C++简化构造函数重载

Question

假设我正在创建一个类client。我想client能够以下几类构成：

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 &) 

它可以假设我的客户类，为简单起见，简单来说，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，什么是这样做的正确的和简单的方法？当然，我这次手动完成了所有的排列组合，但是将来我可以有3个以上的变量进行排列，这会变得很难看，很快。

Answer 1

如何模板：

#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 2

由于uri定义了可以采用string以及const char *的构造函数，因此请删除其参数不包含uri类型的构造函数。这让uri的user-defined conversion为你隐式转换这些类型。

#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)){} 

“我怎么就委托给这个构造特别？” 使用委托构造函数时应该明确。

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 3

可以使能采取所有的这三样东西作为参数的类。您的文章没有包括足够的细节，我写了一个具体的答案，但这里是伪代码：

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);