Question

#include <iostream>
#include <string>
#include "boost/operators.hpp"

using namespace std;

class simple_string : private boost::addable1<simple_string,
                              boost::addable2<simple_string, const char*> > 
{
public:
  simple_string() : m_str("Default")
  {}

  explicit simple_string(const char* s) : m_str(s)
  {}

  simple_string(const simple_string& rhs) : m_str(rhs.m_str)
  {}

  simple_string operator+=(const simple_string& rhs)
  {
    m_str += rhs.m_str;
    return *this;
  }

  simple_string operator+=(const char* rhs)
  {
    this->operator+=(simple_string(rhs));
    return *this;
  }

  friend ostream& operator<<(ostream& os, const simple_string& si)
  {
    os << si.m_str;
    return os;
  }

private:
  string m_str;
};

int main(void)
{
  simple_string s1;
  simple_string s2(s1);
  cout << "[before] s2: " << s2 << endl;

  s2 += s1;
  cout << "s2 += s1: " << s2 << endl;

  simple_string s3 = s1 + s2;
  cout << "s3: " << s3 << endl;

  simple_string s4 = s3 + "Why";
  simple_string s5 = "Now" + s3;
  cout << "s4: " << s4 << endl;
  cout << "s5: " << s5 << endl;
  cout << "Now" + simple_string() << endl;
  return 0;
}

输出

~/Documents/C++/boost $ g++ -o p123 p123.cpp
~/Documents/C++/boost $ ./p123
[before] s2: Default
s2 += s1: DefaultDefault
s3: DefaultDefaultDefault
s4: DefaultDefaultDefaultWhy
s5: DefaultDefaultDefaultNow
DefaultNow
~/Documents/C++/boost $

问题＆GT;为什么生成的operator+(const char*, const simple_string&)没有返回正确的结果？

例如，s5的预期结果应为s5: NowDefaultDefaultDefault

"Now" + simple_string()的结果应为NowDefault

Answer 1

问题是字符串operator+不是可交换的，但是boost实现假设它是。有关详细信息，请参阅http://www.boost.org/doc/libs/1_32_0/libs/utility/operators.htm#symmetry处的对称性说明。

编辑：在阅读实际的增强代码后，我认为我错了。你不能对operator+不可交换的类使用boost运算符生成，因为这是boost中的隐式假设。你必须自己实现它。

基于（我相信）错误阅读文档的原始答案：

您可以强制可能效率较低的不假设对称的代码，如下所示：

...if your code relies on the function signature or a strict symmetric behaviour, you should set BOOST_FORCE_SYMMETRIC_OPERATORS in your user-config. This will force the NRVO-friendly implementation to be used even for compilers that don't implement the NRVO.

boost :: addable2的用法 - 为什么从operator +返回的结果被反转

1 个答案: