为什么boost :: call_traits <T> :: param_type是枚举类型的引用？

Question

一个基本的C++ 03枚举类型只是一个带有花哨名称的整数值,因此我希望通过值传递它....

出于这个原因,我也希望boost::call_traits<T>::param_type通过T=SomeEnum确定最有效的传递方式T是按值.

从提升文档中可以看到呼叫特征:

定义一种类型,表示将类型T的参数传递给函数的"最佳"方式.

当我用boost::call_traits<T>::param_type用T=SomeEnum它确定SomeEnum应通过引用传递.

我也希望C++11 class enums也能通过价值传递.

测试代码:

#include <string>
#include <typeinfo>
#include <boost/call_traits.hpp>
#include <boost/type_traits/is_reference.hpp>

enum SomeEnum
{
    EN1_ZERO = 0,
    EN1_ONE,
    EN1_TWO,
    EN1_THREE
};

struct SomeStruct
{};

template<typename T>
void DisplayCallTraits( const std::string& desc )
{
    typedef typename boost::call_traits<T>::param_type param_type;
    std::cout << "-----------------------------------------------------\n";
    std::cout << "Call traits for: " << desc << "\n";
    std::cout << "\ttypeof T : " << typeid(T).name() << "\n";
    std::cout << "\ttypeof param_type : " << typeid(param_type).name() << "\n";
    std::cout << "\tis_reference<param_type> : " << std::boolalpha 
              << boost::is_reference<param_type>::value << "\n";
}

int main( int, char** )
{
    DisplayCallTraits< unsigned >( "unsigned" );     // pass by value, as expected
    DisplayCallTraits< SomeStruct >( "struct" );     // pass by reference, as expected
    DisplayCallTraits< SomeEnum >( "enumeration" );  // pass by reference - why?

    return 0;
}

Answer 1

已更新以包括is_enum对实施的检查。请参阅此错误报告，已于 2 个月前关闭：https ://svn.boost.org/trac/boost/ticket/5790