CRTP：与表达式模板的编译器相关问题 [英] CRTP: Compiler dependent issue with Expression Template

问题描述

我在编译器依赖的问题中遇到以下代码（存储在crtp.cc中）：

I incurred in a compiler dependent issue with the following code (stored in crtp.cc):

#include <vector>
#include <cassert>
#include <iostream>

template < class Derived >
class AlgebraicVectorExpression {
public:
  typedef std::vector<double>::size_type  SizeType;
  typedef std::vector<double>::value_type ValueType;
  typedef std::vector<double>::reference  ReferenceType;

  SizeType size() const {
    return static_cast<const Derived&>(*this).size();
  }

  ValueType operator[](SizeType ii) const {
    return static_cast<const Derived&>(*this)[ii];
  }

  operator Derived&() {
    return static_cast<Derived&>(*this);
  }

  operator const Derived&() const {
    return static_cast< const Derived& >(*this);
  }
};

template< class T1, class T2>
class AlgebraicVectorSum : public AlgebraicVectorExpression< AlgebraicVectorSum<T1,T2> > {
  const T1 & a_;
  const T2 & b_;

  typedef typename AlgebraicVectorExpression< AlgebraicVectorSum<T1,T2> >::SizeType  SizeType;
  typedef typename AlgebraicVectorExpression< AlgebraicVectorSum<T1,T2> >::ValueType ValueType;
public:

  AlgebraicVectorSum(const AlgebraicVectorExpression<T1>& a, const AlgebraicVectorExpression<T1>& b) :
    a_(a), b_(b)  {
    assert(a_.size() == b_.size());
  }

  SizeType size() const {
    return a_.size();
  }

  ValueType operator[](SizeType ii) const {
    return (a_[ii] + b_[ii]);
  }

};

template< class T1, class T2>
const AlgebraicVectorSum<T1,T2>
operator+(const AlgebraicVectorExpression<T1>& a, const AlgebraicVectorExpression<T2>& b) {
  return AlgebraicVectorSum<T1,T2>(a,b);
}

class AlgebraicVector : public AlgebraicVectorExpression<AlgebraicVector>{
  std::vector<double> data_;

public:
  SizeType size() const {
    return data_.size();
  }

  ValueType operator[](SizeType ii) const {
    return data_[ii];
  }

  ValueType& operator[](SizeType ii) {
    return data_[ii];
  }

  AlgebraicVector(SizeType n) : data_(n,0.0) {
  };

  template< class T>
  AlgebraicVector(const AlgebraicVectorExpression<T>& vec) {
    const T& v = vec;
    data_.resize(v.size());
    for( SizeType idx = 0; idx != v.size(); ++idx) {
      data_[idx] = v[idx];
    }
  }
};

int main() {

  AlgebraicVector x(10);
  AlgebraicVector y(10);
  for (int ii = 0; ii != 10; ++ii)
    x[ii] = y[ii] = ii;    

  AlgebraicVector z(10);
  z = x + y;

  for(int ii = 0; ii != 10; ++ii)
    std::cout << z[ii] << std::endl;
  return 0;
}

事实上，当我编译它：

$ g++ --version
g++ (Ubuntu 4.4.3-4ubuntu5) 4.4.3
Copyright (C) 2009 Free Software Foundation, Inc.
This is free software; see the source for copying conditions.  There is NO
warranty; not even for MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.

$ g++ -O0 -g crtp.cc

我获得： / p>

I obtain:

$ ./a.out 
0
2
4
6
8
10
12
14
16
18

这是预期的行为。当我使用icpc时：

which is the expected behavior. When I use icpc:

$ icpc --version
icpc (ICC) 12.1.0 20110811
Copyright (C) 1985-2011 Intel Corporation.  All rights reserved.    
$ icpc -g -O0 crtp.cc

c>分段故障。正在运行

I obtain instead a Segmentation fault. Running

valgrind --tool=memcheck ./a.out

指向源

AlgebraicVectorExpression<AlgebraicVector>::operator AlgebraicVector const&() const (crtp.cc:29)

到C ++和我花了很多时间搜索一个bug没有任何结果，我想问一个更有经验的人的意见，以了解这个问题是否是由于一些错误，我介绍（如我所料）或编译器错误。

As I am quite new to C++ and I spent quite a time searching for a bug without any result, I would like to ask the opinion of someone more experienced to understand if this issue is due to some error I introduced (as I expect) or to a compiler bug.

编辑：
我在Mike Seymour的回答后更改了现在的代码。现在我没有得到编译器警告，但我仍然得到相同的行为，以前（具有相同的valgrind响应）。有没有人试图用英特尔编译？

Edit: I changed the code as it is now, after the answer from Mike Seymour. Now I don't get compiler warnings, but I still get the same behavior as before (with the same valgrind response). Did anybody try to compile with Intel?

编辑：
我试图编译代码在表达式模板页面。我获得了与我提供的示例相同的确切行为。

Edit: I tried to compile the code in the Expression Templates page of Wikipedia. I obtained the same exact behavior as with the example I provided.

编辑：
我进一步调查了问题，与英特尔 icpc 编译操作符

operator const Derived&() const {
    return static_cast< const Derived& >(*this);
  }

递归调用自身。一个解决方法我发现是用一个方法替换这个运算符：

recursively calls itself. One workaround I found is to substitute this operator with a method:

const Derived& get_ref() const {
    return static_cast< const Derived& >(*this);
  }

并相应地修改各种类的构造函数。

and modify the constructors of the various classes accordingly. Can anybody tell which of these two behaviors is right possibly pointing to the standard to explain that?

推荐答案

任何人都应该总是启用编译器警告;他们经常可以发现微妙的问题。在这种情况下：

You should always enable compiler warnings; they can often spot subtle problems. In this case:

g++ -Wall -Wextra test.cpp
test.cpp: In member function ‘const typename AlgebraicVectorExpression<AlgebraicVectorSum<T1, T2> >::ValueType& AlgebraicVectorSum<T1, T2>::operator[](typename AlgebraicVectorExpression<AlgebraicVectorSum<T1, T2> >::SizeType) const [with T1 = AlgebraicVector, T2 = AlgebraicVector]’:
test.cpp:90:   instantiated from ‘AlgebraicVector::AlgebraicVector(const AlgebraicVectorExpression<T1>&) [with T = AlgebraicVectorSum<AlgebraicVector, AlgebraicVector>]’
test.cpp:103:   instantiated from here
test.cpp:52: warning: returning reference to temporary

这会告诉你问题：

const ValueType& operator[](SizeType ii) const {
    return (a_[ii] + b_[ii]);
}

表达式的结果是一个临时的， ，所以函数返回一个悬挂的引用到不存在的对象。这个操作符将不得不返回值，而不应该实现非 - const 重载，因为没有值要修改。

The result of the expression is a temporary, destroyed at the end of that line, so the function is returning a dangling reference to a non-existent object. This operator will have to return by value instead, and you shouldn't implement the non-const overload since there is no value to modify.

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