为什么此函数重载与参数类型转换不明确？

有人可以解释这个函数重载解析度是如何模棱两可的吗？

鉴于：

/// This type *contains* a @c T.
template< typename T >
class User_Type
{
public:
    /// This <em>conversion constructor</em> is a key part of it's API;
    ///   it won't likely change.
    User_Type( T const & ar_data )
      : m_data( ar_data )
    {}
private:
    T m_data;
};

/// @c some_value is just a templated function that generates a @c T.
template< typename T > T some_value();
template<> char some_value();
template<> int some_value();

/// This overloaded, non-templated function represents some third-party
///   code somewhere else; it's API can't be changed.
void other_function( User_Type< char > const& );
void other_function( User_Type< int > const& );

/// This is user-code.  It's contents exercise some aspect of the 'User_Type' API.
///   This code can change.
template< typename T >
void function()
{
    other_function( some_value< T >() ); /* AMBIGUOUS CALL */

    User_Type< T > user_var = some_value< T >(); /* UNAMBIGUOUS CONVERSION */
    other_function( user_var ); /* UNAMBIGUOUS CALL */
}
template void function< char >();
template void function< int >();

并使用进行编译g++-4.9 -Wall -Wextra，我收到以下错误：

In instantiation of ‘void function() [with T = char]’:
    error: call of overloaded ‘other_function(char)’ is ambiguous
    note: candidates are:
    note: void other_function(const User_Type<char>&)
    note: void other_function(const User_Type<int>&)
In instantiation of ‘void function() [with T = int]’:
    error: call of overloaded ‘other_function(int)’ is ambiguous
    note: candidates are:
    note: void other_function(const User_Type<char>&)
    note: void other_function(const User_Type<int>&)

我希望最好的比赛other_function( char )是other_function( User_Type< char > const& )，最好的比赛other_function( int )是other_function( User_Type< int > const& )。

我了解必须为的每个参数进行类型转换other_function。我预计char到User_Type< char >是一个更好的选择，而不是char到User_Type< int >，这可能是被允许char到int推广。我预计int到User_Type< int >是一个更好的选择，而不是int到User_type< char >，这可能是被允许int到char转换。

另外，如果我User_Type< T > user_var从创建本地T，那么我可以明确地致电other_function( user_var )。从语义上讲，这应该等效于第一个原始语句。