CRTP:表达式模板的编译器相关问题

Question

我在编译器相关问题中遇到以下代码(存储在crtp.cc中):

#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;
}

事实上当我编译它时:

$ 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:

$ ./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:

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

我取而代之的是Segmentation fault.运行

valgrind --tool=memcheck ./a.out

指向来源中的第29行

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

由于我对C++很陌生并且我花了很长时间搜索没有任何结果的错误,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我引入的一些错误(正如我所期待的)或编译错误.

编辑:在Mike Seymour的回答之后,我改变了现在的代码.现在我没有得到编译器警告,但我仍然得到与以前相同的行为(使用相同的valgrind响应).有没有人试图与英特尔合作？

编辑:我试图在维基百科的表达式模板页面中编译代码.我获得了与我提供的示例相同的确切行为.

编辑:我已经进一步调查了这个问题,似乎与英特尔编译icpc运算符

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

递归地调用自己.我找到的一个解决方法是用一个方法替换此运算符:

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.任何人都能说出这两种行为中的哪一种是正确的可能指向标准来解释这一点？

Answer 1

您应始终启用编译器警告; 他们经常会发现微妙的问题.在这种情况下:

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

这告诉你问题:

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

表达式的结果是临时的,在该行的末尾被销毁,因此该函数返回对不存在的对象的悬空引用.此运算符必须通过值返回,并且您不应实现非const重载,因为没有值可以修改.