如何从 boost::property_tree 获取枚举?

How to get enum from boost::property_tree?(如何从 boost::property_tree 获取枚举?)

问题描述

如何从 boost::property_tree 获取枚举?

这是我的非工作"示例.

This is my "non-working" example.

<root>
  <fooEnum>EMISSION::EMIT1</fooEnum>
  <fooDouble>42</fooDouble>
</root>

main.cpp

#include <iostream>
#include <boost/property_tree/ptree.hpp>
#include <boost/property_tree/xml_parser.hpp>

int main()
{
  enum class EMISSION { EMIT1, EMIT2 } ;
  enum EMISSION myEmission;

  //Initialize the XML file into property_tree
  boost::property_tree::ptree pt;
  read_xml("config.xml", pt);

  //test enum (SUCCESS)
  myEmission = EMISSION::EMIT1;
  std::cout << (myEmission == EMISSION::EMIT1) << "
";

  //test basic ptree interpreting capability (SUCCESS)
  const double fooDouble = pt.get<double>("root.fooDouble");
  std::cout << fooDouble << "
";

  //read from enum from ptree and assign (FAILURE)
  myEmission = pt.get<enum EMISSION>( "root.fooEnum" );
  std::cout << (myEmission == EMISSION::EMIT1) << "
";

  return 0;
}

编译输出

/usr/include/boost/property_tree/stream_translator.hpp:36:15: 
error: cannot bind 'std::basic_istream<char>' lvalue to 
'std::basic_istream<char>&&'

/usr/include/c++/4.8/istream:872:5: error:   
initializing argument 1 of 'std::basic_istream<_CharT, 
  _Traits>& std::operator>
(std::basic_istream<_CharT, _Traits>&&, _Tp&)
[with _CharT = char; _Traits = std::char_traits<char>;
_Tp = main()::EMISSION]'

推荐答案

C++ 中枚举的名称是一个符号，而不是字符串.除非您通过编写如下方法自己提供该映射，否则无法在字符串和枚举值之间进行映射:

The name of an enum in C++ is a symbol, not a string. There isn't a way to map between a string and an enum value unless you provide that mapping yourself by writing a method such as:

EMISSION emission_to_string(const std::string& name)
{
    if ( name == "EMISSION::EMIT1")
    {
        return EMISSION::EMIT1;
    }
    ... etc
}

然后，您将从 property_tree 中获取字符串形式的值并应用此映射.

You would then get the value as a string from the property_tree and apply this mapping.

有更好的方法来实现这一点，这些方法可以通过许多枚举值更优雅地扩展.我已经使用 boost::bimap 来启用从 enum->string 或从 string->enum 的映射，当然这也为您提供了一个映射而不是一个愚蠢的大 if 语句.如果您这样做，请考虑使用 boost::assign 来初始化您的静态地图，因为它看起来比其他方法更干净.

There are nicer ways to implement this which scale more elegantly with many enum values. I have done this using boost::bimap to enable a mapping from enum->string OR from string->enum, and of course this also gives you a map instead of a silly big if statement. If you do this, look into using boost::assign to initialise your static map, as it looks cleaner than other methods.

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