produce identical
assembly.

I don't know how difficult this would be to change, but it could certainly
speed up a lot of c++ code.  For instance, this currently prevents boost ublas
from vectorizing.

#include<iostream>

template<typename E, typename Tp> class CRTP_base {
  public:
   typedef E& reference;
   typedef Tp value_type;

   reference operator()() { return *static_cast<E*>(this); }
   value_type square() { return (*this)().x() * (*this)().x(); }
  protected:
   CRTP_base() {}
   ~CRTP_base() {}
};

template<typename Tp> class CRTP_child : public CRTP_base<CRTP_child<Tp>,Tp> {
   Tp xval;
   typedef CRTP_base<CRTP_child<Tp>,Tp> parent;
  public:
   CRTP_child(Tp xv = Tp()) : xval(xv) {}
   Tp x() { return xval; }
   using parent::square;
};

int main() {
  const int N = 100;
  double A[N] __attribute__((aligned(16)));
  double B[N] __attribute__((aligned(16)));
  double sum1=0.0;

  for(int i = 0; i < N; ++i) { A[i] = i; }

  for(int i = 0; i < N; ++i) { B[i] = A[i]*A[i]; }
  for(int i = 0; i < N; ++i) { sum1 += B[i]; }
  std::cout << "Sum of method 1: " << sum1;

  for(int i = 0; i < N; ++i) { B[i] = CRTP_child<double>(A[i]).square(); }
  for(int i = 0; i < N; ++i) { sum1 += B[i]; }
  std::cout << "\nSum of method 2: " << sum2 << std::endl;
  return 0;
}


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