C ++保护的typedef作为返回值

Question

I'm not sure if this even is possible, but basically i have this pointer typedef that is protected in a class that i share among my threads and i want to make a function that returns a pointer of that typedef.

I can declare the function in my .h with no problem but the .cpp says it doesn't recognize it. I can also use my typedef within a function. Only thing as i said i can't do is have it as a return value.

I doubt you guys have to even see the code but i'm posting it just to make it complete.

Shared header:

#pragma once


class CR
{


public:


private:

public:

    int c_outPut(std::string &output);

protected:
    typedef std::unordered_map<in_addr, SOCKET>::const_iterator cit;
};

Short version of the other .h

#pragma once
#include "commonRec.h"

#include <unordered_map>
#include <WinSock2.h>
#include <iterator>

class CH : CR
{

    std::string sendToCon(cit &const_it, const std::string &command);
    cit findHost(std::string &searchHost);
};

.cpp function declaration - error "cit is undeclared identifier"

cit CC::_translateCommand(string &command)
{
}

Answer 1

I can also use my typedef within a function.

The key word here is within .

Only thing as i said i can't do is have it as a return value.

Your return type declaration is outside of the scope of the function. Outside of the scope of CR , cit can only refer to ::cit . But you did not define the type ::cit ; you defined CR::cit . So, when you're outside the scope of CR , such as when you declare a member function out-of-line, you must resolve the scope explicitly:

CR::cit CC::_translateCommand(string &command)

A trailing return type declaration is within the scope of the function, so if you used that, then you wouldn't need explicit resolution:

auto CC::_translateCommand(string &command) -> cit