1

我使用以下代码(存储在 crtp.cc 中)遇到了与编译器相关的问题:

#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

我得到:

$ ./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

我获得了一个Segmentation fault. 跑步

valgrind --tool=memcheck ./a.out

指向源代码中的第 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寻找一个没有任何结果的错误,我想请教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(如我所料)或到编译器错误。

编辑:在 Mike Seymour 的回答之后,我改变了现在的代码。现在我没有收到编译器警告,但我仍然得到与以前相同的行为(具有相同的 valgrind 响应)。有人尝试用 Intel 编译吗?

编辑:我试图在Wikipedia的Expression Templates页面中编译代码。我获得了与我提供的示例相同的行为。

编辑:我已经进一步调查了这个问题,似乎与英特尔一起编译icpc运营商

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

递归调用自身。我发现的一种解决方法是用一种方法替换此运算符:

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。谁能指出这两种行为中哪一种是正确的,可能指向解释这一点的标准?

4

1 回答 1

6

您应该始终启用编译器警告;他们经常能发现细微的问题。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]);
}

表达式的结果是一个临时的,在该行的末尾被销毁,因此该函数返回一个对不存在的对象的悬空引用。该运算符将不得不按值返回,并且您不应该实现非const重载,因为没有要修改的值。

于 2012-03-16T14:32:41.743 回答