C++簡化構造函數重載

Question

假設我正在創建一個類client。我想client能夠以下幾類構成：

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 &) 

它可以假設我的客戶類，爲簡單起見，簡單來說，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)); 

}; 

，目前定義爲：

#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)){} 

所以我的問題是，什麼是這樣做的正確的和簡單的方法？當然，我這次手動完成了所有的排列組合，但是將來我可以有3個以上的變量進行排列，這會變得很難看，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

由於uri定義了可以採用string以及const char *的構造函數，因此請刪除其參數不包含uri類型的構造函數。這讓uri的user-defined conversion爲你隱式轉換這些類型。

#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)){} 

「我怎麼就委託給這個構造特別？」 使用委託構造函數時應該明確。

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

可以使能採取所有的這三樣東西作爲參數的類。您的文章沒有包括足夠的細節，我寫了一個具體的答案，但這裏是僞代碼：

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);