C++ 简化构造函数重载

Question

假设我正在创建一个类 client。我希望能够使用以下类型构造 client:

client(const boost::network::uri::uri &, const boost::network::uri::uri &)
client(const std::string &, const std::string &)
client(const char *, const char *)

但是……我也想要所有的排列……

client(const boost::network::uri::uri &, const boost::network::uri::uri &)
client(const std::string &, const std::string &)
client(const char * &, const char * &)
client(const boost::network::uri::uri &, const std::string &)
client(const std::string &, const boost::network::uri::uri &)
client(const boost::network::uri::uri &, const char * &)
client(const char * &, const boost::network::uri::uri &)
client(const std::string &, const char * &)
client(const char * &, const std::string &)

可以假设我的客户端类，为了简单起见，看起来像下面这样。

#include <string>
#include <boost/network.hpp>

#define HOST_URI "..."
#define AUTH_URI HOST_URI"..."

namespace bn = boost::network;

class client
{

private:

  const bn::uri::uri host_;

  const bn::uri::uri auth_;

public:

  client(const bn::uri::uri & host = const bn::uri::uri(HOST_URI),
         const bn::uri::uri & auth = const bn::uri::uri(AUTH_URI));

  client(const std::string & host = const std::string(HOST_URI),
         const std::string & auth = const std::string(AUTH_URI));

  client(const char * & host = HOST_URI,
         const char * & auth = AUTH_URI);

  client(const bn::uri::uri & host = const bn::uri::uri(HOST_URI),
         const std::string & auth = const std::string(AUTH_URI));

  client(const std::string & host = const std::string(HOST_URI),
         const bn::uri::uri & auth = const bn::uri::uri(AUTH_URI));

  client(const bn::uri::uri & host = const bn::uri::uri(HOST_URI),
         const char * & auth = AUTH_URI);

  client(const char * & host = HOST_URI,
         const bn::uri::uri & auth = const bn::uri::uri(AUTH_URI));

  client(const std::string && host = const std::string(HOST_URI),
         const char * & auth = AUTH_URI);

  client(const char * & host = HOST_URI,
         const std::string && auth = const std::string(AUTH_URI));

};

目前定义为:

#include <string>
#include <boost/network.hpp>

namespace bn = boost::network;

client::client(const bn::uri::uri & host,
               const bn::uri::uri & auth)
: host_(host), auth_(auth)
{
    ...
};

client::client(const std::string & host,
               const std::string & auth)
: client(bn::uri::uri(host), bn::uri::uri(auth)){}

client::client(const char * & host,
               const char * & auth)
: client(bn::uri::uri(host), bn::uri::uri(auth)){}

client::client(const bn::uri::uri & host,
               const std::string & auth)
: client(host, bn::uri::uri(auth)){}

client::client(const std::string & host,
               const bn::uri::uri & auth)
: client(bn::uri::uri(host), auth){}

client::client(const bn::uri::uri & host,
               const char * & auth)
: client(host, bn::uri::uri(auth)){}

client::client(const char * & host,
               const bn::uri::uri & auth)
: client(bn::uri::uri(host), auth){}

client::client(const std::string & host,
               const char * & auth)
: client(bn::uri::uri(host), bn::uri::uri(auth)){}

client::client(const char * & host,
               const std::string & auth)
: client(bn::uri::uri(host), bn::uri::uri(auth)){}

所以我的问题是，执行此操作的正确且简单的方法是什么？诚然，这次我手动完成了所有排列，但将来我可能有 3 个以上的变量要排列，这会很快变得丑陋。

Answer 1

模板怎么样:

#include <type_traits>

class client
{
    uri host_;
    uri auth_;

public:
    template <typename U, typename V,
              typename = typename std::enable_if<
                  std::is_constructible<uri, U&&>::value &&
                  std::is_constructible<uri, V&&>::value>::type>
    client(U && u, V && v)
    : host_(std::forward<U>(u))
    , auth_(std::forward<V>(v))
    { }

    // ...
};