C++ 简化构造函数重载

Question

假设我正在制作 class client。我希望 client 能够用以下类型构造：

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 class 为简单起见被简化后如下所示。

#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？当然，这次我手动完成了所有排列，但将来我可以有 3 个以上的变量来排列，这会变得很丑陋，很快。

Answer 1

由于uri定义了既可以带string又可以带const char *的构造函数，剔除参数不包含uri类型参数的构造函数。这让 uri 的 user-defined conversion 隐式地为你转换这些类型。

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

class client
{
private:
  uri host_;

  uri auth_;

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

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

-

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

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

"How do I delegate it to that constructor specifically?" 使用委托构造函数时要明确。

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

或

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

Answer 2

模板如何：

#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))
    { }

    // ...
};

Answer 3

您可以制作一个 class，它可以将所有这三样东西作为参数。你的 post 没有包含足够的细节让我写一个具体的答案，但这是伪代码：

struct input_helper
{
    input_helper(uri &u);
    input_helper(string &u);
    input_helper(char *u);

    // data members to hold the inputs, maybe other processing to bring them to a common type
};

// the constructor
client(input_helper host, input_helper auth);

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