在构造函数初始化列表中使用 std::variant

Question

I am trying to use std::variant to generalize some of my code. However, I am running issues into when calling a constructor.

I define a class TCPServer as follows:

class TCPServer {
public:
    TCPServer(aio::io_context &io_context, std::variant<PlainServer, PIRServer>);

    std::variant<PlainServer, PIRServer> server_;

private:
    ...
};

And I define classes PIRServer and PlainServer as follows:

class PlainServer : public TCPServer {
public:
    explicit PlainServer(aio::io_context& io_context);
    ...
private:
    ...
};

PlainServer::PlainServer(aio::io_context& io_context) :
    server_config_(server_config),
    TCPServer(io_context, this) {}

I omit PIRServer because it doesn't contribute to the understanding of the problem.

My IDE underlines the initialization of TCPServer in the PlainServer constructor and says: "No matching constructor for initialization of 'TCPServer". Am I using std::variant incorrectly?

Answer 1

Your variant holds a PlainServer . this is a pointer to a PlainServer . You probably want:

class TCPServer {
public:
    TCPServer(aio::io_context &io_context, std::variant<PlainServer*, PIRServer*>);
...
};


PlainServer::PlainServer(aio::io_context& io_context) :
    server_config_(server_config),
    TCPServer(io_context, std::variant<PlainServer*, PIRServer*>(this)) {}

The explicit construction of the variant is required because its single arg constructors are explicit .